# Probabilities, Gaussians, and Bayes' Theorem¶

In [1]:
%matplotlib inline
In [2]:
#format the book
import book_format
book_format.set_style()
Out[2]:

## Introduction¶

The last chapter ended by discussing some of the drawbacks of the Discrete Bayesian filter. For many tracking and filtering problems our desire is to have a filter that is unimodal and continuous. That is, we want to model our system using floating point math (continuous) and to have only one belief represented (unimodal). For example, we want to say an aircraft is at (12.34, -95.54, 2389.5) where that is latitude, longitude, and altitude. We do not want our filter to tell us "it might be at (1.65, -78.01, 2100.45) or it might be at (34.36, -98.23, 2543.79)." That doesn't match our physical intuition of how the world works, and as we discussed, it can be prohibitively expensive to compute the multimodal case. And, of course, multiple position estimates makes navigating impossible.

We desire a unimodal, continuous way to represent probabilities that models how the real world works, and that is computationally efficient to calculate. Gaussian distributions provide all of these features.

## Mean, Variance, and Standard Deviations¶

Most of you will have had exposure to statistics, but allow me to cover this material anyway. I ask that you read the material even if you are sure you know it well. I ask for two reasons. First, I want to be sure that we are using terms in the same way. Second, I strive to form an intuitive understanding of statistics that will serve you well in later chapters. It's easy to go through a stats course and only remember the formulas and calculations, and perhaps be fuzzy on the implications of what you have learned.

### Random Variables¶

Each time you roll a die the outcome will be between 1 and 6. If we rolled a fair die a million times we'd expect to get a one 1/6 of the time. Thus we say the probability, or odds of the outcome 1 is 1/6. Likewise, if I asked you the chance of 1 being the result of the next roll you'd reply 1/6.

This combination of values and associated probabilities is called a random variable. Here random does not mean the process is nondeterministic, only that we lack information about the outcome. The result of a die toss is deterministic, but we lack enough information to compute the result. We don't know what will happen, except probabilistically.

While we are defining terms, the range of values is called the sample space. For a die the sample space is {1, 2, 3, 4, 5, 6}. For a coin the sample space is {H, T}. Space is a mathematical term which means a set with structure. The sample space for the die is a subset of the natural numbers in the range of 1 to 6.

Another example of a random variable is the heights of students in a university. Here the sample space is a range of values in the real numbers between two limits defined by biology.

Random variables such as coin tosses and die rolls are discrete random variables. This means their sample space is represented by either a finite number of values or a countably infinite number of values such as the natural numbers. Heights of humans are called continuous random variables since they can take on any real value between two limits.

Do not confuse the measurement of the random variable with the actual value. If we can only measure the height of a person to 0.1 meters we would only record values from 0.1, 0.2, 0.3...2.7, yielding 27 discrete choices. Nonetheless a person's height can vary between any arbitrary real value between those ranges, and so height is a continuous random variable.

In statistics capital letters are used for random variables, usually from the latter half of the alphabet. So, we might say that $X$ is the random variable representing the die toss, or $Y$ are the heights of the students in the freshmen poetry class. Later chapters use linear algebra to solve these problems, and so there we will follow the convention of using lower case for vectors, and upper case for matrices. Unfortunately these conventions clash, and you will have to determine which an author is using from context. I always use bold symbols for vectors and matrices, which helps distinguish between the two.

## Probability Distribution¶

The probability distribution gives the probability for the random variable to take any value in a sample space. For example, for a fair six sided die we might say:

Value Probability
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6

We denote this distribution with a lower case p: $p(x)$. Using ordinary function notation, we would write:

$$P(X{=}4) = p(4) = \frac{1}{6}$$

This states that the probability of the die landing on 4 is $\frac{1}{6}$. $P(X{=}x_k)$ is notation for "the probability of $X$ being $x_k$". Note the subtle notational difference. The capital $P$ denotes the probability of a single event, and the lower case $p$ is the probability distribution function. This can lead you astray if you are not observent. Some texts use $Pr$ instead of $P$ to ameliorate this.

Another example is a fair coin. It has the sample space {H, T}. The coin is fair, so the probability for heads (H) is 50%, and the probability for tails (T) is 50%. We write this as

$$\begin{gathered}P(X{=}H) = 0.5\\P(X{=}T)=0.5\end{gathered}$$

Sample spaces are not unique. One sample space for a die is {1, 2, 3, 4, 5, 6}. Another valid sample space would be {even, odd}. Another might be {dots in all corners, not dots in all corners}. A sample space is valid so long as it covers all possibilities, and any single event is described by only one element. {even, 1, 3, 4, 5} is not a valid sample space for a die since a value of 4 is matched both by 'even' and '4'.

The probabilities for all values of a discrete random value is known as the discrete probability distribution and the probabilities for all values of a continuous random value is known as the continuous probability distribution.

To be a probability distribution the probability of each value $x_i$ must be $x_i \ge 0$, since no probability can be less than zero. Secondly, the sum of the probabilities for all values must equal one. This should be intuitively clear for a coin toss: if the odds of getting heads is 70%, then the odds of getting tails must be 30%. We formulize this requirement as

$$\sum\limits_u P(X{=}u)= 1$$

for discrete distributions, and as

$$\int\limits_u P(X{=}u) \,du= 1$$

for continuous distributions.

In the previous chapter we used probability distributions to estimate the position of a dog in a hallway. For example:

In [3]:
import numpy as np
import kf_book.book_plots as book_plots

belief = np.array([1, 4, 2, 0, 8, 2, 2, 35, 4, 3, 2])
belief = belief / np.sum(belief)
with book_plots.figsize(y=2):
book_plots.bar_plot(belief)
print('sum = ', np.sum(belief))
sum =  1.0

Each position has a probability between 0 and 1, and the sum of all equals one, so this makes it a probability distribution. Each probability is discrete, so we can more precisely call this a discrete probability distribution. In practice we leave out the terms discrete and continuous unless we have a particular reason to make that distinction.

### The Mean, Median, and Mode of a Random Variable¶

Given a set of data we often want to know a representative or average value for that set. There are many measures for this, and the concept is called a measure of central tendency. For example we might want to know the average height of the students in a class. We all know how to find the average of a set of data, but let me belabor the point so I can introduce more formal notation and terminology. Another word for average is the mean. We compute the mean by summing the values and dividing by the number of values. If the heights of the students in meters is

$$X = \{1.8, 2.0, 1.7, 1.9, 1.6\}$$

we compute the mean as

$$\mu = \frac{1.8 + 2.0 + 1.7 + 1.9 + 1.6}{5} = 1.8$$

It is traditional to use the symbol $\mu$ (mu) to denote the mean.

We can formalize this computation with the equation

$$\mu = \frac{1}{n}\sum^n_{i=1} x_i$$

NumPy provides numpy.mean() for computing the mean.

In [4]:
x = [1.8, 2.0, 1.7, 1.9, 1.6]
np.mean(x)
Out[4]:
1.8

As a convenience NumPy arrays provide the method mean().

In [5]:
x = np.array([1.8, 2.0, 1.7, 1.9, 1.6])
x.mean()
Out[5]:
1.8

The mode of a set of numbers is the number that occurs most often. If only one number occurs most often we say it is a unimodal set, and if two or more numbers occur the most with equal frequency than the set is multimodal. For example the set {1, 2, 2, 2, 3, 4, 4, 4} has modes 2 and 4, which is multimodal, and the set {5, 7, 7, 13} has the mode 7, and so it is unimodal. We will not be computing the mode in this manner in this book, but we do use the concepts of unimodal and multimodal in a more general sense. For example, in the Discrete Bayes chapter we talked about our belief in the dog's position as a multimodal distribution because we assigned different probabilities to different positions.

Finally, the median of a set of numbers is the middle point of the set so that half the values are below the median and half are above the median. Here, above and below is in relation to the set being sorted. If the set contains an even number of values then the two middle numbers are averaged together.

Numpy provides numpy.median() to compute the median. As you can see the median of {1.8, 2.0, 1.7, 1.9, 1.6} is 1.8, because 1.8 is the third element of this set after being sorted. In this case the median equals the mean, but that is not generally true.

In [6]:
np.median(x)
Out[6]:
1.8

## Expected Value of a Random Variable¶

The expected value of a random variable is the average value it would have if we took an infinite number of samples of it and then averaged those samples together. Let's say we have $x=[1,3,5]$ and each value is equally probable. What value would we expect $x$ to have, on average?

It would be the average of 1, 3, and 5, of course, which is 3. That should make sense; we would expect equal numbers of 1, 3, and 5 to occur, so $(1+3+5)/3=3$ is clearly the average of that infinite series of samples. In other words, here the expected value is the mean of the sample space.

Now suppose that each value has a different probability of happening. Say 1 has an 80% chance of occurring, 3 has an 15% chance, and 5 has only a 5% chance. In this case we compute the expected value by multiplying each value of $x$ by the percent chance of it occurring, and summing the result. For this case we could compute

$$\mathbb E[X] = (1)(0.8) + (3)(0.15) + (5)(0.05) = 1.5$$

Here I have introduced the notation $\mathbb E[X]$ for the expected value of $x$. Some texts use $E(x)$. The value 1.5 for $x$ makes intuitive sense because $x$ is far more likely to be 1 than 3 or 5, and 3 is more likely than 5 as well.

We can formalize this by letting $x_i$ be the $i^{th}$ value of $X$, and $p_i$ be the probability of its occurrence. This gives us

$$\mathbb E[X] = \sum_{i=1}^n p_ix_i$$

A trivial bit of algebra shows that if the probabilities are all equal, the expected value is the same as the mean:

$$\mathbb E[X] = \sum_{i=1}^n p_ix_i = \frac{1}{n}\sum_{i=1}^n x_i = \mu_x$$

If $x$ is continuous we substitute the sum for an integral, like so

$$\mathbb E[X] = \int_{a}^b\, xf(x) \,dx$$

where $f(x)$ is the probability distribution function of $x$. We won't be using this equation yet, but we will be using it in the next chapter.

We can write a bit of Python to simulate this. Here I take 1,000,000 samples and compute the expected value of the distribution we just computed analytically.

In [7]:
total = 0
N = 1000000
for r in np.random.rand(N):
if r <= .80: total += 1
elif r < .95: total += 3
else: total += 5

total / N
Out[7]:
1.499952

You can see that the computed value is close to the analytically derived value. It is not exact because getting an exact values requires an infinite sample size.

### Exercise¶

What is the expected value of a die roll?

### Solution¶

Each side is equally likely, so each has a probability of 1/6. Hence \begin{aligned} \mathbb E[X] &= 1/6\times1 + 1/6\times 2 + 1/6\times 3 + 1/6\times 4 + 1/6\times 5 + 1/6\times6 \\ &= 1/6(1+2+3+4+5+6)\\&= 3.5\end{aligned}

### Exercise¶

Given the uniform continuous distribution

$$f(x) = \frac{1}{b - a}$$

compute the expected value for $a=0$ and $b=20$.

### Solution¶

\begin{aligned} \mathbb E[X] &= \int_0^{20}\, x\frac{1}{20} \,dx \\ &= \bigg[\frac{x^2}{40}\bigg]_0^{20} \\ &= 10 - 0 \\ &= 10 \end{aligned}

### Variance of a Random Variable¶

The computation above tells us the average height of the students, but it doesn't tell us everything we might want to know. For example, suppose we have three classes of students, which we label $X$, $Y$, and $Z$, with these heights:

In [8]:
X = [1.8, 2.0, 1.7, 1.9, 1.6]
Y = [2.2, 1.5, 2.3, 1.7, 1.3]
Z = [1.8, 1.8, 1.8, 1.8, 1.8]

Using NumPy we see that the mean height of each class is the same.

In [9]:
print(np.mean(X), np.mean(Y), np.mean(Z))
1.8 1.8 1.8

The mean of each class is 1.8 meters, but notice that there is a much greater amount of variation in the heights in the second class than in the first class, and that there is no variation at all in the third class.

The mean tells us something about the data, but not the whole story. We want to be able to specify how much variation there is between the heights of the students. You can imagine a number of reasons for this. Perhaps a school district needs to order 5,000 desks, and they want to be sure they buy sizes that accommodate the range of heights of the students.

Statistics has formalized this concept of measuring variation into the notion of standard deviation and variance. The equation for computing the variance is

$$\mathit{VAR}(X) = \mathbb E[(X - \mu)^2]$$

Ignoring the square for a moment, you can see that the variance is the expected value for how much the sample space $X$ varies from the mean $\mu:$ ($X-\mu)$. I will explain the purpose of the squared term later. The formula for the expected value is $\mathbb E[X] = \sum\limits_{i=1}^n p_ix_i$ so we can substitute that into the equation above to get

$$\mathit{VAR}(X) = \frac{1}{n}\sum_{i=1}^n (x_i - \mu)^2$$

Let's compute the variance of the three classes to see what values we get and to become familiar with this concept.

The mean of $X$ is 1.8 ($\mu_x = 1.8$) so we compute

\begin{aligned} \mathit{VAR}(X) &=\frac{(1.8-1.8)^2 + (2-1.8)^2 + (1.7-1.8)^2 + (1.9-1.8)^2 + (1.6-1.8)^2} {5} \\ &= \frac{0 + 0.04 + 0.01 + 0.01 + 0.04}{5} \\ \mathit{VAR}(X)&= 0.02 \, m^2 \end{aligned}

NumPy provides the function var() to compute the variance:

In [10]:
print(f"{np.var(X):.2f} meters squared")
0.02 meters squared

This is perhaps a bit hard to interpret. Heights are in meters, yet the variance is meters squared. Thus we have a more commonly used measure, the standard deviation, which is defined as the square root of the variance:

$$\sigma = \sqrt{\mathit{VAR}(X)}=\sqrt{\frac{1}{n}\sum_{i=1}^n(x_i - \mu)^2}$$

It is typical to use $\sigma$ for the standard deviation and $\sigma^2$ for the variance. In most of this book I will be using $\sigma^2$ instead of $\mathit{VAR}(X)$ for the variance; they symbolize the same thing.

For the first class we compute the standard deviation with

\begin{aligned} \sigma_x &=\sqrt{\frac{(1.8-1.8)^2 + (2-1.8)^2 + (1.7-1.8)^2 + (1.9-1.8)^2 + (1.6-1.8)^2} {5}} \\ &= \sqrt{\frac{0 + 0.04 + 0.01 + 0.01 + 0.04}{5}} \\ \sigma_x&= 0.1414 \end{aligned}

We can verify this computation with the NumPy method numpy.std() which computes the standard deviation. 'std' is a common abbreviation for standard deviation.

In [11]:
print(f"std {np.std(X):.4f}")
print(f"var {np.std(X)**2:.4f}")
std 0.1414
var 0.0200

And, of course, $0.1414^2 = 0.02$, which agrees with our earlier computation of the variance.

What does the standard deviation signify? It tells us how much the heights vary amongst themselves. "How much" is not a mathematical term. We will be able to define it much more precisely once we introduce the concept of a Gaussian in the next section. For now I'll say that for many things 68% of all values lie within one standard deviation of the mean. In other words we can conclude that for a random class 68% of the students will have heights between 1.66 (1.8-0.1414) meters and 1.94 (1.8+0.1414) meters.

We can view this in a plot:

In [12]:
from kf_book.gaussian_internal import plot_height_std
import matplotlib.pyplot as plt

plot_height_std(X)

For only 5 students we obviously will not get exactly 68% within one standard deviation. We do see that 3 out of 5 students are within $\pm1\sigma$, or 60%, which is as close as you can get to 68% with only 5 samples. Let's look at the results for a class with 100 students.

We write one standard deviation as $1\sigma$, which is pronounced "one standard deviation", not "one sigma". Two standard deviations is $2\sigma$, and so on.

In [13]:
from numpy.random import randn
data = 1.8 + randn(100)*.1414
mean, std = data.mean(), data.std()

plot_height_std(data, lw=2)
print(f'mean = {mean:.3f}')
print(f'std  = {std:.3f}')
mean = 1.811
std  = 0.130

By eye roughly 68% of the heights lie within $\pm1\sigma$ of the mean 1.8, but we can verify this with code.

In [14]:
np.sum((data > mean-std) & (data < mean+std)) / len(data) * 100.
Out[14]:
69.0

We'll discuss this in greater depth soon. For now let's compute the standard deviation for

$$Y = [2.2, 1.5, 2.3, 1.7, 1.3]$$

The mean of $Y$ is $\mu=1.8$ m, so

\begin{aligned} \sigma_y &=\sqrt{\frac{(2.2-1.8)^2 + (1.5-1.8)^2 + (2.3-1.8)^2 + (1.7-1.8)^2 + (1.3-1.8)^2} {5}} \\ &= \sqrt{0.152} = 0.39 \ m \end{aligned}

We will verify that with NumPy with

In [15]:
print(f'std of Y is {np.std(Y):.2f} m')
std of Y is 0.39 m

This corresponds with what we would expect. There is more variation in the heights for $Y$, and the standard deviation is larger.

Finally, let's compute the standard deviation for $Z$. There is no variation in the values, so we would expect the standard deviation to be zero.

\begin{aligned} \sigma_z &=\sqrt{\frac{(1.8-1.8)^2 + (1.8-1.8)^2 + (1.8-1.8)^2 + (1.8-1.8)^2 + (1.8-1.8)^2} {5}} \\ &= \sqrt{\frac{0+0+0+0+0}{5}} \\ \sigma_z&= 0.0 \ m \end{aligned}
In [16]:
print(np.std(Z))
0.0

Before we continue I need to point out that I'm ignoring that on average men are taller than women. In general the height variance of a class that contains only men or women will be smaller than a class with both sexes. This is true for other factors as well. Well nourished children are taller than malnourished children. Scandinavians are taller than Italians. When designing experiments statisticians need to take these factors into account.

I suggested we might be performing this analysis to order desks for a school district. For each age group there are likely to be two different means - one clustered around the mean height of the females, and a second mean clustered around the mean heights of the males. The mean of the entire class will be somewhere between the two. If we bought desks for the mean of all students we are likely to end up with desks that fit neither the males or females in the school!

We will not to consider these issues in this book. Consult any standard probability text if you need to learn techniques to deal with these issues.

### Why the Square of the Differences¶

Why are we taking the square of the differences for the variance? I could go into a lot of math, but let's look at this in a simple way. Here is a chart of the values of $X$ plotted against the mean for $X=[3,-3,3,-3]$

In [17]:
X = [3, -3, 3, -3]
mean = np.average(X)
for i in range(len(X)):
plt.plot([i ,i], [mean, X[i]], color='k')
plt.axhline(mean)
plt.xlim(-1, len(X))
plt.tick_params(axis='x', labelbottom=False)

If we didn't take the square of the differences the signs would cancel everything out:

$$\frac{(3-0) + (-3-0) + (3-0) + (-3-0)}{4} = 0$$

This is clearly incorrect, as there is more than 0 variance in the data.

Maybe we can use the absolute value? We can see by inspection that the result is $12/4=3$ which is certainly correct — each value varies by 3 from the mean. But what if we have $Y=[6, -2, -3, 1]$? In this case we get $12/4=3$. $Y$ is clearly more spread out than $X$, but the computation yields the same variance. If we use the formula using squares we get a variance of 3.5 for $Y$, which reflects its larger variation.

This is not a proof of correctness. Indeed, Carl Friedrich Gauss, the inventor of the technique, recognized that it is somewhat arbitrary. If there are outliers then squaring the difference gives disproportionate weight to that term. For example, let's see what happens if we have:

In [18]:
X = [1, -1, 1, -2, -1, 2, 1, 2, -1, 1, -1, 2, 1, -2, 100]
print(f'Variance of X with outlier    = {np.var(X):6.2f}')
print(f'Variance of X without outlier = {np.var(X[:-1]):6.2f}')
Variance of X with outlier    = 621.45
Variance of X without outlier =   2.03

Is this "correct"? You tell me. Without the outlier of 100 we get $\sigma^2=2.03$, which accurately reflects how $X$ is varying absent the outlier. The one outlier swamps the variance computation. Do we want to swamp the computation so we know there is an outlier, or robustly incorporate the outlier and still provide an estimate close to the value absent the outlier? Again, you tell me. Obviously it depends on your problem.

I will not continue down this path; if you are interested you might want to look at the work that James Berger has done on this problem, in a field called Bayesian robustness, or the excellent publications on robust statistics by Peter J. Huber [4]. In this book we will always use variance and standard deviation as defined by Gauss.

The point to gather from this is that these summary statistics always tell an incomplete story about our data. In this example variance as defined by Gauss does not tell us we have a single large outlier. However, it is a powerful tool, as we can concisely describe a large data set with a few numbers. If we had 1 billion data points we would not want to inspect plots by eye or look at lists of numbers; summary statistics give us a way to describe the shape of the data in a useful way.

## Gaussians¶

We are now ready to learn about Gaussians. Let's remind ourselves of the motivation for this chapter.

We desire a unimodal, continuous way to represent probabilities that models how the real world works, and that is computationally efficient to calculate.

Let's look at a graph of a Gaussian distribution to get a sense of what we are talking about.

In [19]:
from filterpy.stats import plot_gaussian_pdf
plot_gaussian_pdf(mean=1.8, variance=0.1414**2,
xlabel='Student Height', ylabel='pdf');

This curve is a probability density function or pdf for short. It shows the relative likelihood for the random variable to take on a value. We can tell from the chart student is somewhat more likely to have a height near 1.8 m than 1.7 m, and far more likely to have a height of 1.9 m vs 1.4 m. Put another way, many students will have a height near 1.8 m, and very few students will have a height of 1.4 m or 2.2 meters. Finally, notice that the curve is centered over the mean of 1.8 m.

I explain how to plot Gaussians, and much more, in the Notebook Computing_and_Plotting_PDFs in the Supporting_Notebooks folder. You can read it online here [1].

This may be recognizable to you as a 'bell curve'. This curve is ubiquitous because under real world conditions many observations are distributed in such a manner. I will not use the term 'bell curve' to refer to a Gaussian because many probability distributions have a similar bell curve shape. Non-mathematical sources might not be as precise, so be judicious in what you conclude when you see the term used without definition.

This curve is not unique to heights — a vast amount of natural phenomena exhibits this sort of distribution, including the sensors that we use in filtering problems. As we will see, it also has all the attributes that we are looking for — it represents a unimodal belief or value as a probability, it is continuous, and it is computationally efficient. We will soon discover that it also has other desirable qualities which we may not realize we desire.

To further motivate you, recall the shapes of the probability distributions in the Discrete Bayes chapter:

In [20]:
import kf_book.book_plots as book_plots
belief = [0., 0., 0., 0.1, 0.15, 0.5, 0.2, .15, 0, 0]
book_plots.bar_plot(belief)

They were not perfect Gaussian curves, but they were similar. We will be using Gaussians to replace the discrete probabilities used in that chapter!

## Nomenclature¶

A bit of nomenclature before we continue - this chart depicts the probability density of a random variable having any value between ($-\infty..\infty)$. What does that mean? Imagine we take an infinite number of infinitely precise measurements of the speed of automobiles on a section of highway. We could then plot the results by showing the relative number of cars going past at any given speed. If the average was 120 kph, it might look like this:

In [21]:
plot_gaussian_pdf(mean=120, variance=17**2, xlabel='speed(kph)');

The y-axis depicts the probability density — the relative amount of cars that are going the speed at the corresponding x-axis. I will explain this further in the next section.

The Gaussian model is imperfect. Though these charts do not show it, the tails of the distribution extend out to infinity. Tails are the far ends of the curve where the values are the lowest. Of course human heights or automobile speeds cannot be less than zero, let alone $-\infty$ or $\infty$. “The map is not the territory” is a common expression, and it is true for Bayesian filtering and statistics. The Gaussian distribution above models the distribution of the measured automobile speeds, but being a model it is necessarily imperfect. The difference between model and reality will come up again and again in these filters. Gaussians are used in many branches of mathematics, not because they perfectly model reality, but because they are easier to use than any other relatively accurate choice. However, even in this book Gaussians will fail to model reality, forcing us to use computationally expensive alternatives.

You will hear these distributions called Gaussian distributions or normal distributions. Gaussian and normal both mean the same thing in this context, and are used interchangeably. I will use both throughout this book as different sources will use either term, and I want you to be used to seeing both. Finally, as in this paragraph, it is typical to shorten the name and talk about a Gaussian or normal — these are both typical shortcut names for the Gaussian distribution.

## Gaussian Distributions¶

Let's explore how Gaussians work. A Gaussian is a continuous probability distribution that is completely described with two parameters, the mean ($\mu$) and the variance ($\sigma^2$). It is defined as:

$$f(x, \mu, \sigma) = \frac{1}{\sigma\sqrt{2\pi}} \exp\big [{-\frac{(x-\mu)^2}{2\sigma^2} }\big ]$$

$\exp[x]$ is notation for $e^x$.

Don't be dissuaded by the equation if you haven't seen it before; you will not need to memorize or manipulate it. The computation of this function is stored in stats.py with the function gaussian(x, mean, var, normed=True).

Shorn of the constants, you can see it is a simple exponential:

$$f(x)\propto e^{-x^2}$$

which has the familiar bell curve shape

In [22]:
x = np.arange(-3, 3, .01)
plt.plot(x, np.exp(-x**2));

Let's remind ourselves how to look at the code for a function. In a cell, type the function name followed by two question marks and press CTRL+ENTER. This will open a popup window displaying the source. Uncomment the next cell and try it now.

In [23]:
from filterpy.stats import gaussian
#gaussian??

Let's plot a Gaussian with a mean of 22 $(\mu=22)$, with a variance of 4 $(\sigma^2=4)$.

In [24]:
plot_gaussian_pdf(22, 4, mean_line=True, xlabel='$^{\circ}C$');

What does this curve mean? Assume we have a thermometer which reads 22°C. No thermometer is perfectly accurate, and so we expect that each reading will be slightly off the actual value. However, a theorem called Central Limit Theorem states that if we make many measurements that the measurements will be normally distributed. When we look at this chart we can see it is proportional to the probability of the thermometer reading a particular value given the actual temperature of 22°C.

Recall that a Gaussian distribution is continuous. Think of an infinitely long straight line - what is the probability that a point you pick randomly is at 2. Clearly 0%, as there is an infinite number of choices to choose from. The same is true for normal distributions; in the graph above the probability of being exactly 2°C is 0% because there are an infinite number of values the reading can take.

What is this curve? It is something we call the probability density function. The area under the curve at any region gives you the probability of those values. So, for example, if you compute the area under the curve between 20 and 22 the resulting area will be the probability of the temperature reading being between those two temperatures.

Here is another way to understand it. What is the density of a rock, or a sponge? It is a measure of how much mass is compacted into a given space. Rocks are dense, sponges less so. So, if you wanted to know how much a rock weighed but didn't have a scale, you could take its volume and multiply by its density. This would give you its mass. In practice density varies in most objects, so you would integrate the local density across the rock's volume.

$$M = \iiint_R p(x,y,z)\, dV$$

We do the same with probability density. If you want to know the temperature being between 20°C and 21°C you would integrate the curve above from 20 to 21. As you know the integral of a curve gives you the area under the curve. Since this is a curve of the probability density, the integral of the density is the probability.

What is the probability of the temperature being exactly 22°C? Intuitively, 0. These are real numbers, and the odds of 22°C vs, say, 22.00000000000017°C is infinitesimal. Mathematically, what would we get if we integrate from 22 to 22? Zero.

Thinking back to the rock, what is the weight of an single point on the rock? An infinitesimal point must have no weight. It makes no sense to ask the weight of a single point, and it makes no sense to ask about the probability of a continuous distribution having a single value. The answer for both is obviously zero.

In practice our sensors do not have infinite precision, so a reading of 22°C implies a range, such as 22 $\pm$ 0.1°C, and we can compute the probability of that range by integrating from 21.9 to 22.1.

We can think of this in Bayesian terms or frequentist terms. As a Bayesian, if the thermometer reads exactly 22°C, then our belief is described by the curve - our belief that the actual (system) temperature is near 22°C is very high, and our belief that the actual temperature is near 18 is very low. As a frequentist we would say that if we took 1 billion temperature measurements of a system at exactly 22°C, then a histogram of the measurements would look like this curve.

How do you compute the probability, or area under the curve? You integrate the equation for the Gaussian

$$\int^{x_1}_{x_0} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}{(x-\mu)^2}/\sigma^2 } dx$$

This is called the cumulative probability distribution, commonly abbreviated cdf.

I wrote filterpy.stats.norm_cdf which computes the integral for you. For example, we can compute

In [25]:
from filterpy.stats import norm_cdf
print('Cumulative probability of range 21.5 to 22.5 is {:.2f}%'.format(
norm_cdf((21.5, 22.5), 22,4)*100))
print('Cumulative probability of range 23.5 to 24.5 is {:.2f}%'.format(
norm_cdf((23.5, 24.5), 22,4)*100))
Cumulative probability of range 21.5 to 22.5 is 19.74%
Cumulative probability of range 23.5 to 24.5 is 12.10%

The mean ($\mu$) is what it sounds like — the average of all possible probabilities. Because of the symmetric shape of the curve it is also the tallest part of the curve. The thermometer reads 22°C, so that is what we used for the mean.

The notation for a normal distribution for a random variable $X$ is $X \sim\ \mathcal{N}(\mu,\sigma^2)$ where $\sim$ means distributed according to. This means I can express the temperature reading of our thermometer as

$$\text{temp} \sim \mathcal{N}(22,4)$$

This is an extremely important result. Gaussians allow me to capture an infinite number of possible values with only two numbers! With the values $\mu=22$ and $\sigma^2=4$ I can compute the distribution of measurements over any range.

Some sources use $\mathcal N (\mu, \sigma)$ instead of $\mathcal N (\mu, \sigma^2)$. Either is fine, they are both conventions. You need to keep in mind which form is being used if you see a term such as $\mathcal{N}(22,4)$. In this book I always use $\mathcal N (\mu, \sigma^2)$, so $\sigma=2$, $\sigma^2=4$ for this example.

## The Variance and Belief¶

Since this is a probability density distribution it is required that the area under the curve always equals one. This should be intuitively clear — the area under the curve represents all possible outcomes, something happened, and the probability of something happening is one, so the density must sum to one. We can prove this ourselves with a bit of code. (If you are mathematically inclined, integrate the Gaussian equation from $-\infty$ to $\infty$)

In [26]:
print(norm_cdf((-1e8, 1e8), mu=0, var=4))
1.0

This leads to an important insight. If the variance is small the curve will be narrow. this is because the variance is a measure of how much the samples vary from the mean. To keep the area equal to 1, the curve must also be tall. On the other hand if the variance is large the curve will be wide, and thus it will also have to be short to make the area equal to 1.

Let's look at that graphically. We will use the aforementioned filterpy.stats.gaussian which can take either a single value or array of values.

In [27]:
from filterpy.stats import gaussian

print(gaussian(x=3.0, mean=2.0, var=1))
print(gaussian(x=[3.0, 2.0], mean=2.0, var=1))
0.24197072451914337
[0.378 0.622]

By default gaussian normalizes the output, which turns the output back into a probability distribution. Use the argumentnormed to control this.

In [28]:
print(gaussian(x=[3.0, 2.0], mean=2.0, var=1, normed=False))
[0.242 0.399]

If the Gaussian is not normalized it is called a Gaussian function instead of Gaussian distribution.

In [29]:
xs = np.arange(15, 30, 0.05)
plt.plot(xs, gaussian(xs, 23, 0.2**2), label='$\sigma^2=0.2^2$')
plt.plot(xs, gaussian(xs, 23, .5**2), label='$\sigma^2=0.5^2$', ls=':')
plt.plot(xs, gaussian(xs, 23, 1**2), label='$\sigma^2=1^2$', ls='--')
plt.legend();

What is this telling us? The Gaussian with $\sigma^2=0.2^2$ is very narrow. It is saying that we believe $x=23$, and that we are very sure about that: within $\pm 0.2$ std. In contrast, the Gaussian with $\sigma^2=1^2$ also believes that $x=23$, but we are much less sure about that. Our belief that $x=23$ is lower, and so our belief about the likely possible values for $x$ is spread out — we think it is quite likely that $x=20$ or $x=26$, for example. $\sigma^2=0.2^2$ has almost completely eliminated $22$ or $24$ as possible values, whereas $\sigma^2=1^2$ considers them nearly as likely as $23$.

If we think back to the thermometer, we can consider these three curves as representing the readings from three different thermometers. The curve for $\sigma^2=0.2^2$ represents a very accurate thermometer, and curve for $\sigma^2=1^2$ represents a fairly inaccurate one. Note the very powerful property the Gaussian distribution affords us — we can entirely represent both the reading and the error of a thermometer with only two numbers — the mean and the variance.

An equivalent formation for a Gaussian is $\mathcal{N}(\mu,1/\tau)$ where $\mu$ is the mean and $\tau$ the precision. $1/\tau = \sigma^2$; it is the reciprocal of the variance. While we do not use this formulation in this book, it underscores that the variance is a measure of how precise our data is. A small variance yields large precision — our measurement is very precise. Conversely, a large variance yields low precision — our belief is spread out across a large area. You should become comfortable with thinking about Gaussians in these equivalent forms. In Bayesian terms Gaussians reflect our belief about a measurement, they express the precision of the measurement, and they express how much variance there is in the measurements. These are all different ways of stating the same fact.

I'm getting ahead of myself, but in the next chapters we will use Gaussians to express our belief in things like the estimated position of the object we are tracking, or the accuracy of the sensors we are using.

## The 68-95-99.7 Rule¶

It is worth spending a few words on standard deviation now. The standard deviation is a measure of how much the data deviates from the mean. For Gaussian distributions, 68% of all the data falls within one standard deviation ($\pm1\sigma$) of the mean, 95% falls within two standard deviations ($\pm2\sigma$), and 99.7% within three ($\pm3\sigma$). This is often called the 68-95-99.7 rule. If you were told that the average test score in a class was 71 with a standard deviation of 9.4, you could conclude that 95% of the students received a score between 52.2 and 89.8 if the distribution is normal (that is calculated with $71 \pm (2 * 9.4)$).

Finally, these are not arbitrary numbers. If the Gaussian for our position is $\mu=22$ meters, then the standard deviation also has units meters. Thus $\sigma=0.2$ implies that 68% of the measurements range from 21.8 to 22.2 meters. Variance is the standard deviation squared, thus $\sigma^2 = .04$ meters$^2$. As you saw in the last section, writing $\sigma^2 = 0.2^2$ can make this somewhat more meaningful, since the 0.2 is in the same units as the data.

The following graph depicts the relationship between the standard deviation and the normal distribution.

In [30]:
from kf_book.gaussian_internal import display_stddev_plot
display_stddev_plot()

## Interactive Gaussians¶

For those that are reading this in a Jupyter Notebook, here is an interactive version of the Gaussian plots. Use the sliders to modify $\mu$ and $\sigma^2$. Adjusting $\mu$ will move the graph to the left and right because you are adjusting the mean, and adjusting $\sigma^2$ will make the bell curve thicker and thinner.

In [31]:
import math
from ipywidgets import interact, FloatSlider

def plt_g(mu,variance):
plt.figure()
xs = np.arange(2, 8, 0.01)
ys = gaussian(xs, mu, variance)
plt.plot(xs, ys)
plt.ylim(0, 0.04)

interact(plt_g, mu=FloatSlider(value=5, min=3, max=7),
variance=FloatSlider(value = .03, min=.01, max=1.));

Finally, if you are reading this online, here is an animation of a Gaussian. First, the mean is shifted to the right. Then the mean is centered at $\mu=5$ and the variance is modified.

## Computational Properties of Gaussians¶

The discrete Bayes filter works by multiplying and adding arbitrary probability distributions. The Kalman filter uses Gaussians instead of arbitrary distributions, but the rest of the algorithm remains the same. This means we will need to multiply and add Gaussians.

A remarkable property of Gaussians is that the sum of two independent independent normal variables (https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables) is also normally distributed! The product is not Gaussian, but proportional to a Gaussian. There we can say that the result of multipying two Gaussian distributions is a Gaussian function (recall function in this context means that the property that the values sum to one is not guaranteed).

Before we do the math, let's test this visually.

In [32]:
x = np.arange(-1, 3, 0.01)
g1 = gaussian(x, mean=0.8, var=.1)
g2 = gaussian(x, mean=1.3, var=.2)
plt.plot(x, g1, x, g2)

g = g1 * g2  # element-wise multiplication
g = g / sum(g)  # normalize
plt.plot(x, g, ls='-.');

Here I created two Gaussians, g1=$\mathcal N(0.8, 0.1)$ and g2=$\mathcal N(1.3, 0.2)$ and plotted them. Then I multiplied them together and normalized the result. As you can see the result looks like a Gaussian distribution.

Gaussians are nonlinear functions. Typically, if you multiply a nonlinear equations you end up with a different type of function. For example, the shape of multiplying two sins is very different from sin(x).

In [33]:
x = np.arange(0, 4*np.pi, 0.01)
plt.plot(np.sin(1.2*x))
plt.plot(np.sin(1.2*x) * np.sin(2*x));

But the result of multiplying two Gaussians distributions is a Gaussian function. This is a key reason why Kalman filters are computationally feasible. Said another way, Kalman filters use Gaussians because they are computationally nice.

The product of two independent Gaussians is given by:

\begin{aligned}\mu &=\frac{\sigma_1^2\mu_2 + \sigma_2^2\mu_1}{\sigma_1^2+\sigma_2^2}\\ \sigma^2 &=\frac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \end{aligned}

The sum of two Gaussians is given by

$$\begin{gathered}\mu = \mu_1 + \mu_2 \\ \sigma^2 = \sigma^2_1 + \sigma^2_2 \end{gathered}$$

At the end of the chapter I derive these equations. However, understanding the deriviation is not very important.

## Putting it all Together¶

Now we are ready to talk about how Gaussians can be used in filtering. In the next chapter we will implement a filter using Gaussins. Here I will explain why we would want to use Gaussians.

In the previous chapter we represented probability distributions with an array. We performed the update computation by computing the element-wise product of that distribution with another distribution representing the likelihood of the measurement at each point, like so:

In [34]:
def normalize(p):
return p / sum(p)

def update(likelihood, prior):
return normalize(likelihood * prior)

prior =      normalize(np.array([4, 2, 0, 7, 2, 12, 35, 20, 3, 2]))
likelihood = normalize(np.array([3, 4, 1, 4, 2, 38, 20, 18, 1, 16]))
posterior = update(likelihood, prior)
book_plots.bar_plot(posterior)

In other words, we have to compute 10 multiplications to get this result. For a real filter with large arrays in multiple dimensions we'd require billions of multiplications, and vast amounts of memory.

But this distribution looks like a Gaussian. What if we use a Gaussian instead of an array? I'll compute the mean and variance of the posterior and plot it against the bar chart.

In [35]:
xs = np.arange(0, 10, .01)

def mean_var(p):
x = np.arange(len(p))
mean = np.sum(p * x,dtype=float)
var = np.sum((x - mean)**2 * p)
return mean, var

mean, var = mean_var(posterior)
book_plots.bar_plot(posterior)
plt.plot(xs, gaussian(xs, mean, var, normed=False), c='r');
print('mean: %.2f' % mean, 'var: %.2f' % var)
mean: 5.88 var: 1.24

This is impressive. We can describe an entire distribution of numbers with only two numbers. Perhaps this example is not persuasive, given there are only 10 numbers in the distribution. But a real problem could have millions of numbers, yet still only require two numbers to describe it.

Next, recall that our filter implements the update function with

def update(likelihood, prior):
return normalize(likelihood * prior)

If the arrays contain a million elements, that is one million multiplications. However, if we replace the arrays with a Gaussian then we would perform that calculation with

\begin{aligned}\mu &=\frac{\sigma_1^2\mu_2 + \sigma_2^2\mu_1}{\sigma_1^2+\sigma_2^2}\\ \sigma^2 &=\frac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \end{aligned}

which is three multiplications and two divisions.

### Bayes Theorem¶

In the last chapter we developed an algorithm by reasoning about the information we have at each moment, which we expressed as discrete probability distributions. In the process we discovered Bayes' Theorem. Bayes theorem tells us how to compute the probability of an event given prior information.

We implemented the update() function with this probability calculation:

$$\mathtt{posterior} = \frac{\mathtt{likelihood}\times \mathtt{prior}}{\mathtt{normalization}}$$

It turns out that this is Bayes' theorem. In a second I will develop the mathematics, but in many ways that obscures the simple idea expressed in this equation. We read this as:

$$updated\,knowledge = \big\|likelihood\,of\,new\,knowledge\times prior\, knowledge \big\|$$

where $\| \cdot\|$ expresses normalizing the term.

We came to this with simple reasoning about a dog walking down a hallway. Yet, as we will see, the same equation applies to a universe of filtering problems. We will use this equation in every subsequent chapter.

To review, the prior is the probability of something happening before we include the probability of the measurement (the likelihood) and the posterior is the probability we compute after incorporating the information from the measurement.

Bayes theorem is

$$P(A \mid B) = \frac{P(B \mid A)\, P(A)}{P(B)}$$

$P(A \mid B)$ is called a conditional probability. That is, it represents the probability of $A$ happening if $B$ happened. For example, it is more likely to rain today compared to a typical day if it also rained yesterday because rain systems usually last more than one day. We'd write the probability of it raining today given that it rained yesterday as $P$(rain today $\mid$ rain yesterday).

I've glossed over an important point. In our code above we are not working with single probabilities, but an array of probabilities - a probability distribution. The equation I just gave for Bayes uses probabilities, not probability distributions. However, it is equally valid with probability distributions. We use a lower case $p$ for probability distributions

$$p(A \mid B) = \frac{p(B \mid A)\, p(A)}{p(B)}$$

In the equation above $B$ is the evidence, $p(A)$ is the prior, $p(B \mid A)$ is the likelihood, and $p(A \mid B)$ is the posterior. By substituting the mathematical terms with the corresponding words you can see that Bayes theorem matches our update equation. Let's rewrite the equation in terms of our problem. We will use $x_i$ for the position at i, and $z$ for the measurement. Hence, we want to know $P(x_i \mid z)$, that is, the probability of the dog being at $x_i$ given the measurement $z$.

So, let's plug that into the equation and solve it.

$$p(x_i \mid z) = \frac{p(z \mid x_i) p(x_i)}{p(z)}$$

That looks ugly, but it is actually quite simple. Let's figure out what each term on the right means. First is $p(z \mid x_i)$. This is the likelihood, or the probability for the measurement at every cell $x_i$. $p(x_i)$ is the prior - our belief before incorporating the measurements. We multiply those together. This is just the unnormalized multiplication in the update() function:

def update(likelihood, prior):
posterior = prior * likelihood   # p(z|x) * p(x)
return normalize(posterior)

The last term to consider is the denominator $p(z)$. This is the probability of getting the measurement $z$ without taking the location into account. It is often called the evidence. We compute that by taking the sum of $x$, or sum(belief) in the code. That is how we compute the normalization! So, the update() function is doing nothing more than computing Bayes' theorem.

The literature often gives you these equations in the form of integrals. After all, an integral is just a sum over a continuous function. So, you might see Bayes' theorem written as

$$p(A \mid B) = \frac{p(B \mid A)\, p(A)}{\int p(B \mid A_j) p(A_j) \,\, \mathtt{d}A_j}\cdot$$

This denominator is usually impossible to solve analytically; when it can be solved the math is fiendishly difficult. A recent opinion piece for the Royal Statistical Society called it a "dog's breakfast" [8]. Filtering textbooks that take a Bayesian approach are filled with integral laden equations with no analytic solution. Do not be cowed by these equations, as we trivially handled this integral by normalizing our posterior. We will learn more techniques to handle this in the Particle Filters chapter. Until then, recognize that in practice it is just a normalization term over which we can sum. What I'm trying to say is that when you are faced with a page of integrals, just think of them as sums, and relate them back to this chapter, and often the difficulties will fade. Ask yourself "why are we summing these values", and "why am I dividing by this term". Surprisingly often the answer is readily apparent. Surprisingly often the author neglects to mention this interpretation.

It's probable that the strength of Bayes' theorem is not yet fully apparent to you. We want to compute $p(x_i \mid Z)$. That is, at step i, what is our probable state given a measurement. That's an extraordinarily difficult problem in general. Bayes' Theorem is general. We may want to know the probability that we have cancer given the results of a cancer test, or the probability of rain given various sensor readings. Stated like that the problems seem unsolvable.

But Bayes' Theorem lets us compute this by using the inverse $p(Z\mid x_i)$, which is often straightforward to compute

$$p(x_i \mid Z) \propto p(Z\mid x_i)\, p(x_i)$$

That is, to compute how likely it is to rain given specific sensor readings we only have to compute the likelihood of the sensor readings given that it is raining! That's a much easier problem! Well, weather prediction is still a difficult problem, but Bayes makes it tractable.

Likewise, as you saw in the Discrete Bayes chapter, we computed the likelihood that Simon was in any given part of the hallway by computing how likely a sensor reading is given that Simon is at position x. A hard problem becomes easy.

### Total Probability Theorem¶

We now know the formal mathematics behind the update() function; what about the predict() function? predict() implements the total probability theorem. Let's recall what predict() computed. It computed the probability of being at any given position given the probability of all the possible movement events. Let's express that as an equation. The probability of being at any position $i$ at time $t$ can be written as $P(X_i^t)$. We computed that as the sum of the prior at time $t-1$ $P(X_j^{t-1})$ multiplied by the probability of moving from cell $x_j$ to $x_i$. That is

$$P(X_i^t) = \sum_j P(X_j^{t-1}) P(x_i | x_j)$$

That equation is called the total probability theorem. Quoting from Wikipedia [6] "It expresses the total probability of an outcome which can be realized via several distinct events". I could have given you that equation and implemented predict(), but your chances of understanding why the equation works would be slim. As a reminder, here is the code that computes this equation

for i in range(N):
for k in range (kN):
index = (i + (width-k) - offset) % N
result[i] += prob_dist[index] * kernel[k]

## Computing Probabilities with scipy.stats¶

In this chapter I used code from FilterPy to compute and plot Gaussians. I did that to give you a chance to look at the code and see how these functions are implemented. However, Python comes with "batteries included" as the saying goes, and it comes with a wide range of statistics functions in the module scipy.stats. So let's walk through how to use scipy.stats to compute statistics and probabilities.

The scipy.stats module contains a number of objects which you can use to compute attributes of various probability distributions. The full documentation for this module is here: http://docs.scipy.org/doc/scipy/reference/stats.html. We will focus on the norm variable, which implements the normal distribution. Let's look at some code that uses scipy.stats.norm to compute a Gaussian, and compare its value to the value returned by the gaussian() function from FilterPy.

In [36]:
from scipy.stats import norm
import filterpy.stats
print(norm(2, 3).pdf(1.5))
print(filterpy.stats.gaussian(x=1.5, mean=2, var=3*3))
0.13114657203397997
0.13114657203397995

The call norm(2, 3) creates what scipy calls a 'frozen' distribution - it creates and returns an object with a mean of 2 and a standard deviation of 3. You can then use this object multiple times to get the probability density of various values, like so:

In [37]:
n23 = norm(2, 3)
print('pdf of 1.5 is       %.4f' % n23.pdf(1.5))
print('pdf of 2.5 is also  %.4f' % n23.pdf(2.5))
print('pdf of 2 is         %.4f' % n23.pdf(2))
pdf of 1.5 is       0.1311
pdf of 2.5 is also  0.1311
pdf of 2 is         0.1330

The documentation for scipy.stats.norm [2] lists many other functions. For example, we can generate $n$ samples from the distribution with the rvs() function.

In [38]:
np.set_printoptions(precision=3, linewidth=50)
print(n23.rvs(size=15))
[ 3.026  7.51  -2.588  6.081 -3.413 -1.11   6.484
5.935  2.313  1.912  1.895  7.964  4.876 -0.841
-0.174]

We can get the cumulative distribution function (CDF), which is the probability that a randomly drawn value from the distribution is less than or equal to $x$.

In [39]:
# probability that a random value is less than the mean 2
print(n23.cdf(2))
0.5

We can get various properties of the distribution:

In [40]:
print('variance is', n23.var())
print('standard deviation is', n23.std())
print('mean is', n23.mean())
variance is 9.0
standard deviation is 3.0
mean is 2.0

## Limitations of Using Gaussians to Model the World¶

Earlier I mentioned the central limit theorem, which states that under certain conditions the arithmetic sum of any independent random variable will be normally distributed, regardless of how the random variables are distributed. This is important to us because nature is full of distributions which are not normal, but when we apply the central limit theorem over large populations we end up with normal distributions.

However, a key part of the proof is “under certain conditions”. These conditions often do not hold for the physical world. For example, a kitchen scale cannot read below zero, but if we represent the measurement error as a Gaussian the left side of the curve extends to negative infinity, implying a very small chance of giving a negative reading.

This is a broad topic which I will not treat exhaustively.

Let's consider a trivial example. We think of things like test scores as being normally distributed. If you have ever had a professor “grade on a curve” you have been subject to this assumption. But of course test scores cannot follow a normal distribution. This is because the distribution assigns a nonzero probability distribution for any value, no matter how far from the mean. So, for example, say your mean is 90 and the standard deviation is 13. The normal distribution assumes that there is a large chance of somebody getting a 90, and a small chance of somebody getting a 40. However, it also implies that there is a tiny chance of somebody getting a grade of -10, or 150. It assigns an extremely small chance of getting a score of $-10^{300}$ or $10^{32986}$. The tails of a Gaussian distribution are infinitely long.

But for a test we know this is not true. Ignoring extra credit, you cannot get less than 0, or more than 100. Let's plot this range of values using a normal distribution to see how poorly this represents real test scores distributions.

In [41]:
xs = np.arange(10, 100, 0.05)
ys = [gaussian(x, 90, 30) for x in xs]
plt.plot(xs, ys, label='var=0.2')
plt.xlim(0, 120)
plt.ylim(-0.02, 0.09);

The area under the curve cannot equal 1, so it is not a probability distribution. What actually happens is that more students than predicted by a normal distribution get scores nearer the upper end of the range (for example), and that tail becomes “fat”. Also, the test is probably not able to perfectly distinguish minute differences in skill in the students, so the distribution to the left of the mean is also probably a bit bunched up in places.

Sensors measure the world. The errors in a sensor's measurements are rarely truly Gaussian. It is far too early to be talking about the difficulties that this presents to the Kalman filter designer. It is worth keeping in the back of your mind the fact that the Kalman filter math is based on an idealized model of the world. For now I will present a bit of code that I will be using later in the book to form distributions to simulate various processes and sensors. This distribution is called the Student's $t$-distribution.

Let's say I want to model a sensor that has some white noise in the output. For simplicity, let's say the signal is a constant 10, and the standard deviation of the noise is 2. We can use the function numpy.random.randn() to get a random number with a mean of 0 and a standard deviation of 1. I can simulate this with:

In [42]:
from numpy.random import randn
def sense():
return 10 + randn()*2

Let's plot that signal and see what it looks like.

In [43]:
zs = [sense() for i in range(5000)]
plt.plot(zs, lw=1);

That looks like what I would expect. The signal is centered around 10. A standard deviation of 2 means that 68% of the measurements will be within $\pm$ 2 of 10, and 99% will be within $\pm$ 6 of 10, and that looks like what is happening.

Now let's look at distribution generated with the Student's $t$-distribution. I will not go into the math, but just give you the source code for it and then plot a distribution using it.

In [44]:
import random
import math

def rand_student_t(df, mu=0, std=1):
"""return random number distributed by Student's t
distribution with df degrees of freedom with the
specified mean and standard deviation.
"""
x = random.gauss(0, std)
y = 2.0*random.gammavariate(0.5*df, 2.0)
return x / (math.sqrt(y / df)) + mu
In [45]:
def sense_t():
return 10 + rand_student_t(7)*2

zs = [sense_t() for i in range(5000)]
plt.plot(zs, lw=1);

We can see from the plot that while the output is similar to the normal distribution there are outliers that go far more than 3 standard deviations from the mean (7 to 13).

It is unlikely that the Student's $t$-distribution is an accurate model of how your sensor (say, a GPS or Doppler) performs, and this is not a book on how to model physical systems. However, it does produce reasonable data to test your filter's performance when presented with real world noise. We will be using distributions like these throughout the rest of the book in our simulations and tests.

This is not an idle concern. The Kalman filter equations assume the noise is normally distributed, and perform sub-optimally if this is not true. Designers for mission critical filters, such as the filters on spacecraft, need to master a lot of theory and empirical knowledge about the performance of the sensors on their spacecraft. For example, a presentation I saw on a NASA mission stated that while theory states that they should use 3 standard deviations to distinguish noise from valid measurements in practice they had to use 5 to 6 standard deviations. This was something they determined by experiments.

The code for rand_student_t is included in filterpy.stats. You may use it with

from filterpy.stats import rand_student_t

While I'll not cover it here, statistics has defined ways of describing the shape of a probability distribution by how it varies from an exponential distribution. The normal distribution is shaped symmetrically around the mean - like a bell curve. However, a probability distribution can be asymmetrical around the mean. The measure of this is called skew. The tails can be shortened, fatter, thinner, or otherwise shaped differently from an exponential distribution. The measure of this is called kurtosis. the scipy.stats module contains the function describe which computes these statistics, among others.

In [46]:
import scipy
scipy.stats.describe(zs)
Out[46]:
DescribeResult(nobs=5000, minmax=(1.7707248080804092, 17.824961707265643), mean=10.004201301507253, variance=2.721973140216924, skewness=-0.03534037712630867, kurtosis=1.583968849611276)

Let's examine two normal populations, one small, one large:

In [47]:
print(scipy.stats.describe(np.random.randn(10)))
print()
print(scipy.stats.describe(np.random.randn(300000)))
DescribeResult(nobs=10, minmax=(-1.1170056026895896, 1.0389413316321683), mean=-0.14317693801053674, variance=0.63137454114842, skewness=0.23966670696111078, kurtosis=-1.3702119526058378)

DescribeResult(nobs=300000, minmax=(-4.262303362805642, 4.603650808299195), mean=-0.00040808441709135014, variance=0.998953112076118, skewness=-0.0010979017079029859, kurtosis=-0.00030025117171517124)

The small sample has very non-zero skew and kurtosis because the small number of samples is not well distributed around the mean of 0. You can see this also by comparing the computed mean and variance with the theoretical mean of 0 and variance 1. In comparison the large sample's mean and variance are very close to the theoretical values, and both the skew and kurtosis are near zero.

## Product of Gaussians (Optional)¶

It is not important to read this section. Here I derive the equations for the product of two Gaussians.

You can find this result by multiplying the equation for two Gaussians together and combining terms. The algebra gets messy. I will derive it using Bayes theorem. We can state the problem as: let the prior be $N(\bar\mu, \bar\sigma^2)$, and measurement be $z \propto N(z, \sigma_z^2)$. What is the posterior x given the measurement z?

Write the posterior as $p(x \mid z)$. Now we can use Bayes Theorem to state

$$p(x \mid z) = \frac{p(z \mid x)p(x)}{p(z)}$$

$p(z)$ is a normalizing constant, so we can create a proportinality

$$p(x \mid z) \propto p(z|x)p(x)$$

Now we subtitute in the equations for the Gaussians, which are

$$p(z \mid x) = \frac{1}{\sqrt{2\pi\sigma_z^2}}\exp \Big[-\frac{(z-x)^2}{2\sigma_z^2}\Big]$$$$p(x) = \frac{1}{\sqrt{2\pi\bar\sigma^2}}\exp \Big[-\frac{(x-\bar\mu)^2}{2\bar\sigma^2}\Big]$$

We can drop the leading terms, as they are constants, giving us

\begin{aligned} p(x \mid z) &\propto \exp \Big[-\frac{(z-x)^2}{2\sigma_z^2}\Big]\exp \Big[-\frac{(x-\bar\mu)^2}{2\bar\sigma^2}\Big]\\ &\propto \exp \Big[-\frac{(z-x)^2}{2\sigma_z^2}-\frac{(x-\bar\mu)^2}{2\bar\sigma^2}\Big] \\ &\propto \exp \Big[-\frac{1}{2\sigma_z^2\bar\sigma^2}[\bar\sigma^2(z-x)^2+\sigma_z^2(x-\bar\mu)^2]\Big] \end{aligned}

Now we multiply out the squared terms and group in terms of the posterior $x$.

\begin{aligned} p(x \mid z) &\propto \exp \Big[-\frac{1}{2\sigma_z^2\bar\sigma^2}[\bar\sigma^2(z^2 -2xz + x^2) + \sigma_z^2(x^2 - 2x\bar\mu+\bar\mu^2)]\Big ] \\ &\propto \exp \Big[-\frac{1}{2\sigma_z^2\bar\sigma^2}[x^2(\bar\sigma^2+\sigma_z^2)-2x(\sigma_z^2\bar\mu + \bar\sigma^2z) + (\bar\sigma^2z^2+\sigma_z^2\bar\mu^2)]\Big ] \end{aligned}

The last parentheses do not contain the posterior $x$, so it can be treated as a constant and discarded.

$$p(x \mid z) \propto \exp \Big[-\frac{1}{2}\frac{x^2(\bar\sigma^2+\sigma_z^2)-2x(\sigma_z^2\bar\mu + \bar\sigma^2z)}{\sigma_z^2\bar\sigma^2}\Big ]$$

Divide numerator and denominator by $\bar\sigma^2+\sigma_z^2$ to get

$$p(x \mid z) \propto \exp \Big[-\frac{1}{2}\frac{x^2-2x(\frac{\sigma_z^2\bar\mu + \bar\sigma^2z}{\bar\sigma^2+\sigma_z^2})}{\frac{\sigma_z^2\bar\sigma^2}{\bar\sigma^2+\sigma_z^2}}\Big ]$$

Proportionality allows us create or delete constants at will, so we can factor this into

$$p(x \mid z) \propto \exp \Big[-\frac{1}{2}\frac{(x-\frac{\sigma_z^2\bar\mu + \bar\sigma^2z}{\bar\sigma^2+\sigma_z^2})^2}{\frac{\sigma_z^2\bar\sigma^2}{\bar\sigma^2+\sigma_z^2}}\Big ]$$

A Gaussian is

$$N(\mu,\, \sigma^2) \propto \exp\Big [-\frac{1}{2}\frac{(x - \mu)^2}{\sigma^2}\Big ]$$

So we can see that $p(x \mid z)$ has a mean of

$$\mu_\mathtt{posterior} = \frac{\sigma_z^2\bar\mu + \bar\sigma^2z}{\bar\sigma^2+\sigma_z^2}$$

and a variance of $$\sigma_\mathtt{posterior} = \frac{\sigma_z^2\bar\sigma^2}{\bar\sigma^2+\sigma_z^2}$$

I've dropped the constants, and so the result is not a normal, but proportional to one. Bayes theorem normalizes with the $p(z)$ divisor, ensuring that the result is normal. We normalize in the update step of our filters, ensuring the filter estimate is Gaussian.

$$\mathcal N_1 = \| \mathcal N_2\cdot \mathcal N_3\|$$

## Sum of Gaussians (Optional)¶

Likewise, this section is not important to read. Here I derive the equations for the sum of two Gaussians.

The sum of two Gaussians is given by

$$\begin{gathered}\mu = \mu_1 + \mu_2 \\ \sigma^2 = \sigma^2_1 + \sigma^2_2 \end{gathered}$$

There are several proofs for this. I will use convolution since we used convolution in the previous chapter for the histograms of probabilities.

To find the density function of the sum of two Gaussian random variables we sum the density functions of each. They are nonlinear, continuous functions, so we need to compute the sum with an integral. If the random variables $p$ and $z$ (e.g. prior and measurement) are independent we can compute this with

$p(x) = \int\limits_{-\infty}^\infty f_p(x-z)f_z(z)\, dx$

This is the equation for a convolution. Now we just do some math:

$p(x) = \int\limits_{-\infty}^\infty f_2(x-x_1)f_1(x_1)\, dx$

$= \int\limits_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma_z}\exp\left[-\frac{(x - z - \mu_z)^2}{2\sigma^2_z}\right] \frac{1}{\sqrt{2\pi}\sigma_p}\exp\left[-\frac{(x - \mu_p)^2}{2\sigma^2_p}\right] \, dx$

$= \int\limits_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sqrt{\sigma_p^2 + \sigma_z^2}} \exp\left[ -\frac{(x - (\mu_p + \mu_z)))^2}{2(\sigma_z^2+\sigma_p^2)}\right] \frac{1}{\sqrt{2\pi}\frac{\sigma_p\sigma_z}{\sqrt{\sigma_p^2 + \sigma_z^2}}} \exp\left[ -\frac{(x - \frac{\sigma_p^2(x-\mu_z) + \sigma_z^2\mu_p}{}))^2}{2\left(\frac{\sigma_p\sigma_x}{\sqrt{\sigma_z^2+\sigma_p^2}}\right)^2}\right] \, dx$

$= \frac{1}{\sqrt{2\pi}\sqrt{\sigma_p^2 + \sigma_z^2}} \exp\left[ -\frac{(x - (\mu_p + \mu_z)))^2}{2(\sigma_z^2+\sigma_p^2)}\right] \int\limits_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\frac{\sigma_p\sigma_z}{\sqrt{\sigma_p^2 + \sigma_z^2}}} \exp\left[ -\frac{(x - \frac{\sigma_p^2(x-\mu_z) + \sigma_z^2\mu_p}{}))^2}{2\left(\frac{\sigma_p\sigma_x}{\sqrt{\sigma_z^2+\sigma_p^2}}\right)^2}\right] \, dx$

The expression inside the integral is a normal distribution. The sum of a normal distribution is one, hence the integral is one. This gives us

$$p(x) = \frac{1}{\sqrt{2\pi}\sqrt{\sigma_p^2 + \sigma_z^2}} \exp\left[ -\frac{(x - (\mu_p + \mu_z)))^2}{2(\sigma_z^2+\sigma_p^2)}\right]$$

This is in the form of a normal, where

$$\begin{gathered}\mu_x = \mu_p + \mu_z \\ \sigma_x^2 = \sigma_z^2+\sigma_p^2\, \square\end{gathered}$$

## Summary and Key Points¶

This chapter is a poor introduction to statistics in general. I've only covered the concepts that needed to use Gaussians in the remainder of the book, no more. What I've covered will not get you very far if you intend to read the Kalman filter literature. If this is a new topic to you I suggest reading a statistics textbook. I've always liked the Schaum series for self study, and Alan Downey's Think Stats [5] is also very good and freely available online.

The following points must be understood by you before we continue:

• Normals express a continuous probability distribution
• They are completely described by two parameters: the mean ($\mu$) and variance ($\sigma^2$)
• $\mu$ is the average of all possible values
• The variance $\sigma^2$ represents how much our measurements vary from the mean
• The standard deviation ($\sigma$) is the square root of the variance ($\sigma^2$)
• Many things in nature approximate a normal distribution, but the math is not perfect.
• In filtering problems computing $p(x\mid z)$ is nearly impossible, but computing $p(z\mid x)$ is straightforward. Bayes' lets us compute the former from the latter.

The next several chapters will be using Gaussians with Bayes' theorem to help perform filtering. As noted in the last section, sometimes Gaussians do not describe the world very well. Latter parts of the book are dedicated to filters which work even when the noise or system's behavior is very non-Gaussian.

## References¶

[4] Huber, Peter J. Robust Statistical Procedures, Second Edition. Society for Industrial and Applied Mathematics, 1996.

[5] Downey, Alan. Think Stats, Second Edition. O'Reilly Media.

https://github.com/AllenDowney/ThinkStats2

http://greenteapress.com/thinkstats/