# Estimation of rates of random events¶

Antonino Ingargiola -- tritemio AT gmail.com
ORCID 0000-0002-9348-1397

Date: April 2016

## Abstract¶

In this notebook I explore properties of different estimators of "rate" for random events generated by a Poisson process. The initial reason was computing the rate of photon detection events (for example in single-molecule spectroscopy experiments), but the treatment is general and valid for any Poisson process.

# The Model¶

Let's assume that $\{d_i\}$ is a set of $n$ samples distributed as $D \sim\operatorname{Exp}(\lambda)$.

Then, the can construct $m = n + 1$ times of random events $\{t_i\}$ as:

$$t_0 = 0, \quad t_i = \sum_{j=0}^{i-1} d_j \quad\textrm{for}\quad i > 1$$

These events are stochastic, statistically independent and have a fixed rate $\lambda$.

# Problem: how to estimate $\lambda$ from $\{t_i\}$?¶

Given $\{t_i\}$ we can trivially compute $\{d_i\}$ as:

$$d_i = t_{i+1} - t_i$$

Instead of $\lambda$, we can directly estimate $\tau = 1/\lambda$:

$$\hat{\tau} = \frac{1}{n}\sum_{i=0}^{n-1} d_i = \frac{T_n}{n}$$

The estimator $\hat{\tau} \sim \frac{1}{n}\operatorname{Erlang}(n, \lambda)$. The Erlang distribution has mean $n/\lambda$ and variance $n/\lambda^2$.

Note that:

$$\hat{\tau} \sim \frac{1}{n}\operatorname{Erlang}(n, \lambda) = \operatorname{Erlang}(n, n\,\lambda)$$

thus, $\hat{\tau}$ has mean $1/\lambda$ and variance $1/(n \lambda)^2$, i.e. it is an unbiased estimator of $1/\lambda$.

However, the fact the $\hat{\tau}$ is an unbiased estimator of $1/\lambda$ does not mean that $1/\hat{\tau}$ is an unbiased estimator of $\lambda$. Indeed, as shown in the next section, the unbiased estimator of the rate is:

$$\hat\lambda_u = \frac{n-1}{T_n}$$

It can be shown that $\hat{\lambda}_u$ has also the minimum variance among the unbiased estimators (i.e. it is a minimum variance unbiased estimator, MVUE) of the rate (see Lehmann-Scheffé theorem and Rao–Blackwell theorem).

## MLE estimator of the rate¶

For our sample of $n$ observations from an Exponential distribution, the likelihood function is

$$L(\lambda) = \prod_{i=1}^n f(d_i;\lambda) = \prod_{i=1}^n \lambda\, e^{-\lambda d_i}$$

$$\log L(\lambda) = \sum_{i=1}^n \log (\lambda\, e^{-\lambda d_i} ) = n\log(\lambda) - \lambda \sum_{i=1}^n d_i = n\log(\lambda) - \lambda T_n$$

Taking the first derivative and setting it to 0 we obtain:

$$\frac{\rm d}{{\rm d}\lambda} \log L(\lambda) = \frac{n}{\lambda} - T_n = 0$$

$$\Rightarrow\quad \hat{\lambda}_{MLE} = \frac{n}{T_n} = \frac{n}{\sum_{i=1}^n d_i}$$

## Minimum MSE estimator¶

Given an estimator $\Lambda_n$ (a R.V.), we define the mean square error (MSE) as:

$$\epsilon(\Lambda_n) = \mathrm{E}\{ (\Lambda_n - \lambda)^2 \} = \mathrm{Var}\{ \Lambda_n\} + (\mathrm{E}\{\Lambda_n\} - \lambda)^2$$

In general the minimum MSE (MMSE) estimator is defined as:

$$\hat{\lambda}_{MSE} = \operatorname*{arg\,min}_{\Lambda_n} \epsilon(\Lambda_n)$$

If $\Lambda_n$ has the form:

$$\Lambda_{n,c} = \frac{n - c}{T_n}$$

we need to find $c$ that minimizes $\epsilon(\Lambda_{n,c})$. It can be shown that the minimum is achieved for $c=2$

To derive this result use the expression for $\mathrm{Var}\{ \Lambda_{n,c}\}$ derived in Appendix 2.

## Family of estimators¶

We consider the family of estimators

$$\Lambda_{n,c} = \frac{n - c}{T_n}$$

where $T_n = \sum d_i \sim \textrm{Erlang}(\lambda, n)$ and $c < n$. We can write:

$$\textrm{Erlang PDF:}\quad f(x \:|\: n, \lambda) = \frac{\lambda^n x^{n-1} e^{-\lambda x}}{(n-1)!\,}$$

$$\mathrm{E}\{\Lambda_{n,c}\} = \int \frac{n - c}{x} \, f(x \:|\: n, \lambda)\, dx = (n - c) \int_0^\infty \frac{1}{x}\,f(x \:|\: n, \lambda)\, dx = (n - c) \int_0^\infty \frac{\lambda^n x^{n-2} e^{-\lambda x}}{(n-1)!\,}\, dx = \frac{(n - c)\lambda^n}{(n-1)!} \int_0^\infty x^{n-2} e^{-\lambda x}\, dx$$

Knowing that (see Appendix 1):

$$\int_0^{+\infty} x^{n-2} e^{-\lambda x}\, dx = \frac{(n-2)!}{\lambda^{n-1}}\qquad\qquad \textrm{(A)}$$

we find that:

$$\mathrm{E}\{\Lambda_{n,c}\} = (n - c)\frac{\lambda}{n-1} = \overline{\Lambda}_{n,c}$$

Note that

• for $c=1$ we obtain the unbiased estimator ($\overline{\Lambda}_{n,c} = \lambda$).
• for $c=0$ we obtain the MLE estimator derived in a previous section.
• for $c=2$ we obtain the minimum MSE (MMSE) estimator introduced in a previous section.

Furthermore, from simulations (see below), I found that for $c=1/3$ the estimator has median equal to the rate $\lambda$. This result should not be difficult to prove analytically.

# Simulation of rate estimators¶

In [1]:
%matplotlib inline
import matplotlib.pylab as plt
import numpy as np

In [2]:
n = 5                  # number of delays
num_times = n + 1      # number of timestamps
λ = 1000               # simulated rate
num_iter = 100 * 1000

In [3]:
np.random.seed(5)
d1 = np.random.exponential(scale=1/λ, size=(num_iter * n)).reshape(num_iter, n)
t = np.cumsum(d1, axis=1)

assert np.allclose(np.cumsum(d1, axis=1), t)


We $n$ delays, exponentially distributed with a fixed rate. Their cumulative sum (the times) will not have a sharp cut-off:

In [4]:
plt.hist(t.ravel(), bins=50);


The delays are exponential by construction:

In [5]:
plt.hist(d1.ravel(), bins=40, histtype='step');
plt.yscale('log')
plt.title('$\{d_i\}\quad$ Mean = %.3e  Std.Dev. = %.3e ' % (d1.mean(), d1.std()));


Here we compare a set of estimators:

$$\hat{\Lambda}_{n,c} = \frac{n - c}{T_n}$$

In [6]:
kws = dict(bins=np.arange(0, 4, 0.05), histtype='step', lw=1.5, normed=True)
print('{:>8} {:>6} {:>8} {:>8}  {:>8}  {:>8} {:>8}'
.format('c', 'n', 'Mean', 'Std', 'MSE', 'Median', '% > λ'))
for c in (-1, 0, 1/3, 1, 2, 3):
r_hc = (n - c)  / d1.sum(axis=1) / λ
r_hc_err_rms = np.sqrt(np.mean(((r_hc - 1)**2)))
plt.hist(r_hc, **kws, label = 'c = %.1f' % c);
print('%8.2f %6d %8.5f %8.2f%% %8.3f%% %8.3f %8.2f%%' %
(c, n, r_hc.mean(), r_hc.std()*100, r_hc_err_rms*100, np.median(r_hc),
(r_hc > 1).sum() * 100 / num_iter))

r_hm = 1/np.median(d1, axis=1) / λ
r_hm_err_rms = np.sqrt(np.mean(((r_hm - 1)**2)))
plt.hist(r_hm, **kws, label = 'median');
print('%8s %6d %8.5f %8.2f%% %8.3f%% %8.3f %8.2f%%' %
('median', n, r_hm.mean(), r_hm.std()*100, r_hm_err_rms*100, np.median(r_hm),
(r_hm > 1).sum() * 100 / num_iter))
plt.xlabel('Normalized Rate')
plt.axvline(1, color='k');
plt.legend()
plt.text(0.35, 0.75, r'$\Lambda_{n,c} = \frac{n - c}{T_n}$',
fontsize=24, transform=plt.gcf().transFigure);

       c      n     Mean      Std       MSE    Median    % > λ
-1.00      5  1.49712    85.90%   99.247%    1.282    71.34%
0.00      5  1.24760    71.58%   75.744%    1.068    55.67%
0.33      5  1.16442    66.81%   68.804%    0.997    49.76%
1.00      5  0.99808    57.27%   57.267%    0.854    36.93%
2.00      5  0.74856    42.95%   49.769%    0.641    18.41%
3.00      5  0.49904    28.63%   57.702%    0.427     5.27%
median      5  1.92114   181.83%  203.832%    1.440    73.49%


We can observe that:

• with $c = 2$ we obtain the mean square error (MSE) in estimating the rate,
• with $c = 1$ we obtained the unbiased estimator of the rate (but higher MSE error),
• with $c = 0$, we obtain the MLE estimator,
• with $c = 1/3$, the estimator has median equal to the true rate ($\lambda$).

By contrast, when estimating $\tau = \lambda^{-1}$, the MVUE is the empirical mean of the delays ($\hat{\tau}$). The distribution of $\hat{\tau}$ is shown below:

In [7]:
kws = dict(bins=np.arange(0, 4, 0.05), histtype='step', lw=1.5, normed=True)
print('{:>6} {:>8} {:>8}  {:>8}  {:>8} {:>8}'
.format('n', 'Mean', 'Std', 'Err.RMS', 'Median', '% > 1/λ'))
d_h = np.mean(d1, axis=1) * λ
d_h_err_rms = np.sqrt(np.mean(((d_h - 1)**2)))
plt.hist(d_h, **kws, label = r'$\hat\tau$');
print('%6d %8.5f %8.2f%% %8.3f%% %8.3f %8.2f%%' %
(n, d_h.mean(), d_h.std()*100, d_h_err_rms*100, np.median(d_h),
(d_h > 1).sum() * 100 / num_iter))
plt.legend(fontsize=18)
plt.xlabel('Normalized Delays')
plt.axvline(1, color='k');

     n     Mean      Std   Err.RMS    Median  % > 1/λ
5  1.00177    44.74%   44.742%    0.936    44.33%


Since both $\hat\tau = T_n / n$ and $\hat\lambda_u = (n-1)/T_n$ are unbiased estimators their empirical mean computed on $N$ (num_iter) sets of n delays, will converge to $\lambda^{-1}$ and $\lambda$ respectively. In symbols:

$$\textrm{E}[\hat\lambda_u] = \lambda$$

$$\textrm{E}[\hat\tau] = \lambda^{-1} \quad\Rightarrow\quad \frac{1}{\textrm{E}[\hat\tau]} = \lambda$$

Below we plot the convergence of the expected value, approximated by an empirical average with number of "trials" $N \to \infty$.

Let's start plotting $1/\textrm{E}[\hat\tau]$ for $N \to \infty$ first:

In [8]:
rate_t = 1 / ((d1.mean(axis=1)).cumsum() / np.arange(1, num_iter+1)) / λ

plt.plot(rate_t[100:])
plt.ylim(0.98, 1.02)
plt.axhline(1, color='k');


And now, let's plot the same for $\textrm{E}[\hat\lambda_u]$:

In [9]:
rate_t2 = ((n - 1) / d1.sum(axis=1)).cumsum() / np.arange(1, num_iter+1) / λ

plt.plot(rate_t2[100:])
plt.ylim(0.98, 1.02)
plt.axhline(1, color='k');


In both cases there is convergence to $\lambda$.

# Appendix 1. Computation of integral "A"¶

Here we want to compute the integral:

$$\int_0^{\infty} x^{n-2} e^{-\lambda x}\, dx$$

The $\Gamma$ function is defined as:

$$\Gamma(s) = \int_0^{\infty} t^{s-1}\,e^{-t}\,{\rm d}t$$

Therefore, with the variable substitution $t = \lambda x$:

$${\rm d}t = \lambda {\rm d} x$$

$$\frac{1}{\lambda^{n-2}} \int_0^\infty t^{n-2} e^{-t}\,\frac{{\rm d}t}{\lambda} = \frac{1}{\lambda^{n-1}} \int_0^\infty t^{n-2} e^{-t}\,{\rm d}t = \frac{1}{\lambda^{n-1}} \Gamma(n-1) = \frac{(n-2)!}{\lambda^{n-1}}$$

So we obtain:

$$\int_0^{\infty} x^{n-2} e^{-\lambda x}\, dx = \frac{(n-2)!}{\lambda^{n-1}}\qquad\qquad \textrm{(A)}$$

Finally, note that $n$ is the number of delays, while the number of times/events is $m = n+1$.

# Appendix 2. Variance of the rate estimators¶

We want to compute the variance for the family of rate estimators $\Lambda_{n,c}$ defined as

$$\Lambda_{n,c} = \frac{n-c}{T_n}$$

where $T_n = \sum d_i$ is the sum of the $n$ delays, and $c$ is a real parameter with $c < n$.

We start by writing:

$$\mathrm{Var}\{\Lambda_{n,c}\} = \mathrm{E}\left\{(\Lambda_{n,c} - \mathrm{E}\{\Lambda_{n,c}\})^2 \right\} = \int \left( \Lambda_{n,c} - \overline{\Lambda}_{n,c} \right)^2 f(x \:|\: n, \lambda)\, dx = \int \left(\frac{n - c}{x} - (n - c)\frac{\lambda}{n-1} \right)^2 f(x \:|\: n, \lambda)\, dx =$$

$$= (n-c)^2 \int \left(\frac{1}{x} - \frac{\lambda}{n-1} \right)^2 f(x \:|\: n, \lambda)\, dx$$

For brevity, let's define:

$$\left(\frac{1}{x} - \frac{\lambda}{n-1} \right)^2 = \left(\frac{1}{x^2} + \left(\frac{\lambda}{n-1}\right)^2 -2 \frac{\lambda}{x(n-1)})\right) = \left(\frac{1}{x^2} + a + \frac{b}{x} \right)$$

From term 2:

$$(n-c)^2 \int a \, f(x \:|\: n, \lambda)\, dx = a (n-c)^2 = (n-c)^2\left( \frac{\lambda}{n-1} \right)^2$$

From term 1:

$$(n-c)^2 \int \frac{1}{x^2} \, f(x \:|\: n, \lambda)\, dx = (n - c)^2 \int \frac{\lambda^n x^{n-3} e^{-\lambda x}}{(n-1)!\,}\, dx = \frac{(n - c)^2\lambda^n}{(n-1)!} \int x^{n-3} e^{-\lambda x}\, dx = \frac{(n - c)^2\lambda^n}{(n-1)!} \frac{(n-3)!}{\lambda^{n-2}} =$$

$$= \frac{(n - c)^2\lambda^2}{(n-1)(n-2)}$$

And from term 3:

$$(n-c)^2 \int \frac{b}{x} \, f(x \:|\: n, \lambda)\, dx = b(n - c)^2 \int \frac{\lambda^n x^{n-2} e^{-\lambda x}}{(n-1)!\,}\, dx = \frac{b(n - c)^2\lambda^n}{(n-1)!} \int x^{n-2} e^{-\lambda x}\, dx = \frac{b(n - c)^2\lambda^n}{(n-1)!} \frac{(n-2)!}{\lambda^{n-1}} =$$

$$= \frac{b(n - c)^2\lambda}{(n-1)} = -\frac{2\lambda}{n-1} \frac{(n - c)^2\lambda}{(n-1)} = -2 \frac{(n - c)^2\lambda^2}{(n-1)^2}$$

Summing these 3 terms we obtain the variance:

$$\mathrm{Var}\{\Lambda_{n,c}\} = \lambda^2(n-c)^2\left[ \frac{1}{n-1} + \frac{1}{n-2} - 2 \frac{1}{n-1}\right] = \lambda^2(n-c)^2\left[\frac{1}{n-2} - \frac{1}{n-1}\right]$$