**Abstract**: In this session we review the *probabilistic* approach to machine
learning. We start with a review of probability, and introduce the
concepts of probabilistic modelling. We then apply the approach in
practice to Naive Bayesian classification. In this session we review the
probabilistic formulation of a classification model, reviewing initially
maximum likelihood and the naive Bayes model.

$$ \newcommand{\Amatrix}{\mathbf{A}} \newcommand{\KL}[2]{\text{KL}\left( #1\,\|\,#2 \right)} \newcommand{\Kaast}{\kernelMatrix_{\mathbf{ \ast}\mathbf{ \ast}}} \newcommand{\Kastu}{\kernelMatrix_{\mathbf{ \ast} \inducingVector}} \newcommand{\Kff}{\kernelMatrix_{\mappingFunctionVector \mappingFunctionVector}} \newcommand{\Kfu}{\kernelMatrix_{\mappingFunctionVector \inducingVector}} \newcommand{\Kuast}{\kernelMatrix_{\inducingVector \bf\ast}} \newcommand{\Kuf}{\kernelMatrix_{\inducingVector \mappingFunctionVector}} \newcommand{\Kuu}{\kernelMatrix_{\inducingVector \inducingVector}} \newcommand{\Kuui}{\Kuu^{-1}} \newcommand{\Qaast}{\mathbf{Q}_{\bf \ast \ast}} \newcommand{\Qastf}{\mathbf{Q}_{\ast \mappingFunction}} \newcommand{\Qfast}{\mathbf{Q}_{\mappingFunctionVector \bf \ast}} \newcommand{\Qff}{\mathbf{Q}_{\mappingFunctionVector \mappingFunctionVector}} \newcommand{\aMatrix}{\mathbf{A}} \newcommand{\aScalar}{a} \newcommand{\aVector}{\mathbf{a}} \newcommand{\acceleration}{a} 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\newcommand{\weightedAdjacencyVector}{\mathbf{ \weightedAdjacencyScalar}} \newcommand{\onesVector}{\mathbf{1}} \newcommand{\zerosVector}{\mathbf{0}} $$

What is machine learning? At its most basic level machine learning is a combination of

$$\text{data} + \text{model} \xrightarrow{\text{compute}} \text{prediction}$$

where *data* is our observations. They can be actively or passively
acquired (meta-data). The *model* contains our assumptions, based on
previous experience. That experience can be other data, it can come from
transfer learning, or it can merely be our beliefs about the
regularities of the universe. In humans our models include our inductive
biases. The *prediction* is an action to be taken or a categorization or
a quality score. The reason that machine learning has become a mainstay
of artificial intelligence is the importance of predictions in
artificial intelligence. The data and the model are combined through
computation.

In practice we normally perform machine learning using two functions. To combine data with a model we typically make use of:

**a prediction function** a function which is used to make the
predictions. It includes our beliefs about the regularities of the
universe, our assumptions about how the world works, e.g. smoothness,
spatial similarities, temporal similarities.

**an objective function** a function which defines the cost of
misprediction. Typically it includes knowledge about the world's
generating processes (probabilistic objectives) or the costs we pay for
mispredictions (empiricial risk minimization).

The combination of data and model through the prediction function and
the objectie function leads to a *learning algorithm*. The class of
prediction functions and objective functions we can make use of is
restricted by the algorithms they lead to. If the prediction function or
the objective function are too complex, then it can be difficult to find
an appropriate learning algorithm. Much of the acdemic field of machine
learning is the quest for new learning algorithms that allow us to bring
different types of models and data together.

A useful reference for state of the art in machine learning is the UK Royal Society Report, Machine Learning: Power and Promise of Computers that Learn by Example.

You can also check my blog post on "What is Machine Learning?"

We are now going to do some simple review of probabilities and use this review to explore some aspects of our data.

A probability distribution expresses uncertainty about the outcome of an event. We often encode this uncertainty in a variable. So if we are considering the outcome of an event, $Y$, to be a coin toss, then we might consider $Y=1$ to be heads and $Y=0$ to be tails. We represent the probability of a given outcome with the notation: $$ P(Y=1) = 0.5 $$ The first rule of probability is that the probability must normalize. The sum of the probability of all events must equal 1. So if the probability of heads ($Y=1$) is 0.5, then the probability of tails (the only other possible outcome) is given by $$ P(Y=0) = 1-P(Y=1) = 0.5 $$

Probabilities are often defined as the limit of the ratio between the
number of positive outcomes (e.g. *heads*) given the number of trials.
If the number of positive outcomes for event $y$ is denoted by $n$ and
the number of trials is denoted by $N$ then this gives the ratio $$
P(Y=y) = \lim_{N\rightarrow
\infty}\frac{n_y}{N}.
$$ In practice we never get to observe an event infinite times, so
rather than considering this we often use the following estimate $$
P(Y=y) \approx \frac{n_y}{N}.
$$

As an example data set we will use Nigerian NMIS Health Facility data from openAFRICA. It can be found here https://africaopendata.org/dataset/nigeria-nmis-health-facility-data-2014

Taking from the information on the site,

The Nigeria MDG (Millennium Development Goals) Information System – NMIS health facility data is collected by the Office of the Senior Special Assistant to the President on the Millennium Development Goals (OSSAP-MDGs) in partner with the Sustainable Engineering Lab at Columbia University. A rigorous, geo-referenced baseline facility inventory across Nigeria is created spanning from 2009 to 2011 with an additional survey effort to increase coverage in 2014, to build Nigeria’s first nation-wide inventory of health facility. The database includes 34,139 health facilities info in Nigeria.

The goal of this database is to make the data collected available to planners, government officials, and the public, to be used to make strategic decisions for planning relevant interventions.

For data inquiry, please contact Ms. Funlola Osinupebi, Performance Monitoring & Communications, Advisory Power Team, Office of the Vice President at [email protected]

To learn more, please visit http://csd.columbia.edu/2014/03/10/the-nigeria-mdg-information-system-nmis-takes-open-data-further/

Suggested citation: Nigeria NMIS facility database (2014), the Office of the Senior Special Assistant to the President on the Millennium Development Goals (OSSAP-MDGs) & Columbia University

In [243]:

```
import urllib.request
```

In [244]:

```
urllib.request.urlretrieve('https://energydata.info/dataset/f85d1796-e7f2-4630-be84-79420174e3bd/resource/6e640a13-cab4-457b-b9e6-0336051bac27/download/healthmopupandbaselinenmisfacility.csv', 'healthmopupandbaselinenmisfacility.csv')
```

Out[244]:

In [245]:

```
import pandas as pd
```

In [246]:

```
data = pd.read_csv('healthmopupandbaselinenmisfacility.csv')
```

Once it is loaded in the data can be summarized using the `describe`

method in pandas.

In [247]:

```
data.describe()
```

Out[247]:

In python and jupyter notebook it is possible to see a list of all
possible functions and attributes by typing the name of the object
followed by .`num_nurses_fulltime`

, and also functions, such as
`.describe()`

.

For functions we can also see the documentation about the function by following the name with a question mark. This will open a box with documentation at the bottom which can be closed with the x button.

In [248]:

```
data.describe?
```

The NMIS facility data is stored in an object known as a 'data frame'.
Data frames come from the statistical family of programming languages
based on `S`

, the most widely used of which is
`R`

). The data
frame gives us a convenient object for manipulating data. The describe
method summarizes which columns there are in the data frame and gives us
counts, means, standard deviations and percentiles for the values in
those columns. To access a column directly we can write

In [249]:

```
print(data['num_doctors_fulltime'])
#print(data['num_nurses_fulltime'])
```

This shows the number of doctors per facility. We can plot the data against the number of nurses as follows.

In [250]:

```
# this ensures the plot appears in the web browser
%matplotlib inline
import matplotlib.pyplot as plt # this imports the plotting library in python
```

In [251]:

```
_ = plt.plot(data['num_doctors_fulltime'], data['num_nurses_fulltime'], 'rx')
```

You may be curious what the arguments we give to `plt.plot`

are for, now
is the perfect time to look at the documentation

In [252]:

```
plt.plot?
```

We immediately note that some facilities have a lot of nurses, which prevent's us seeing the detail of the main number of facilities. First lets identify the facilities with the most nurses.

In [253]:

```
data[data['num_nurses_fulltime']>100]
```

Out[253]:

Here we are using the command `data['num_nurses_fulltime']>100`

to index
the facilities in the pandas data frame which have over 100 nurses. To
sort them in order we can also use the `sort`

command. The result of
this command on its own is a data `Series`

of `True`

and `False`

values.
However, when it is passed to the `data`

data frame it returns a new
data frame which contains only those values for which the data series is
`True`

. We can also sort the result. To sort the result by the values in
the `num_nurses_fulltime`

column in *descending* order we use the
following command.

In [254]:

```
data[data['num_nurses_fulltime']>100].sort_values(by='num_nurses_fulltime', ascending=False)
```

Out[254]:

We now see that the 'University of Calabar Teaching Hospital' is a large outlier with 513 nurses. We can try and determine how much of an outlier by histograming the data.

In [255]:

```
data['num_nurses_fulltime'].hist(bins=20) # histogram the data with 20 bins.
plt.title('Histogram of Number of Nurses')
```

Out[255]:

We can't see very much here. Two things are happening. There are so many
facilities with zero or one nurse that we don't see the histogram for
hospitals with many nurses. We can try more bins and using a *log* scale
on the $y$-axis.

In [256]:

```
data['num_nurses_fulltime'].hist(bins=100) # histogram the data with 20 bins.
plt.title('Histogram of Number of Nurses')
ax = plt.gca()
ax.set_yscale('log')
```

In [257]:

```
fig, ax = plt.subplots(figsize=(10, 7))
ax.plot(data['num_doctors_fulltime'], data['num_nurses_fulltime'], 'rx')
ax.set_xscale('log') # use a logarithmic x scale
ax.set_yscale('log') # use a logarithmic Y scale
# give the plot some titles and labels
plt.title('Number of Nurses against Number of Doctors')
plt.ylabel('number of nurses')
plt.xlabel('number of doctors')
```

Out[257]:

Note a few things. We are interacting with our data. In particular, we
are replotting the data according to what we have learned so far. We are
using the progamming language as a *scripting* language to give the
computer one command or another, and then the next command we enter is
dependent on the result of the previous. This is a very different
paradigm to classical software engineering. In classical software
engineering we normally write many lines of code (entire object classes
or functions) before compiling the code and running it. Our approach is
more similar to the approach we take whilst debugging. Historically,
researchers interacted with data using a *console*. A command line
window which allowed command entry. The notebook format we are using is
slightly different. Each of the code entry boxes acts like a separate
console window. We can move up and down the notebook and run each part
in a different order. The *state* of the program is always as we left it
after running the previous part.

Let's use the sum rule to compute the estimate the probability that a facility has more than two nurses.

In [258]:

```
large = (data.num_nurses_fulltime>2).sum() # number of positive outcomes (in sum True counts as 1, False counts as 0)
total_facilities = data.num_nurses_fulltime.count()
prob_large = float(large)/float(total_facilities)
print("Probability of number of nurses being greather than 2 is:", prob_large)
```

When predicting whether a coin turns up head or tails, we might think
that this event is *independent* of the year or time of day. If we
include an observation such as time, then in a probability this is known
as *condtioning*. We use this notation, $P(Y=y|X=x)$, to condition the
outcome on a second variable (in this case the number of doctors). Or,
often, for a shorthand we use $P(y|x)$ to represent this distribution
(the $Y=$ and $X=$ being implicit). If two variables are independent
then we find that $$
P(y|x) = p(y).
$$ However, we might believe that the number of nurses is dependent on
the number of doctors. For this we can try estimating $P(Y>2 | X>1)$ and
compare the result, for example to $P(Y>2|X\leq 1)$ using our empirical
estimate of the probability.

In [259]:

```
large = ((data.num_nurses_fulltime>2) & (data.num_doctors_fulltime>1)).sum()
total_large_doctors = (data.num_doctors_fulltime>1).sum()
prob_both_large = large/total_large_doctors
print("Probability of number of nurses being greater than 2 given number of doctors is greater than 1 is:", prob_both_large)
```

Write code that prints out the probability of nurses being greater than 2 for different numbers of doctors.

*15 marks*

In [260]:

```
# Write your answer to Question 2 here
```

Make sure the plot is included in *this* notebook file (the `IPython`

magic command `%matplotlib inline`

we ran above will do that for you, it
only needs to be run once per file).

Terminology Mathematical notation Description

joint $P(X=x, Y=y)$ prob. that X=x *and* Y=y
marginal $P(X=x)$ prob. that X=x *regardless of* Y
conditional $P(X=x\vert Y=y)$ prob. that X=x *given that* Y=y

[Inspired by lectures from Christopher Bishop]{style="text-align:right"}

Terminology Definition Probability Notation

Joint Probability $\lim_{N\rightarrow\infty}\frac{n_{X=3,Y=4}}{N}$ $P\left(X=3,Y=4\right)$ Marginal Probability $\lim_{N\rightarrow\infty}\frac{n_{X=5}}{N}$ $P\left(X=5\right)$ Conditional Probability $\lim_{N\rightarrow\infty}\frac{n_{X=3,Y=4}}{n_{Y=4}}$ $P\left(X=3\vert Y=4\right)$

Typically we should write out $P\left(X=x,Y=y\right)$, but in practice
we often shorten this to $P\left(x,y\right)$. This looks very much like
we might write a multivariate function, *e.g.* $$
f\left(x,y\right)=\frac{x}{y},
$$ but for a multivariate function $$
f\left(x,y\right)\neq f\left(y,x\right).
$$ However, $$
P\left(x,y\right)=P\left(y,x\right)
$$ because $$
P\left(X=x,Y=y\right)=P\left(Y=y,X=x\right).
$$ Sometimes I think of this as akin to the way in Python we can write
'keyword arguments' in functions. If we use keyword arguments, the
ordering of arguments doesn't matter.

We've now introduced conditioning and independence to the notion of
probability and computed some conditional probabilities on a practical
example The scatter plot of deaths vs year that we created above can be
seen as a *joint* probability distribution. We represent a joint
probability using the notation $P(Y=y, X=x)$ or $P(y, x)$ for short.
Computing a joint probability is equivalent to answering the
simultaneous questions, what's the probability that the number of nurses
was over 2 and the number of doctors was 1? Or any other question that
may occur to us. Again we can easily use pandas to ask such questions.

In [261]:

```
num_doctors = 1
large = (data.num_nurses_fulltime[data.num_doctors_fulltime==num_doctors]>2).sum()
total_facilities = data.num_nurses_fulltime.count() # this is total number of films
prob_large = float(large)/float(total_facilities)
print("Probability of nurses being greater than 2 and number of doctors being", num_doctors, "is:", prob_large)
```

This number is the joint probability, $P(Y, X)$ which is much *smaller*
than the conditional probability. The number can never be bigger than
the conditional probabililty because it is computed using the *product
rule*. $$
p(Y=y, X=x) = p(Y=y|X=x)p(X=x)
$$ and $$p(X=x)$$ is a probability distribution, which is equal or less
than 1, ensuring the joint distribution is typically smaller than the
conditional distribution.

The product rule is a *fundamental* rule of probability, and you must
remember it! It gives the relationship between the two questions: 1)
What's the probability that a facility has over two nurses *and* one
doctor? and 2) What's the probability that a facility has over two
nurses *given that* it has one doctor?

In our shorter notation we can write the product rule as $$ p(y, x) = p(y|x)p(x) $$ We can see the relation working in practice for our data above by computing the different values for $x=1$.

In [262]:

```
num_doctors=1
num_nurses=2
p_x = float((data.num_doctors_fulltime==num_doctors).sum())/float(data.num_nurses_fulltime.count())
p_y_given_x = float((data.num_nurses_fulltime[data.num_doctors_fulltime==num_doctors]>num_nurses).sum())/float((data.num_doctors_fulltime==num_doctors).sum())
p_y_and_x = float((data.num_nurses_fulltime[data.num_doctors_fulltime==num_doctors]>num_nurses).sum())/float(data.num_nurses_fulltime.count())
print("P(x) is", p_x)
print("P(y|x) is", p_y_given_x)
print("P(y,x) is", p_y_and_x)
```

The other *fundamental rule* of probability is the *sum rule* this tells
us how to get a *marginal* distribution from the joint distribution.
Simply put it says that we need to sum across the value we'd like to
remove. $$
P(Y=y) = \sum_{x} P(Y=y, X=x)
$$ Or in our shortened notation $$
P(y) = \sum_{x} P(y, x)
$$

Write code that computes $P(y)$ by adding $P(y, x)$ for all values of $x$.

*10 marks*

In [263]:

```
# Write your answer to Question 3 here
```

Bayes rule is a very simple rule, it's hardly worth the name of a rule
at all. It follows directly from the product rule of probability.
Because $P(y, x) = P(y|x)P(x)$ and by symmetry
$P(y,x)=P(x,y)=P(x|y)P(y)$ then by equating these two equations and
dividing through by $P(y)$ we have $$
P(x|y) =
\frac{P(y|x)P(x)}{P(y)}
$$ which is known as Bayes' rule (or Bayes's rule, it depends how you
choose to pronounce it). It's not difficult to derive, and its
importance is more to do with the semantic operation that it enables.
Each of these probability distributions represents the answer to a
question we have about the world. Bayes rule (via the product rule)
tells us how to *invert* the probability.

What use is all this probability in data science? Let's think about how we might use the probabilities to do some decision making. Let's look at the information data.

In [264]:

```
data.columns
```

Out[264]:

Now we see we have several additional features. Let's assume we want to
predict `maternal_health_delivery_services`

. How would we go about doing
it?

Using what you've learnt about joint, conditional and marginal probabilities, as well as the sum and product rule, how would you formulate the question you want to answer in terms of probabilities? Should you be using a joint or a conditional distribution? If it's conditional, what should the distribution be over, and what should it be conditioned on?

*20 marks*

This Bayesian approach is designed to deal with uncertainty arising from fitting our prediction function to the data we have, a reduced data set.

The Bayesian approach can be derived from a broader understanding of what our objective is. If we accept that we can jointly represent all things that happen in the world with a probability distribution, then we can interogate that probability to make predictions. So, if we are interested in predictions, $\dataScalar_*$ at future points input locations of interest, $\inputVector_*$ given previously training data, $\dataVector$ and corresponding inputs, $\inputMatrix$, then we are really interogating the following probability density, $$ p(\dataScalar_*|\dataVector, \inputMatrix, \inputVector_*), $$ there is nothing controversial here, as long as you accept that you have a good joint model of the world around you that relates test data to training data, $p(\dataScalar_*, \dataVector, \inputMatrix, \inputVector_*)$ then this conditional distribution can be recovered through standard rules of probability ($\text{data} + \text{model} \rightarrow \text{prediction}$).

We can construct this joint density through the use of the following decomposition: $$ p(\dataScalar_*|\dataVector, \inputMatrix, \inputVector_*) = \int p(\dataScalar_*|\inputVector_*, \mappingMatrix) p(\mappingMatrix | \dataVector, \inputMatrix) \text{d} \mappingMatrix $$

where, for convenience, we are assuming *all* the parameters of the
model are now represented by $\parameterVector$ (which contains
$\mappingMatrix$ and $\mappingMatrixTwo$) and
$p(\parameterVector | \dataVector, \inputMatrix)$ is recognised as the
posterior density of the parameters given data and
$p(\dataScalar_*|\inputVector_*, \parameterVector)$ is the *likelihood*
of an individual test data point given the parameters.

The likelihood of the data is normally assumed to be independent across the parameters, $$ p(\dataVector|\inputMatrix, \mappingMatrix) \prod_{i=1}^\numData p(\dataScalar_i|\inputVector_i, \mappingMatrix),$$

and if that is so, it is easy to extend our predictions across all future, potential, locations, $$ p(\dataVector_*|\dataVector, \inputMatrix, \inputMatrix_*) = \int p(\dataVector_*|\inputMatrix_*, \parameterVector) p(\parameterVector | \dataVector, \inputMatrix) \text{d} \parameterVector. $$

The likelihood is also where the *prediction function* is incorporated.
For example in the regression case, we consider an objective based
around the Gaussian density, $$
p(\dataScalar_i | \mappingFunction(\inputVector_i)) = \frac{1}{\sqrt{2\pi \dataStd^2}} \exp\left(-\frac{\left(\dataScalar_i - \mappingFunction(\inputVector_i)\right)^2}{2\dataStd^2}\right)
$$

In short, that is the classical approach to probabilistic inference, and
all approaches to Bayesian neural networks fall within this path. For a
deep probabilistic model, we can simply take this one stage further and
place a probability distribution over the input locations, $$
p(\dataVector_*|\dataVector) = \int p(\dataVector_*|\inputMatrix_*, \parameterVector) p(\parameterVector | \dataVector, \inputMatrix) p(\inputMatrix) p(\inputMatrix_*) \text{d} \parameterVector \text{d} \inputMatrix \text{d}\inputMatrix_*
$$ and we have *unsupervised learning* (from where we can get deep
generative models).

One way of representing a joint distribution is to consider conditional dependencies between data. Conditional dependencies allow us to factorize the distribution. For example, a Markov chain is a factorization of a distribution into components that represent the conditional relationships between points that are neighboring, often in time or space. It can be decomposed in the following form. $$p(\dataVector) = p(\dataScalar_\numData | \dataScalar_{\numData-1}) p(\dataScalar_{\numData-1}|\dataScalar_{\numData-2}) \dots p(\dataScalar_{2} | \dataScalar_{1})$$

By specifying conditional independencies we can reduce the parameterization required for our data, instead of directly specifying the parameters of the joint distribution, we can specify each set of parameters of the conditonal independently. This can also give an advantage in terms of interpretability. Understanding a conditional independence structure gives a structured understanding of data. If developed correctly, according to causal methodology, it can even inform how we should intervene in the system to drive a desired result [@Pearl:causality95].

However, a challenge arise when the data becomes more complex. Consider
the graphical model shown below, used to predict the perioperative risk
of *C Difficile* infection following colon surgery
[@Steele:predictive12].

To capture the complexity in the interelationship between the data the graph becomes more complex, and less interpretable.

Machine learning problems normally involve a prediction function and an
objective function. So far in the course we've mainly focussed on the
case where the prediction function was over the real numbers, so the
codomain of the functions, $\mappingFunction(\inputMatrix)$ was the real
numbers or sometimes real vectors. The classification problem consists
of predicting whether or not a particular example is a member of a
particular class. So we may want to know if a particular image
represents a digit 6 or if a particular user will click on a given
advert. These are classification problems, and they require us to map to
*yes* or *no* answers. That makes them naturally discrete mappings.

In classification we are given an input vector, $\inputVector$, and an
associated label, $\dataScalar$ which either takes the value $-1$ to
represent *no* or $1$ to represent *yes*.

- Classifiying hand written digits from binary images (automatic zip code reading)
- Detecting faces in images (e.g. digital cameras).
- Who a detected face belongs to (e.g. Picasa, Facebook, DeepFace, GaussianFace)
- Classifying type of cancer given gene expression data.
- Categorization of document types (different types of news article on the internet)

Our focus has been on models where the objective function is inspired by a probabilistic analysis of the problem. In particular we've argued that we answer questions about the data set by placing probability distributions over the various quantities of interest. For the case of binary classification this will normally involve introducing probability distributions for discrete variables. Such probability distributions, are in some senses easier than those for continuous variables, in particular we can represent a probability distribution over $\dataScalar$, where $\dataScalar$ is binary, with one value. If we specify the probability that $\dataScalar=1$ with a number that is between 0 and 1, i.e. let's say that $P(\dataScalar=1) = \pi$ (here we don't mean $\pi$ the number, we are setting $\pi$ to be a variable) then we can specify the probability distribution through a table.

Mathematically we can use a trick to implement this same table. We can use the value $\dataScalar$ as a mathematical switch and write that $$ P(\dataScalar) = \pi^\dataScalar (1-\pi)^{(1-\dataScalar)} $$ where our probability distribution is now written as a function of $\dataScalar$. This probability distribution is known as the Bernoulli distribution. The Bernoulli distribution is a clever trick for mathematically switching between two probabilities if we were to write it as code it would be better described as

In [265]:

```
def bernoulli(y_i, pi):
if y_i == 1:
return pi
else:
return 1-pi
```

If we insert $\dataScalar=1$ then the function is equal to $\pi$, and if we insert $\dataScalar=0$ then the function is equal to $1-\pi$. So the function recreates the table for the distribution given above.

The probability distribution is named for Jacob
Bernoulli, the swiss
mathematician. In his book Ars Conjectandi he considered the
distribution and the result of a number of 'trials' under the Bernoulli
distribution to form the *binomial* distribution. Below is the page
where he considers Pascal's triangle in forming combinations of the
Bernoulli distribution to realise the binomial distribution for the
outcome of positive trials.

In [266]:

```
import pods
pods.notebook.display_google_book(id='CF4UAAAAQAAJ', page='PA87')
```

Thomas Bayes also described the Bernoulli distribution, only he didn't
refer to Jacob Bernoulli's work, so he didn't call it by that name. He
described the distribution in terms of a table (think of a *billiard
table*) and two balls. Bayes suggests that each ball can be rolled
across the table such that it comes to rest at a position that is
*uniformly distributed* between the sides of the table.

Let's assume that the first ball is rolled, and that it comes to reset at a position that is $\pi$ times the width of the table from the left hand side.

Now, we roll the second ball. We are interested if the second ball ends up on the left side (+ve result) or the right side (-ve result) of the first ball. We use the Bernoulli distribution to determine this.

For this reason in Bayes's distribution there is considered to be
*aleatoric* uncertainty about the distribution parameter.

In [267]:

```
import pods
from ipywidgets import IntSlider
```

In [268]:

```
pods.notebook.display_plots('bayes-billiard{counter:0>3}.svg',
directory='../slides/diagrams/ml',
counter=IntSlider(0,0,9,1))
```

Maximum likelihood in the Bernoulli distribution is straightforward. Let's assume we have data, $\dataVector$ which consists of a vector of binary values of length $n$. If we assume each value was sampled independently from the Bernoulli distribution, conditioned on the parameter $\pi$ then our joint probability density has the form $$ p(\dataVector|\pi) = \prod_{i=1}^{\numData} \pi^{\dataScalar_i} (1-\pi)^{1-\dataScalar_i}. $$ As normal in maximum likelihood we consider the negative log likelihood as our objective, $$\begin{align*} \errorFunction(\pi)& = -\log p(\dataVector|\pi)\\ & = -\sum_{i=1}^{\numData} \dataScalar_i \log \pi - \sum_{i=1}^{\numData} (1-\dataScalar_i) \log(1-\pi), \end{align*}$$

and we can derive the gradient with respect to the parameter $\pi$. $$\frac{\text{d}\errorFunction(\pi)}{\text{d}\pi} = -\frac{\sum_{i=1}^{\numData} \dataScalar_i}{\pi} + \frac{\sum_{i=1}^{\numData} (1-\dataScalar_i)}{1-\pi},$$

and as normal we look for a stationary point for the log likelihood by setting this derivative to zero, $$0 = -\frac{\sum_{i=1}^{\numData} \dataScalar_i}{\pi} + \frac{\sum_{i=1}^{\numData} (1-\dataScalar_i)}{1-\pi},$$ rearranging we form $$(1-\pi)\sum_{i=1}^{\numData} \dataScalar_i = \pi\sum_{i=1}^{\numData} (1-\dataScalar_i),$$ which implies $$\sum_{i=1}^{\numData} \dataScalar_i = \pi\left(\sum_{i=1}^{\numData} (1-\dataScalar_i) + \sum_{i=1}^{\numData} \dataScalar_i\right),$$

and now we recognise that $\sum_{i=1}^{\numData} (1-\dataScalar_i) + \sum_{i=1}^{\numData} \dataScalar_i = \numData$ so we have $$\pi = \frac{\sum_{i=1}^{\numData} \dataScalar_i}{\numData}$$

so in other words we estimate the probability associated with the Bernoulli by setting it to the number of observed positives, divided by the total length of $\dataScalar$. This makes intiutive sense. If I asked you to estimate the probability of a coin being heads, and you tossed the coin 100 times, and recovered 47 heads, then the estimate of the probability of heads should be $\frac{47}{100}$.

Show that the maximum likelihood solution we have found is a *minimum*
for our objective.

In [269]:

```
# Use this box for any code you need for the exercise
```

$$ \text{posterior} = \frac{\text{likelihood}\times\text{prior}}{\text{marginal likelihood}} $$

Four components:

- Prior distribution
- Likelihood
- Posterior distribution
- Marginal likelihood

*Note*: Everything we do below is possible using standard packages like
`scikit-learn`

, our purpose in this session is to help you understand
how those engines are constructed. In practice for an application you
should use a library like `scikit-learn`

.

In probabilistic machine learning we place probability distributions (or
densities) over all the variables of interest, our first classification
algorithm will do just that. We will consider how to form a
classification by making assumptions about the *joint* density of our
observations. We need to make assumptions to reduce the number of
parameters we need to optimise.

In the ideal world, given label data $\dataVector$ and the inputs $\inputMatrix$ we should be able to specify the joint density of all potential values of $\dataVector$ and $\inputMatrix$, $p(\dataVector, \inputMatrix)$. If $\inputMatrix$ and $\dataVector$ are our training data, and we can somehow extend our density to incorporate future test data (by augmenting $\dataVector$ with a new observation $\dataScalar^*$ and $\inputMatrix$ with the corresponding inputs, $\inputVector^*$), then we can answer any given question about a future test point $\dataScalar^*$ given its covariates $\inputVector^*$ by conditioning on the training variables to recover, $$ p(\dataScalar^*|\inputMatrix, \dataVector, \inputVector^*), $$

We can compute this distribution using the product and sum rules.
However, to specify this density we must give the probability associated
with all possible combinations of $\dataVector$ and $\inputMatrix$.
There are $2^{\numData}$ possible combinations for the vector
$\dataVector$ and the probability for each of these combinations must be
jointly specified along with the joint density of the matrix
$\inputMatrix$, as well as being able to *extend* the density for any
chosen test location $\inputVector^*$.

In naive Bayes we make certain simplifying assumptions that allow us to perform all of the above in practice.

If we are given model parameters $\paramVector$ we assume that
conditioned on all these parameters that all data points in the model
are independent. In other words we have, $$
p(\dataScalar^*, \inputVector^*, \dataVector, \inputMatrix|\paramVector) = p(\dataScalar^*, \inputVector^*|\paramVector)\prod_{i=1}^{\numData} p(\dataScalar_i, \inputVector_i | \paramVector).
$$ This is a conditional independence assumption because we are not
assuming our data are purely independent. If we were to assume that,
then there would be nothing to learn about our test data given our
training data. We are assuming that they are independent *given* our
parameters, $\paramVector$. We made similar assumptions for regression,
where our parameter set included $\mappingVector$ and $\dataStd^2$.
Given those parameters we assumed that the density over
$\dataVector, \dataScalar^*$ was *independent*. Here we are going a
little further with that assumption because we are assuming the *joint*
density of $\dataVector$ and $\inputMatrix$ is independent across the
data given the parameters.

Computing posterior distribution in this case becomes easier, this is known as the 'Bayes classifier'.

$$ p(\inputVector_i | \dataScalar_i, \paramVector) = \prod_{j=1}^{\dataDim} p(\inputScalar_{i,j}|\dataScalar_i, \paramVector) $$ where $\dataDim$ is the dimensionality of our inputs.

The assumption that is particular to naive Bayes is to now consider that
the *features* are also conditionally independent, but not only given
the parameters. We assume that the features are independent given the
parameters *and* the label. So for each data point we have
$$p(\inputVector_i | \dataScalar_i, \paramVector) = \prod_{j=1}^{\dataDim} p(\inputScalar_{i,j}|\dataScalar_i,\paramVector)$$
where $\dataDim$ is the dimensionality of our inputs.

$$ p(\inputScalar_{i,j},\dataScalar_i| \paramVector) = p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)p(\dataScalar_i). $$

We now have nearly all of the components we need to specify the full joint density. However, the feature conditional independence doesn't yet give us the joint density over $p(\dataScalar_i, \inputVector_i)$ which is required to subsitute in to our data conditional independence to give us the full density. To recover the joint density given the conditional distribution of each feature, $p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)$, we need to make use of the product rule and combine it with a marginal density for $\dataScalar_i$,

$$p(\inputScalar_{i,j},\dataScalar_i| \paramVector) = p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)p(\dataScalar_i).$$
Because $\dataScalar_i$ is binary the *Bernoulli* density makes a
suitable choice for our prior over $\dataScalar_i$,
$$p(\dataScalar_i|\pi) = \pi^{\dataScalar_i} (1-\pi)^{1-\dataScalar_i}$$
where $\pi$ now has the interpretation as being the *prior* probability
that the classification should be positive.

This allows us to write down the full joint density of the training data, $$ p(\dataVector, \inputMatrix|\paramVector, \pi) = \prod_{i=1}^{\numData} \prod_{j=1}^{\dataDim} p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)p(\dataScalar_i|\pi) $$

which can now be fit by maximum likelihood. As normal we form our objective as the negative log likelihood,

$$\begin{align*}
\errorFunction(\paramVector, \pi)& = -\log p(\dataVector, \inputMatrix|\paramVector, \pi) \\ &= -\sum_{i=1}^{\numData} \sum_{j=1}^{\dataDim} \log p(\inputScalar_{i, j}|\dataScalar_i, \paramVector) - \sum_{i=1}^{\numData} \log p(\dataScalar_i|\pi),
\end{align*}$$ which we note *decomposes* into two objective functions,
one which is dependent on $\pi$ alone and one which is dependent on
$\paramVector$ alone so we have, $$
\errorFunction(\pi, \paramVector) = \errorFunction(\paramVector) + \errorFunction(\pi).
$$ Since the two objective functions are separately dependent on the
parameters $\pi$ and $\paramVector$ we can minimize them independently.
Firstly, minimizing the Bernoulli likelihood over the labels we have, $$
\errorFunction(\pi) = -\sum_{i=1}^{\numData}\log p(\dataScalar_i|\pi) = -\sum_{i=1}^{\numData} \dataScalar_i \log \pi - \sum_{i=1}^{\numData} (1-\dataScalar_i) \log (1-\pi)
$$ which we already minimized above recovering $$
\pi = \frac{\sum_{i=1}^{\numData} \dataScalar_i}{\numData}.
$$

We now need to minimize the objective associated with the conditional distributions for the features, $$ \errorFunction(\paramVector) = -\sum_{i=1}^{\numData} \sum_{j=1}^{\dataDim} \log p(\inputScalar_{i, j} |\dataScalar_i, \paramVector), $$ which necessarily implies making some assumptions about the form of the conditional distributions. The right assumption will depend on the nature of our input data. For example, if we have an input which is real valued, we could use a Gaussian density and we could allow the mean and variance of the Gaussian to be different according to whether the class was positive or negative and according to which feature we were measuring. That would give us the form, $$ p(\inputScalar_{i, j} | \dataScalar_i,\paramVector) = \frac{1}{\sqrt{2\pi \dataStd_{\dataScalar_i,j}^2}} \exp \left(-\frac{(\inputScalar_{i,j} - \mu_{\dataScalar_i, j})^2}{\dataStd_{\dataScalar_i,j}^2}\right), $$ where $\dataStd_{1, j}^2$ is the variance of the density for the $j$th output and the class $\dataScalar_i=1$ and $\dataStd_{0, j}^2$ is the variance if the class is 0. The means can vary similarly. Our parameters, $\paramVector$ would consist of all the means and all the variances for the different dimensions.

As normal we form our objective as the negative log likelihood, $$
\errorFunction(\paramVector, \pi) = -\log p(\dataVector, \inputMatrix|\paramVector, \pi) = -\sum_{i=1}^{\numData} \sum_{j=1}^{\dataDim} \log p(\inputScalar_{i, j}|\dataScalar_i, \paramVector) - \sum_{i=1}^{\numData} \log p(\dataScalar_i|\pi),
$$ which we note *decomposes* into two objective functions, one which is
dependent on $\pi$ alone and one which is dependent on $\paramVector$
alone so we have, $$
\errorFunction(\pi, \paramVector) = \errorFunction(\paramVector) + \errorFunction(\pi).
$$

First we will load in the Nigerian NMIS health data. Our aim will be to predict whether a center has maternal health delivery services given the attributes in the data. We will predict of the number of nurses, the number of doctors, location etc.

Let's first remind ourselves of the data.

In [270]:

```
data.head()
```

Out[270]:

Now we will convert this data into a form which we can use as inputs
`X`

, and labels `y`

.

In [271]:

```
import pandas as pd
import numpy as np
```

In [272]:

```
data = data[~pd.isnull(data['maternal_health_delivery_services'])]
data = data.dropna() # Remove entries with missing values
X = data[['emergency_transport',
'num_chews_fulltime',
'phcn_electricity',
'child_health_measles_immun_calc',
'num_nurses_fulltime',
'num_doctors_fulltime',
'improved_water_supply',
'improved_sanitation',
'antenatal_care_yn',
'family_planning_yn',
'malaria_treatment_artemisinin',
'latitude',
'longitude']].copy()
y = data['maternal_health_delivery_services']==True # set label to be whether there's a maternal health delivery service
# Create series of health center types with the relevant index
s = data['facility_type_display'].apply(pd.Series, 1).stack()
s.index = s.index.droplevel(-1) # to line up with df's index
# Extract from the series the unique list of types.
types = s.unique()
# For each type extract the indices where it is present and add a column to X
type_names = []
for htype in types:
index = s[s==htype].index.tolist()
type_col=htype.replace(' ', '_').replace('/','-').lower()
type_names.append(type_col)
X.loc[:, type_col] = 0.0
X.loc[index, type_col] = 1.0
```

This has given us a new data frame `X`

which contains the different
facility types in different columns.

In [273]:

```
X.describe()
```

Out[273]:

We can now specify the naive Bayes model. For the genres we want to model the data as Bernoulli distributed, and for the year and body count we want to model the data as Gaussian distributed. We set up two data frames to contain the parameters for the rows and the columns below.

In [274]:

```
# assume data is binary or real.
# this list encodes whether it is binary or real (1 for binary, 0 for real)
binary_columns = ['emergency_transport',
'phcn_electricity',
'child_health_measles_immun_calc',
'improved_water_supply',
'improved_sanitation',
'antenatal_care_yn',
'family_planning_yn',
'malaria_treatment_artemisinin'] + type_names
real_columns = ['num_chews_fulltime',
'num_nurses_fulltime',
'num_doctors_fulltime',
'latitude',
'longitude']
Bernoulli = pd.DataFrame(data=np.zeros((2,len(binary_columns))), columns=binary_columns, index=['theta_0', 'theta_1'])
Gaussian = pd.DataFrame(data=np.zeros((4,len(real_columns))), columns=real_columns, index=['mu_0', 'sigma2_0', 'mu_1', 'sigma2_1'])
```

Now we have the data in a form ready for analysis, let's construct our data matrix.

In [275]:

```
num_train = 20000
indices = np.random.permutation(X.shape[0])
train_indices = indices[:num_train]
test_indices = indices[num_train:]
X_train = X.iloc[train_indices]
y_train = y.iloc[train_indices]==True
X_test = X.iloc[test_indices]
y_test = y.iloc[test_indices]==True
```

And we can now train the model. For each feature we can make the fit independently. The fit is given by either counting the number of positives (for binary data) which gives us the maximum likelihood solution for the Bernoulli. Or by computing the empirical mean and variance of the data for the Gaussian, which also gives us the maximum likelihood solution.

In [276]:

```
for column in X_train:
if column in Gaussian:
Gaussian[column]['mu_0'] = X_train[column][~y_train].mean()
Gaussian[column]['mu_1'] = X_train[column][y_train].mean()
Gaussian[column]['sigma2_0'] = X_train[column][~y_train].var(ddof=0)
Gaussian[column]['sigma2_1'] = X_train[column][y_train].var(ddof=0)
if column in Bernoulli:
Bernoulli[column]['theta_0'] = X_train[column][~y_train].sum()/(~y_train).sum()
Bernoulli[column]['theta_1'] = X_train[column][y_train].sum()/(y_train).sum()
```

We can examine the nature of the distributions we've fitted to the model by looking at the entries in these data frames.

In [277]:

```
Bernoulli
```

Out[277]:

In [278]:

```
Gaussian
```

Out[278]:

The final model parameter is the prior probability of the positive class, $\pi$, which is computed by maximum likelihood.

In [293]:

```
prior = float(y_train.sum())/len(y_train)
print("""The prior probability of finding maternal services
without knowing any other characteristics of the hospital is:""", prior)
```

Naive Bayes has given us the class conditional densities:
$p(\inputVector_i | \dataScalar_i, \paramVector)$. To make predictions
with these densities we need to form the distribution given by $$
P(\dataScalar^*| \dataVector, \inputMatrix, \inputVector^*, \paramVector)
$$ This can be computed by using the product rule. We know that $$
P(\dataScalar^*| \dataVector, \inputMatrix, \inputVector^*, \paramVector)p(\dataVector, \inputMatrix, \inputVector^*|\paramVector) = p(\dataScalar*, \dataVector, \inputMatrix, \inputVector^*| \paramVector)
$$ implying that $$
P(\dataScalar^*| \dataVector, \inputMatrix, \inputVector^*, \paramVector) = \frac{p(\dataScalar*, \dataVector, \inputMatrix, \inputVector^*| \paramVector)}{p(\dataVector, \inputMatrix, \inputVector^*|\paramVector)}
$$ and we've already defined
$p(\dataScalar^*, \dataVector, \inputMatrix, \inputVector^*| \paramVector)$
using our conditional independence assumptions above $$
p(\dataScalar^*, \dataVector, \inputMatrix, \inputVector^*| \paramVector) = \prod_{j=1}^{\dataDim} p(\inputScalar^*_{j}|\dataScalar^*, \paramVector)p(\dataScalar^*|\pi)\prod_{i=1}^{\numData} \prod_{j=1}^{\dataDim} p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)p(\dataScalar_i|\pi)
$$ The other required density is $$
p(\dataVector, \inputMatrix, \inputVector^*|\paramVector)
$$ which can be found from
$$p(\dataScalar^*, \dataVector, \inputMatrix, \inputVector^*| \paramVector)$$
using the *sum rule* of probability, $$
p(\dataVector, \inputMatrix, \inputVector^*|\paramVector) = \sum_{\dataScalar^*=0}^1 p(\dataScalar^*, \dataVector, \inputMatrix, \inputVector^*| \paramVector).
$$ Because of our independence assumptions that is simply equal to $$
p(\dataVector, \inputMatrix, \inputVector^*| \paramVector) = \sum_{\dataScalar^*=0}^1 \prod_{j=1}^{\dataDim} p(\inputScalar^*_{j}|\dataScalar^*_i, \paramVector)p(\dataScalar^*|\pi)\prod_{i=1}^{\numData} \prod_{j=1}^{\dataDim} p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)p(\dataScalar_i|\pi).
$$ Substituting both forms in to recover our distribution over the test
label conditioned on the training data we have, $$
P(\dataScalar^*| \dataVector, \inputMatrix, \inputVector^*, \paramVector) = \frac{\prod_{j=1}^{\dataDim} p(\inputScalar^*_{j}|\dataScalar^*_i, \paramVector)p(\dataScalar^*|\pi)\prod_{i=1}^{\numData} \prod_{j=1}^{\dataDim} p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)p(\dataScalar_i|\pi)}{\sum_{\dataScalar^*=0}^1 \prod_{j=1}^{\dataDim} p(\inputScalar^*_{j}|\dataScalar^*_i, \paramVector)p(\dataScalar^*|\pi)\prod_{i=1}^{\numData} \prod_{j=1}^{\dataDim} p(\inputScalar_{i,j}|\dataScalar_i, \paramVector)p(\dataScalar_i|\pi)}
$$ and we notice that all the terms associated with the training data
actually cancel, the test prediction is *conditionally independent* of
the training data *given* the parameters. This is a result of our
conditional independence assumptions over the data points. $$
p(\dataScalar^*| \inputVector^*, \paramVector) = \frac{\prod_{j=1}^{\dataDim} p(\inputScalar^*_{j}|\dataScalar^*_i,
\paramVector)p(\dataScalar^*|\pi)}{\sum_{\dataScalar^*=0}^1 \prod_{j=1}^{\dataDim} p(\inputScalar^*_{j}|\dataScalar^*_i, \paramVector)p(\dataScalar^*|\pi)}
$$ This formula is also fairly straightforward to implement. First we
implement the log probabilities for the Gaussian density.

In [280]:

```
def log_gaussian(x, mu, sigma2):
return -0.5* np.log(2*np.pi*sigma2)-((x-mu)**2)/(2*sigma2)
```

Now for any test point we compute the joint distribution of the Gaussian
features by *summing* their log probabilities. Working in log space can
be a considerable advantage over computing the probabilities directly:
as the number of features we include goes up, because all the
probabilities are less than 1, the joint probability will become smaller
and smaller, and may be difficult to represent accurately (or even
underflow). Working in log space can ameliorate this problem. We can
also compute the log probability for the Bernoulli distribution.

In [281]:

```
def log_bernoulli(x, theta):
return x*np.log(theta) + (1-x)*np.log(1-theta)
```

Before we proceed, let's just pause and think for a moment what will
happen if `theta`

here is either zero or one. This will result in
$\log 0 = -\infty$ and cause numerical problems. This definitely can
happen in practice. If some of the features are rare or very common
across the data set then the maximum likelihood solution could find
values of zero or one respectively. Such values are problematic because
they cause posterior probabilities of class membership of either one or
zero. In practice we deal with this using *Laplace smoothing* (which
actually has an interpretation as a Bayesian fit of the Bernoulli
distribution. Laplace used an example of the sun rising each day, and a
wish to predict the sun rise the following day to describe his idea of
smoothing, which can be found at the bottom of following page from
Laplace's 'Essai Philosophique ...'

In [282]:

```
import pods
pods.notebook.display_google_book(id='1YQPAAAAQAAJ', page='PA16')
```

Laplace suggests that when computing the probability of an event where a success or failure is rare (he uses an example of the sun rising across the last 5,000 years or 1,826,213 days) that even though only successes have been observed (in the sun rising case) that the odds for tomorrow shouldn't be given as $$ \frac{1,826,213}{1,826,213} = 1 $$ but rather by adding one to the numerator and two to the denominator, $$ \frac{1,826,213 + 1}{1,826,213 + 2} = 0.99999945. $$ This technique is sometimes called a 'pseudocount technique' because it has an intepretation of assuming some observations before you start, it's as if instead of observing $\sum_{i}\dataScalar_i$ successes you have an additional success, $\sum_{i}\dataScalar_i + 1$ and instead of having observed $n$ events you've observed $\numData + 2$. So we can think of Laplace's idea saying (before we start) that we have 'two observations worth of belief, that the odds are 50/50', because before we start (i.e. when $\numData=0$) our estimate is 0.5, yet because the effective $n$ is only 2, this estimate is quickly overwhelmed by data. Laplace used ideas like this a lot, and it is known as his 'principle of insufficient reason'. His idea was that in the absence of knowledge (i.e. before we start) we should assume that all possible outcomes are equally likely. This idea has a modern counterpart, known as the principle of maximum entropy. A lot of the theory of this approach was developed by Ed Jaynes, who according to his erstwhile collaborator and friend, John Skilling, learnt French as an undergraduate by reading the works of Laplace. Although John also related that Jaynes's spoken French was not up to the standard of his scientific French. For me Ed Jaynes's work very much carries on the tradition of Laplace into the modern era, in particular his focus on Bayesian approaches. I'm very proud to have met those that knew and worked with him. It turns out that Laplace's idea also has a Bayesian interpretation (as Laplace understood), it comes from assuming a particular prior density for the parameter $\pi$, but we won't explore that interpretation for the moment, and merely choose to estimate the probability as, $$ \pi = \frac{\sum_{i=1}^{\numData} \dataScalar_i + 1}{\numData + 2} $$ to prevent problems with certainty causing numerical issues and misclassifications. Let's refit the Bernoulli features now.

In [283]:

```
# fit the Bernoulli with Laplace smoothing.
for column in X_train:
if column in Bernoulli:
Bernoulli[column]['theta_0'] = (X_train[column][~y_train].sum() + 1)/((~y_train).sum() + 2)
Bernoulli[column]['theta_1'] = (X_train[column][y_train].sum() + 1)/((y_train).sum() + 2)
```

That places us in a position to write the prediction function.

In [284]:

```
import numpy as np
import pandas as pd
```

In [285]:

```
def predict(X_test, Gaussian, Bernoulli, prior):
log_positive = pd.Series(data = np.zeros(X_test.shape[0]), index=X_test.index)
log_negative = pd.Series(data = np.zeros(X_test.shape[0]), index=X_test.index)
for column in X_test.columns:
if column in Gaussian:
log_positive += log_gaussian(X_test[column], Gaussian[column]['mu_1'], Gaussian[column]['sigma2_1'])
log_negative += log_gaussian(X_test[column], Gaussian[column]['mu_0'], Gaussian[column]['sigma2_0'])
elif column in Bernoulli:
log_positive += log_bernoulli(X_test[column], Bernoulli[column]['theta_1'])
log_negative += log_bernoulli(X_test[column], Bernoulli[column]['theta_0'])
v = np.zeros_like(log_positive.values)
for i in range(X_test.shape[0]):
v[i] = np.exp(log_positive.values[i] + np.log(prior))/(np.exp(log_positive.values[i] + np.log(prior))
+ np.exp(log_negative.values[i] + np.log(1-prior)))
return v
#return np.exp(log_positive + np.log(prior))/(np.exp(log_positive + np.log(prior)) + np.exp(log_negative + np.log(1-prior)))
```

Now we are in a position to make the predictions for the test data.

In [286]:

```
p_y = predict(X_test, Gaussian, Bernoulli, prior)
```

We can test the quality of the predictions in the following way. Firstly, we can threshold our probabilities at 0.5, allocating points with greater than 50% probability of membership of the positive class to the positive class. We can then compare to the true values, and see how many of these values we got correct. This is our total number correct.

In [287]:

```
correct = y_test.eq(p_y>0.5)
total_correct = sum(correct)
print("Total correct", total_correct, " out of ", len(y_test), "which is", float(total_correct)/len(y_test), "%")
```

We can also now plot the confusion
matrix. A confusion
matrix tells us where we are making mistakes. Along the diagonal it
stores the *true positives*, the points that were positive class that we
classified correctly, and the *true negatives*, the points that were
negative class and that we classified correctly. The off diagonal terms
contain the false positives and the false negatives. Along the rows of
the matrix we place the actual class, and along the columns we place our
predicted class.

In [288]:

```
confusion_matrix = pd.DataFrame(data=np.zeros((2,2)),
columns=['predicted no maternity', 'predicted maternity'],
index =['actual no maternity','actual maternity'])
confusion_matrix['predicted maternity']['actual maternity'] = (y_test & (p_y>0.5)).sum()
confusion_matrix['predicted maternity']['actual no maternity'] = (~y_test & (p_y>0.5)).sum()
confusion_matrix['predicted no maternity']['actual maternity'] = (y_test & ~(p_y>0.5)).sum()
confusion_matrix['predicted no maternity']['actual no maternity'] = (~y_test & ~(p_y>0.5)).sum()
confusion_matrix
```

Out[288]:

In [289]:

```
# Use this box for any code you need for the exercise
```

We have decided to classify positive if probability of maternity is greater than 0.5. This has led us to accidentally classify some facilities as havien't facilities for maternity when in fact they don't. Imagine you wish to ensure that a facility handles maternity. With your test set how low do you have to set the threshold to avoid all the false negatives (i.e. facilities where you predicted there was no maternity, but in actuality there were?

In [290]:

```
# Use this box for any code you need for the exercise
```

Naive Bayes has given us the class conditional densities: $p(\inputVector_i | \dataScalar_i, \paramVector)$. To make predictions with these densities we need to form the distribution given by $$ P(\dataScalar^*| \dataVector, \inputMatrix, \inputVector^*, \paramVector) $$

Write down the negative log likelihood of the Gaussian density over a vector of variables $\inputVector$. Assume independence between each variable. Minimize this objective to obtain the maximum likelihood solution of the form. $$ \mu = \frac{\sum_{i=1}^{\numData} \inputScalar_i}{\numData} $$ $$ \dataStd^2 = \frac{\sum_{i=1}^{\numData} (\inputScalar_i - \mu)^2}{\numData} $$

In [291]:

```
# Use this box for any code you need for the exercise
```

If the input data was *binary* then we could also make use of the
Bernoulli distribution for the features. For that case we would have the
form, $$
p(\inputScalar_{i, j} | \dataScalar_i,\paramVector) = \theta_{\dataScalar_i, j}^{\inputScalar_{i, j}}(1-\theta_{\dataScalar_i, j})^{(1-\inputScalar_{i,j})},
$$ where $\theta_{1, j}$ is the probability that the $j$th feature is on
if $\dataScalar_i$ is 1.

In either case, maximum likelihood fitting would proceed in the same way. The objective has the form, $$ \errorFunction(\paramVector) = -\sum_{j=1}^{\dataDim} \sum_{i=1}^{\numData} \log p(\inputScalar_{i,j} |\dataScalar_i, \paramVector), $$ and if, as above, the parameters of the distributions are specific to each feature vector (we had means and variances for each continuous feature, and a probability for each binary feature) then we can use the fact that these parameters separate into disjoint subsets across the features to write, $$ \begin{align*} \errorFunction(\paramVector) &= -\sum_{j=1}^{\dataDim} \sum_{i=1}^{\numData} \log p(\inputScalar_{i,j} |\dataScalar_i, \paramVector_j)\\ & \sum_{j=1}^{\dataDim} \errorFunction(\paramVector_j), \end{align*} $$ which means we can minimize our objective on each feature independently.

These characteristics mean that naive Bayes scales very well with big data. To fit the model we consider each feature in turn, we select the positive class and fit parameters for that class, then we select each negative class and fit features for that class. We have code below.

Naive Bayes is making very simple assumptions about the data, in
particular it is modeling the full *joint* probability of the data set,
$p(\dataVector, \inputMatrix | \paramVector, \pi)$ by very strong
assumptions about factorizations that are unlikely to be true in
practice. The data conditional independence assumption is common, and
relies on a rich parameter vector to absorb all the information in the
training data. The additional assumption of naive Bayes is that features
are conditional independent given the class label $\dataScalar_i$ (and
the parameter vector, $\paramVector$. This is quite a strong assumption.
However, it causes the objective function to decompose into parts which
can be independently fitted to the different feature vectors, meaning it
is very easy to fit the model to large data. It is also clear how we
should handle *streaming* data and *missing* data. This means that the
model can be run 'live', adapting parameters and information as it
arrives. Indeed, the model is even capable of dealing with new
*features* that might arrive at run time. Such is the strength of the
modeling the joint probability density. However, the factorization
assumption that allows us to do this efficiently is very strong and may
lead to poor decision boundaries in practice.

- Chapter 5 of @Rogers:book11 up to pg 179 (Section 5.1, and 5.2 up to 5.2.2).