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likelihood/lecture/likelihood.tex
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@ -4,111 +4,132 @@
\label{maximumlikelihoodchapter}
\exercisechapter{Maximum likelihood estimation}
A common problem in statistics is to estimate from a probability
distribution one or more parameters $\theta$ that best describe the
data $x_1, x_2, \ldots x_n$. \enterm[maximum likelihood
estimator]{Maximum likelihood estimators} (\enterm[mle|see{maximum
likelihood estimator}]{mle},
\determ{Maximum-Likelihood-Sch\"atzer}) choose the parameters such
that they maximize the likelihood of the data $x_1, x_2, \ldots x_n$
to originate from the distribution.
The core task of statistics is to infer from measured data some
parameters describing the data. These parameters can be simply a mean,
a standard deviation, or any other parameter needed to describe the
distribution the data a re originating from, a correlation
coefficient, or some parameters of a function describing a particular
dependence between the data. The brain faces exactly the same
problem. Given the activity pattern of some neurons (the data) it
needs to infer some aspects (parameters) of the environment and the
internal state of the body in order to generate some useful
behavior. One possible approach to estimate parameters from data are
\enterm[maximum likelihood estimator]{maximum likelihood estimators}
(\enterm[mle|see{maximum likelihood estimator}]{mle},
\determ{Maximum-Likelihood-Sch\"atzer}). They choose the parameters
such that they maximize the likelihood of the specific data values to
originate from a specific distribution.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Maximum Likelihood}
\section{Maximum likelihood}
Let $p(x|\theta)$ (to be read as ``probability(density) of $x$ given
$\theta$.'') the probability (density) distribution of $x$ given the
parameters $\theta$. This could be the normal distribution
$\theta$.'') the probability (density) distribution of data value $x$
given parameter values $\theta$. This could be the normal distribution
\begin{equation}
\label{normpdfmean}
p(x|\mu, \sigma) = \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
\end{equation}
defined by the mean $\mu$ and the standard deviation $\sigma$ as
parameters $\theta$. If the $n$ independent observations of $x_1,
x_2, \ldots x_n$ originate from the same probability density
distribution (they are \enterm[i.i.d.|see{independent and identically
distributed}]{i.i.d.}, \enterm{independent and identically
distributed}) then the conditional probability $p(x_1,x_2, \ldots
x_n|\theta)$ of observing $x_1, x_2, \ldots x_n$ given some specific
parameter values $\theta$ is given by
parameters $\theta$. If the $n$ observations $x_1, x_2, \ldots, x_n$
are independent of each other and originate from the same probability
density distribution (they are \enterm[i.i.d.|see{independent and
identically distributed}]{i.i.d.}, \enterm{independent and
identically distributed}), then the conditional probability
$p(x_1,x_2, \ldots, x_n|\theta)$ of observing the particular data
values $x_1, x_2, \ldots, x_n$ given some specific parameter values
$\theta$ of the probability density is given by the product of the
probability densities of each data value:
\begin{equation}
p(x_1,x_2, \ldots x_n|\theta) = p(x_1|\theta) \cdot p(x_2|\theta)
\ldots p(x_n|\theta) = \prod_{i=1}^n p(x_i|\theta) \; .
\label{probdata}
p(x_1,x_2, \ldots, x_n|\theta) = p(x_1|\theta) \cdot p(x_2|\theta)
\ldots, p(x_n|\theta) = \prod_{i=1}^n p(x_i|\theta) \; .
\end{equation}
Vice versa, the \entermde{Likelihood}{likelihood} of the parameters $\theta$
given the observed data $x_1, x_2, \ldots x_n$ is
given the observed data $x_1, x_2, \ldots, x_n$ is
\begin{equation}
{\cal L}(\theta|x_1,x_2, \ldots x_n) = p(x_1,x_2, \ldots x_n|\theta) \; .
\label{likelihood}
{\cal L}(\theta|x_1,x_2, \ldots, x_n) = p(x_1,x_2, \ldots, x_n|\theta) \; .
\end{equation}
Note: the likelihood ${\cal L}$ is not a probability in the
Note, that the likelihood ${\cal L}$ is not a probability in the
classic sense since it does not integrate to unity ($\int {\cal
L}(\theta|x_1,x_2, \ldots x_n) \, d\theta \ne 1$).
When applying maximum likelihood estimations we are interested in the
parameter values
L}(\theta|x_1,x_2, \ldots, x_n) \, d\theta \ne 1$). For given
observations $x_1, x_2, \ldots, x_n$, the likelihood
\eqref{likelihood} is a function of the parameters $\theta$. This
function has a global maximum for some specific parameter values. At
this maximum the probability \eqref{probdata} to observe the measured
data values is the largest.
Maximum likelihood estimators just find the parameter values
\begin{equation}
\theta_{mle} = \text{argmax}_{\theta} {\cal L}(\theta|x_1,x_2, \ldots x_n)
\theta_{mle} = \text{argmax}_{\theta} {\cal L}(\theta|x_1,x_2, \ldots, x_n)
\end{equation}
that maximize the likelihood. $\text{argmax}_xf(x)$ is the value of
the argument $x$ for which the function $f(x)$ assumes its global
maximum. Thus, we search for the parameter values $\theta$ at which
the likelihood ${\cal L}(\theta)$ reaches its maximum. For these
paraemter values the measured data most likely originated from the
corresponding distribution.
that maximize the likelihood \eqref{likelihood}.
$\text{argmax}_xf(x)$ is the value of the argument $x$ for which the
function $f(x)$ assumes its global maximum. Thus, we search for the
parameter values $\theta$ at which the likelihood ${\cal L}(\theta)$
reaches its maximum. For these parameter values the measured data most
likely originated from the corresponding distribution.
The position of a function's maximum does not change when the values
of the function are transformed by a strictly monotonously rising
function such as the logarithm. For numerical reasons and reasons that
we will discuss below, we search for the maximum of the logarithm of
the likelihood
(\entermde[likelihood!log-]{Likelihood!Log-}{log-likelihood}):
we discuss below, we instead search for the maximum of the logarithm
of the likelihood
(\entermde[likelihood!log-]{Likelihood!Log-}{log-likelihood})
\begin{eqnarray}
\theta_{mle} & = & \text{argmax}_{\theta}\; {\cal L}(\theta|x_1,x_2, \ldots x_n) \nonumber \\
& = & \text{argmax}_{\theta}\; \log {\cal L}(\theta|x_1,x_2, \ldots x_n) \nonumber \\
\theta_{mle} & = & \text{argmax}_{\theta}\; {\cal L}(\theta|x_1,x_2, \ldots, x_n) \nonumber \\
& = & \text{argmax}_{\theta}\; \log {\cal L}(\theta|x_1,x_2, \ldots, x_n) \nonumber \\
& = & \text{argmax}_{\theta}\; \log \prod_{i=1}^n p(x_i|\theta) \nonumber \\
& = & \text{argmax}_{\theta}\; \sum_{i=1}^n \log p(x_i|\theta) \label{loglikelihood}
\end{eqnarray}
which is the sum of the logarithms of the probabilites of each
observation. Let's illustrate the concept of maximum likelihood
estimation on the arithmetic mean.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example: the arithmetic mean}
Suppose that the measurements $x_1, x_2, \ldots x_n$ originate from a
normal distribution \eqnref{normpdfmean} and we consider the mean
$\mu$ as the only parameter $\theta$. Which value of $\theta$
maximizes the likelihood of the data?
\subsection{Arithmetic mean}
Suppose that the measurements $x_1, x_2, \ldots, x_n$ originate from a
normal distribution \eqref{normpdfmean} and we do not know the
population mean $\mu$ of the normal distribution
(\figrefb{mlemeanfig}). In this setting $\mu$ is the only parameter
$\theta$. Which value of $\mu$ maximizes the likelihood of the data?
\begin{figure}[t]
\includegraphics[width=1\textwidth]{mlemean}
\titlecaption{\label{mlemeanfig} Maximum likelihood estimation of
the mean.}{Top: The measured data (blue dots) together with three
different possible normal distributions with different means
(arrows) the data could have originated from. Bottom left: the
likelihood as a function of $\theta$ i.e. the mean. It is maximal
at a value of $\theta = 2$. Bottom right: the
log-likelihood. Taking the logarithm does not change the position
of the maximum.}
normal distributions differing in their means (arrows) from which
the data could have originated from. Bottom left: the likelihood
as a function of the parameter $\mu$. For the data it is maximal
at a value of $\mu = 2$. Bottom right: the log-likelihood. Taking
the logarithm does not change the position of the maximum.}
\end{figure}
The log-likelihood \eqnref{loglikelihood}
With the normal distribution \eqref{normpdfmean} and applying
logarithmic identities, the log-likelihood \eqref{loglikelihood} reads
\begin{eqnarray}
\log {\cal L}(\theta|x_1,x_2, \ldots x_n)
& = & \sum_{i=1}^n \log \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x_i-\theta)^2}{2\sigma^2}} \nonumber \\
& = & \sum_{i=1}^n - \log \sqrt{2\pi \sigma^2} -\frac{(x_i-\theta)^2}{2\sigma^2} \; .
\log {\cal L}(\mu|x_1,x_2, \ldots, x_n)
& = & \sum_{i=1}^n \log \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x_i-\mu)^2}{2\sigma^2}} \nonumber \\
& = & \sum_{i=1}^n - \log \sqrt{2\pi \sigma^2} -\frac{(x_i-\mu)^2}{2\sigma^2} \; .
\end{eqnarray}
Since the logarithm is the inverse function of the exponential
($\log(e^x)=x$), taking the logarithm removes the exponential from the
normal distribution. To calculate the maximum of the log-likelihood,
we need to take the derivative with respect to $\theta$ and set it to
zero:
normal distribution. This is the second reason why it is useful to
maximize the log-likelihood. To calculate the maximum of the
log-likelihood, we need to take the derivative with respect to $\mu$
and set it to zero:
\begin{eqnarray}
\frac{\text{d}}{\text{d}\theta} \log {\cal L}(\theta|x_1,x_2, \ldots x_n) & = & \sum_{i=1}^n - \frac{2(x_i-\theta)}{2\sigma^2} \;\; = \;\; 0 \nonumber \\
\Leftrightarrow \quad \sum_{i=1}^n x_i - \sum_{i=1}^n \theta & = & 0 \nonumber \\
\Leftrightarrow \quad n \theta & = & \sum_{i=1}^n x_i \nonumber \\
\Leftrightarrow \quad \theta & = & \frac{1}{n} \sum_{i=1}^n x_i \;\; = \;\; \bar x
\frac{\text{d}}{\text{d}\mu} \log {\cal L}(\mu|x_1,x_2, \ldots, x_n) & = & \sum_{i=1}^n - \frac{\text{d}}{\text{d}\mu} \log \sqrt{2\pi \sigma^2} - \frac{\text{d}}{\text{d}\mu} \frac{(x_i-\mu)^2}{2\sigma^2} \;\; = \;\; 0 \nonumber \\
\Leftrightarrow \quad \sum_{i=1}^n - \frac{2(x_i-\mu)}{2\sigma^2} & = & 0 \nonumber \\
\Leftrightarrow \quad \sum_{i=1}^n x_i - \sum_{i=1}^n \mu & = & 0 \nonumber \\
\Leftrightarrow \quad n \mu & = & \sum_{i=1}^n x_i \nonumber \\
\Leftrightarrow \quad \mu & = & \frac{1}{n} \sum_{i=1}^n x_i \;\; = \;\; \bar x
\end{eqnarray}
Thus, the maximum likelihood estimator is the arithmetic mean. That
is, the arithmetic mean maximizes the likelihood that the data
originate from a normal distribution centered at the arithmetic mean
Thus, the maximum likelihood estimator of the population mean of
normally distributed data is the arithmetic mean. That is, the
arithmetic mean maximizes the likelihood that the data originate from
a normal distribution centered at the arithmetic mean
(\figref{mlemeanfig}). Equivalently, the standard deviation computed
from the data, maximizes the likelihood that the data were generated
from a normal distribution with this standard deviation.
@ -123,6 +144,17 @@ from a normal distribution with this standard deviation.
the maxima with the mean calculated from the data.
\end{exercise}
Comparing the values of the likelihood with the ones of the
log-likelihood shown in \figref{mlemeanfig}, shows the numerical
reason for taking the logarithm of the likelihood. The likelihood
values can get very small, because we multiply many, potentially small
probability densities with each other. The likelihood quickly gets
smaller than the samlles number a floating point number of a computer
can represent. Try it by increasing the number of data values in the
exercise. Taking the logarithm avoids this problem. The log-likelihood
assumes well behaving numbers that can be handled well by the
computer.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Fitting probability distributions}
@ -130,32 +162,39 @@ Consider normally distributed data with unknown mean and standard
deviation. From the considerations above we just have seen that a
Gaussian distribution with mean at the arithmetic mean and standard
deviation equal to the standard deviation computed from the data is
the best Gaussian distribution that fits the data best in a maximum
likelihood sense, i.e. the likelihood that the data were generated
from this distribution is the largest. Fitting a Gaussian distribution
to data is very simple: just compute the two parameter of the Gaussian
distribution as the arithmetic mean and a standard deviation directly
from the data.
For non-Gaussian distributions (e.g. a Gamma-distribution), however,
such simple analytical expressions for the parameters of the
distribution do not exist, e.g. the shape parameter of a
\entermde[distribution!Gamma-]{Verteilung!Gamma-}{Gamma-distribution}. How
do we fit such a distribution to some data? That is, how should we
compute the values of the parameters of the distribution, given the
data?
the Gaussian that fits the data best in a maximum likelihood sense,
i.e. the likelihood that the data were generated from this
distribution is the largest. Fitting a Gaussian distribution to data
is very simple: just compute the two parameter of the Gaussian
distribution $\mu$ and $\sigma$ as the arithmetic mean and a standard
deviation, respectively, directly from the data.
For non-Gaussian distributions, for example a
\entermde[distribution!Gamma-]{Verteilung!Gamma-}{Gamma-distribution}
\begin{equation}
\label{gammapdf}
p(x|\alpha,\beta) \sim x^{\alpha-1}e^{-\beta x} \; ,
\end{equation}
however, such simple analytical expressions for the parameters of the
distribution do not exist. This is the case, for example, for the
shape parameter $\alpha$ of the Gamma-distribution. How do we fit such
a distribution to some data? That is, how should we compute the
values of the parameters of the distribution, given the data?
A first guess could be to fit the probability density function by
minimization of the squared difference to a histogram of the measured
data. For several reasons this is, however, not the method of choice:
(i) Probability densities can only be positive which leads, for small
values in particular, to asymmetric distributions. (ii) The values of
a histogram estimating the density are not independent because the
integral over a density is unity. The two basic assumptions of
normally distributed and independent samples, which are a prerequisite
make the minimization of the squared difference \eqnref{chisqmin} to a
maximum likelihood estimation, are violated. (iii) The histogram
strongly depends on the chosen bin size \figref{mlepdffig}).
data in the same way as we fit a a function to some data. For several
reasons this is, however, not the method of choice: (i) Probability
densities can only be positive which leads, for small values in
particular, to asymmetric distributions of the estimated histogram
around the true density. (ii) The values of a histogram estimating the
density are not independent because the integral over a density is
unity. The two basic assumptions of normally distributed and
independent samples, which are a prerequisite for making the
minimization of the squared difference to a maximum likelihood
estimation (see next section), are violated. (iii) The estimation of
the probability density by means of a histogram strongly depends on
the chosen bin size \figref{mlepdffig}).
\begin{figure}[t]
\includegraphics[width=1\textwidth]{mlepdf}
@ -173,11 +212,10 @@ Instead we should stay with maximum-likelihood estimation. Exactly in
the same way as we estimated the mean value of a Gaussian distribution
above, we can numerically fit the parameter of any type of
distribution directly from the data by means of maximizing the
likelihood. We simply search for the parameter $\theta$ of the
desired probability density function that maximizes the
log-likelihood. In general this is a non-linear optimization problem
that is solved with numerical methods such as the gradient descent
\matlabfun{mle()}.
likelihood. We simply search for the parameter values of the desired
probability density function that maximize the log-likelihood. In
general this is a non-linear optimization problem that is solved with
numerical methods such as the gradient descent \matlabfun{mle()}.
\begin{exercise}{mlegammafit.m}{mlegammafit.out}
Generate a sample of gamma-distributed random numbers and apply the
@ -191,12 +229,16 @@ that is solved with numerical methods such as the gradient descent
When fitting a function of the form $f(x;\theta)$ to data pairs
$(x_i|y_i)$ one tries to adapt the parameter $\theta$ such that the
function best describes the data. With maximum likelihood we search
for the parameter value $\theta$ for which the likelihood that the data
were drawn from the corresponding function is maximal. If we assume
that the $y_i$ values are normally distributed around the function
values $f(x_i;\theta)$ with a standard deviation $\sigma_i$, the
log-likelihood is
function best describes the data. In
chapter~\ref{gradientdescentchapter} we simply assumed that ``best''
means minimizing the squared distance between the data and the
function. With maximum likelihood we search for the parameter value
$\theta$ for which the likelihood that the data were drawn from the
corresponding function is maximal.
If we assume that the $y_i$ values are normally distributed around the
function values $f(x_i;\theta)$ with a standard deviation $\sigma_i$,
the log-likelihood is
\begin{eqnarray}
\log {\cal L}(\theta|(x_1,y_1,\sigma_1), \ldots, (x_n,y_n,\sigma_n))
& = & \sum_{i=1}^n \log \frac{1}{\sqrt{2\pi \sigma_i^2}}e^{-\frac{(y_i-f(x_i;\theta))^2}{2\sigma_i^2}} \nonumber \\
@ -218,18 +260,17 @@ the position of the minimum:
\theta_{mle} = \text{argmin}_{\theta} \; \sum_{i=1}^n \left( \frac{y_i-f(x_i;\theta)}{\sigma_i} \right)^2 \;\; = \;\; \text{argmin}_{\theta} \; \chi^2
\end{equation}
The sum of the squared differences normalized by the standard
deviation is also called $\chi^2$. The parameter $\theta$ which
minimizes the squared differences is thus the one that maximizes the
likelihood that the data actually originate from the given
function. Minimizing $\chi^2$ therefore is a maximum likelihood
deviation is also called $\chi^2$ (chi squared). The parameter
$\theta$ which minimizes the squared differences is thus the one that
maximizes the likelihood of the data to actually originate from the
given function. Therefore, minimizing $\chi^2$ is a maximum likelihood
estimation.
From the mathematical considerations above we can see that the
minimization of the squared difference is a maximum-likelihood
estimation only if the data are normally distributed around the
function. In case of other distributions, the log-likelihood
\eqnref{loglikelihood} needs to be adapted accordingly and be
maximized respectively.
\eqnref{loglikelihood} needs to be adapted accordingly.
\begin{figure}[t]
\includegraphics[width=1\textwidth]{mlepropline}
@ -377,7 +418,7 @@ orientation $\phi$ of an edge is given by
The log-likelihood of the edge orientation $\phi$ given the
activity pattern in the population $r_1$, $r_2$, ... $r_n$ is thus
\begin{equation}
{\cal L}(\phi|r_1, r_2, \ldots r_n) = \sum_{i=1}^n \log p_i(r_i|\phi)
{\cal L}(\phi|r_1, r_2, \ldots, r_n) = \sum_{i=1}^n \log p_i(r_i|\phi)
\end{equation}
The angle $\phi$ that maximizes this likelihood is then an estimate of
the orientation of the edge.

4
likelihood/lecture/mlemean.py
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@ -52,7 +52,7 @@ for i, theta in enumerate(thetas) :
p=np.prod(ps,axis=0)
# plot it:
ax = fig.add_subplot(spec[1, 0])
ax.set_xlabel(r'Parameter $\theta$')
ax.set_xlabel(r'Parameter $\mu$')
ax.set_ylabel('Likelihood')
ax.set_xticks(np.arange(1.6, 2.5, 0.4))
ax.annotate('Maximum',
@ -68,7 +68,7 @@ ax.annotate('',
ax.plot(thetas, p, **lsAm)
ax = fig.add_subplot(spec[1, 1])
ax.set_xlabel(r'Parameter $\theta$')
ax.set_xlabel(r'Parameter $\mu$')
ax.set_ylabel('Log-Likelihood')
ax.set_ylim(-50,-20)
ax.set_xticks(np.arange(1.6, 2.5, 0.4))

1
regression/lecture/regression.tex
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@ -1,4 +1,5 @@
\chapter{Optimization and gradient descent}
\label{gradientdescentchapter}
\exercisechapter{Optimization and gradient descent}
Optimization problems arise in many different contexts. For example,