From 454c7178c7f4dc81cc9a5d375b8786f09ee65caa Mon Sep 17 00:00:00 2001
From: Jan Benda <jan.benda@uni-tuebingen.de>
Date: Sat, 9 Jan 2021 19:48:59 +0100
Subject: [PATCH] [likelihood] improved text
---
likelihood/lecture/likelihood-chapter.tex | 7 +-
likelihood/lecture/likelihood.tex | 334 +++++++++++++---------
likelihood/lecture/mlecoding.py | 43 +--
3 files changed, 232 insertions(+), 152 deletions(-)
diff --git a/likelihood/lecture/likelihood-chapter.tex b/likelihood/lecture/likelihood-chapter.tex
index 736d9eb..40021f4 100644
--- a/likelihood/lecture/likelihood-chapter.tex
+++ b/likelihood/lecture/likelihood-chapter.tex
@@ -20,12 +20,11 @@
\section{TODO}
\begin{itemize}
-\item Fitting psychometric functions:
+\item Fitting psychometric functions, logistic regression:
Variable $x_i$, responses $r_i$ either 0 or 1.
- $p(x_i, \theta)$ is Weibull or Boltzmann function.
+ $p(x_i, \theta)$ is Weibull or Boltzmann or logistic function.
Likelihood is $L = \prod p(x_i, \theta)^{r_i} (1-p(x_i, \theta))^{1-r_i}$.
- Use fminsearch for fitting.
-\item GLM model fitting?
+ Use fminsearch for fitting?
\end{itemize}
\end{document}
diff --git a/likelihood/lecture/likelihood.tex b/likelihood/lecture/likelihood.tex
index cae9ff0..ea0cfd4 100644
--- a/likelihood/lecture/likelihood.tex
+++ b/likelihood/lecture/likelihood.tex
@@ -7,15 +7,15 @@
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},
+distribution the data are 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.
@@ -124,7 +124,7 @@ and set it to zero:
\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
+ \Leftrightarrow \quad \mu & = & \frac{1}{n} \sum_{i=1}^n x_i \;\; = \;\; \bar x \label{arithmeticmean}
\end{eqnarray}
Thus, the maximum likelihood estimator of the population mean of
normally distributed data is the arithmetic mean. That is, the
@@ -173,28 +173,28 @@ 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} \; ,
+ 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 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}).
+with a shape parameter $\alpha$ and a rate parameter $\beta$
+(\figrefb{mlepdffig}), in general no such simple analytical
+expressions for estimating the parameters directly from the data do
+not exist. How do we fit such a distribution to the 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 a
+\enterm{least squares} fit to a normalized histogram of the measured data in
+the same way as we fit 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, in particular for small values, 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.
\begin{figure}[t]
\includegraphics[width=1\textwidth]{mlepdf}
@@ -203,9 +203,9 @@ the chosen bin size \figref{mlepdffig}).
order Gamma-distribution. The maximum likelihood estimation of the
probability density function is shown in orange, the true pdf is
shown in red. Right: normalized histogram of the data together
- with the real (red) and the fitted probability density
- functions. The fit was done by minimizing the squared difference
- to the histogram.}
+ with the true probability density (red) and the probability
+ density function obtained by a least squares fit to the
+ histogram.}
\end{figure}
Instead we should stay with maximum-likelihood estimation. Exactly in
@@ -228,13 +228,13 @@ numerical methods such as the gradient descent \matlabfun{mle()}.
\section{Curve fitting}
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
+$(x_i|y_i)$ one tries to adapt the parameters $\theta$ such that the
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.
+function. With maximum likelihood we search for those parameter
+values $\theta$ that maximize the likelihood of the data to be drawn
+from the corresponding function.
If we assume that the $y_i$ values are normally distributed around the
function values $f(x_i;\theta)$ with a standard deviation $\sigma_i$,
@@ -242,35 +242,49 @@ 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 \\
- & = & \sum_{i=1}^n - \log \sqrt{2\pi \sigma_i^2} -\frac{(y_i-f(x_i;\theta))^2}{2\sigma_i^2} \\
+ & = & \sum_{i=1}^n - \log \sqrt{2\pi \sigma_i^2} -\frac{(y_i-f(x_i;\theta))^2}{2\sigma_i^2}
\end{eqnarray}
-The only difference to the previous example is that the averages in
-the equations above are now given as the function values
-$f(x_i;\theta)$. The parameter $\theta$ should be the one that
-maximizes the log-likelihood. The first part of the sum is independent
-of $\theta$ and can thus be ignored when computing the the maximum:
+The only difference to the previous example of the arithmetic mean is
+that the means $\mu$ in the equations above are given by the function
+values $f(x_i;\theta)$. The parameters $\theta$ should maximize the
+log-likelihood. The first term in the sum is independent of $\theta$
+and can be ignored when computing the the maximum. From the second
+term we pull out the constant factor $-\frac{1}{2}$:
\begin{eqnarray}
& = & - \frac{1}{2} \sum_{i=1}^n \left( \frac{y_i-f(x_i;\theta)}{\sigma_i} \right)^2
\end{eqnarray}
We can further simplify by inverting the sign and then search for the
-minimum. Also the factor $1/2$ can be ignored since it does not affect
-the position of the minimum:
+minimum. Also the factor $\frac{1}{2}$ can be ignored since it does
+not affect the position of the minimum:
\begin{equation}
\label{chisqmin}
\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 $\chi^2$ (chi squared). The parameter
+The sum of the squared differences between the $y$-data values and the
+function values, normalized by the standard deviation of the data
+around the function, is 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.
+given function.
+
+Whether minimizing $\chi^2$ or the \enterm{mean squared error}
+\eqref{meansquarederror} introduced in
+chapter~\ref{gradientdescentchapter} does not matter. The latter is
+the mean and $\chi^2$ is the sum of the squared differences. They
+simply differ by the constant factor $n$, the number of data pairs,
+which does not affect the position of the minimum. $\chi^2$ is more
+general in that it allows for different standard deviations for each
+data pair. If they are all the same ($\sigma_i = \sigma$), the common
+standard deviation can be pulled out of the sum and also does not
+influence the position of the minimum. Both \enterm{least squares} and
+minimizing $\chi^2$ are maximum likelihood estimators of the
+parameters $\theta$ of a function.
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.
+\eqref{loglikelihood} needs to be adapted accordingly.
\begin{figure}[t]
\includegraphics[width=1\textwidth]{mlepropline}
@@ -283,26 +297,32 @@ function. In case of other distributions, the log-likelihood
the normal distribution of the data around the line (right histogram).}
\end{figure}
+Let's go on and calculate the minimum \eqref{chisqmin} of $\chi^2$
+analytically for a simple function.
-\subsection{Example: simple proportionality}
-The function of a line with slope $\theta$ through the origin is
-\[ f(x) = \theta x \; . \]
-The $\chi^2$-sum is thus
-\[ \chi^2 = \sum_{i=1}^n \left( \frac{y_i-\theta x_i}{\sigma_i} \right)^2 \; . \]
-To estimate the minimum we again take the first derivative with
-respect to $\theta$ and equate it to zero:
-\begin{eqnarray}
- \frac{\text{d}}{\text{d}\theta}\chi^2 & = & \frac{\text{d}}{\text{d}\theta} \sum_{i=1}^n \left( \frac{y_i-\theta x_i}{\sigma_i} \right)^2 \nonumber \\
- & = & \sum_{i=1}^n \frac{\text{d}}{\text{d}\theta} \left( \frac{y_i-\theta x_i}{\sigma_i} \right)^2 \nonumber \\
- & = & -2 \sum_{i=1}^n \frac{x_i}{\sigma_i} \left( \frac{y_i-\theta x_i}{\sigma_i} \right) \nonumber \\
- & = & -2 \sum_{i=1}^n \left( \frac{x_i y_i}{\sigma_i^2} - \theta \frac{x_i^2}{\sigma_i^2} \right) \;\; = \;\; 0 \nonumber \\
-\Leftrightarrow \quad \theta \sum_{i=1}^n \frac{x_i^2}{\sigma_i^2} & = & \sum_{i=1}^n \frac{x_i y_i}{\sigma_i^2} \nonumber
-\end{eqnarray}
+
+\subsection{Straight line trough the origin}
+The function of a straight line with slope $m$ through the origin
+is
+\[ f(x) = m x \; . \]
+With this specific function, $\chi^2$ reads
+\[ \chi^2 = \sum_{i=1}^n \left( \frac{y_i-m x_i}{\sigma_i} \right)^2
+\; . \] To calculate the minimum we take the first derivative with
+respect to $m$ and equate it to zero:
\begin{eqnarray}
-\Leftrightarrow \quad \theta & = & \frac{\sum_{i=1}^n \frac{x_i y_i}{\sigma_i^2}}{ \sum_{i=1}^n \frac{x_i^2}{\sigma_i^2}} \label{mleslope}
+ \frac{\text{d}}{\text{d}m}\chi^2 & = & \sum_{i=1}^n \frac{\text{d}}{\text{d}m} \left( \frac{y_i-m x_i}{\sigma_i} \right)^2 \nonumber \\
+ & = & -2 \sum_{i=1}^n \frac{x_i}{\sigma_i} \left( \frac{y_i-m x_i}{\sigma_i} \right) \nonumber \\
+ & = & -2 \sum_{i=1}^n \left( \frac{x_i y_i}{\sigma_i^2} - m \frac{x_i^2}{\sigma_i^2} \right) \;\; = \;\; 0 \nonumber \\
+\Leftrightarrow \quad m \sum_{i=1}^n \frac{x_i^2}{\sigma_i^2} & = & \sum_{i=1}^n \frac{x_i y_i}{\sigma_i^2} \nonumber \\
+\Leftrightarrow \quad m & = & \frac{\sum_{i=1}^n \frac{x_i y_i}{\sigma_i^2}}{ \sum_{i=1}^n \frac{x_i^2}{\sigma_i^2}} \label{mleslope}
\end{eqnarray}
-This is an analytical expression for the estimation of the slope
-$\theta$ of the regression line (\figref{mleproplinefig}).
+This is an analytical expression for the estimation of the slope $m$
+of the regression line (\figref{mleproplinefig}). We do not need to
+apply numerical methods like the gradient descent to find the slope
+that minimizes the squared differences. Instead, we can compute the
+slope directly from the data by means of \eqnref{mleslope}, very much
+like we calculate the mean of some data by means of the arithmetic
+mean \eqref{arithmeticmean}.
\subsection{Linear and non-linear fits}
A gradient descent, as we have done in the previous chapter, is not
@@ -317,47 +337,68 @@ or the coefficients $a_k$ of a polynomial
\matlabfun{polyfit()}.
Parameters that are non-linearly combined can not be calculated
-analytically. Consider for example the rate $\lambda$ of the
-exponential decay
+analytically. Consider, for example, the factor $c$ and the rate
+$\lambda$ of the exponential decay
\[ y = c \cdot e^{\lambda x} \quad , \quad c, \lambda \in \reZ \; . \]
Such cases require numerical solutions for the optimization of the
cost function, e.g. the gradient descent \matlabfun{lsqcurvefit()}.
\subsection{Relation between slope and correlation coefficient}
-Let us have a closer look on \eqnref{mleslope}. If the standard
-deviation of the data $\sigma_i$ is the same for each data point,
-i.e. $\sigma_i = \sigma_j \; \forall \; i, j$, the standard deviation drops
-out of \eqnref{mleslope} and we get
+Let us have a closer look on \eqnref{mleslope} for the slope of a line
+through the origin. If the standard deviation of the data $\sigma_i$
+is the same for each data point, i.e. $\sigma_i = \sigma_j \; \forall
+\; i, j$, the standard deviation drops out and \eqnref{mleslope}
+simplifies to
\begin{equation}
\label{whitemleslope}
- \theta = \frac{\sum_{i=1}^n x_i y_i}{\sum_{i=1}^n x_i^2}
+ m = \frac{\sum_{i=1}^n x_i y_i}{\sum_{i=1}^n x_i^2}
\end{equation}
-To see what this expression is, we need to standardize the data. We
-make the data mean free and normalize them to their standard
-deviation, i.e. $x \mapsto (x - \bar x)/\sigma_x$. The resulting
-numbers are also called \entermde[z-values]{z-Wert}{$z$-values} or
-$z$-scores and they have the property $\bar x = 0$ and $\sigma_x =
-1$. $z$-scores are often used in Biology to make quantities that
-differ in their units comparable. For standardized data the variance
+To see what the nominator and the denominator of this expression
+describe, we need to subtract from the data their mean value. We make
+the data mean free, i.e. $x \mapsto x - \bar x$ and $y \mapsto y -
+\bar y$ . For mean-free data the variance
\begin{equation}
- \sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar x)^2 = \frac{1}{n} \sum_{i=1}^n x_i^2 = 1
+ \sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar x)^2 = \frac{1}{n} \sum_{i=1}^n x_i^2
\end{equation}
-is given by the mean squared data and equals one.
-The covariance between $x$ and $y$ also simplifies to
+is given by the mean squared data. In the same way, the covariance
+between $x$ and $y$ simplifies to
\begin{equation}
\text{cov}(x, y) = \frac{1}{n} \sum_{i=1}^n (x_i - \bar x)(y_i -
- \bar y) =\frac{1}{n} \sum_{i=1}^n x_i y_i
+ \bar y) =\frac{1}{n} \sum_{i=1}^n x_i y_i \; ,
+\end{equation}
+the averaged product between pairs of $x$ and $y$ values. Expanding
+the fraction in \eqnref{whitemleslope} by $\frac{1}{n}$ we get
+\begin{equation}
+ \label{meanfreeslope}
+ m = \frac{\frac{1}{n}\sum_{i=1}^n x_i y_i}{\frac{1}{n}\sum_{i=1}^n x_i^2}
+ = \frac{\text{cov}(x, y)}{\sigma_x^2}
\end{equation}
-the averaged product between pairs of $x$ and $y$ values. Recall that
-the correlation coefficient $r_{x,y}$,
-\eqnref{correlationcoefficient}, is the covariance normalized by the
-product of the standard deviations of $x$ and $y$,
-respectively. Therefore, in case the standard deviations equal one,
-the correlation coefficient is identical to the covariance.
-Consequently, for standardized data the slope of the regression line
-\eqnref{whitemleslope} simplifies to
+
+Recall that the correlation coefficient $r_{x,y}$ is the covariance
+normalized by the product of the standard deviations of $x$ and $y$:
+\begin{equation}
+ \label{meanfreecorrcoef}
+ r_{x,y} = \frac{\text{cov}(x, y)}{\sigma_x\sigma_y}
+\end{equation}
+If furthermore the standard deviations of $x$ and $y$ are the same,
+i.e. $\sigma_x = \sigma_y$, the slope of a line trough the origin is
+identical to the correlation coefficient.
+
+This relation between slope and correlation coefficient in particular
+holds for standardized data that have been made mean free and have
+been normalized by their standard deviation, i.e. $x \mapsto (x - \bar
+x)/\sigma_x$ and $y \mapsto (y - \bar x)/\sigma_y$. The resulting
+numbers are called \entermde[z-value]{z-Wert}{$z$-values} or
+\enterm[z-score]{$z$-scores}. Their mean equals zero and their
+standard deviation one. $z$-scores are often used to make quantities
+that differ in their units comparable. For standardized data the
+denominators for both the slope \eqref{meanfreeslope} and the
+correlation coefficient \eqref{meanfreecorrcoef} equal one. For
+standardized data, both the slope of the regression line and the
+corelation coefficient reduce to the covariance between the $x$ and
+$y$ data:
\begin{equation}
- \theta = \frac{1}{n} \sum_{i=1}^n x_i y_i = \text{cov}(x,y) = r_{x,y}
+ m = \frac{1}{n} \sum_{i=1}^n x_i y_i = \text{cov}(x,y) = r_{x,y}
\end{equation}
For standardized data the slope of the regression line is the same as the
correlation coefficient!
@@ -365,63 +406,100 @@ correlation coefficient!
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Neural coding}
+Maximum likelihood estimators are not only a central concept for data
+analysis. Neural systems face the very same problem. They also need to
+estimate parameters of the internal and external environment based on
+the activity of neurons.
+
In sensory systems certain aspects of the environment are encoded in
the neuronal activity of populations of neurons. One example of such a
population code is the tuning of neurons in the primary visual cortex
-(V1) to the orientation of an edge or bar in the visual
-stimulus. Different neurons respond best to different edge
-orientations. Traditionally, such a tuning is measured by analyzing
-the neuronal response strength (e.g. the firing rate) as a function of
-the orientation of a black bar and is illustrated and summarized
-with the so called \enterm{tuning-curve} (\determ{Abstimmkurve},
+(V1) to the orientation of a bar in the visual stimulus. Different
+neurons respond best to different bar orientations. Traditionally,
+such a tuning is measured by analyzing the neuronal response strength
+(e.g. the firing rate) as a function of the orientation of a black bar
+and is illustrated and summarized with the so called
+\enterm{tuning-curve} (\determ{Abstimmkurve},
figure~\ref{mlecodingfig}, top).
\begin{figure}[tp]
\includegraphics[width=1\textwidth]{mlecoding}
\titlecaption{\label{mlecodingfig} Maximum likelihood estimation of
- a stimulus parameter from neuronal activity.}{Top: Tuning curve of
- an individual neuron as a function of the stimulus orientation (a
- dark bar in front of a white background). The stimulus that evokes
- the strongest activity in that neuron is the bar with the vertical
- orientation (arrow, $\phi_i=90$\,\degree). The red area indicates
- the variability of the neuronal activity $p(r|\phi)$ around the
- tuning curve. Center: In a population of neurons, each neuron may
- have a different tuning curve (colors). A specific stimulus (the
- vertical bar) activates the individual neurons of the population
- in a specific way (dots). Bottom: The log-likelihood of the
- activity pattern will be maximized close to the real stimulus
+ a stimulus parameter from neuronal activity.}{Top: Tuning curve
+ $r_i(\phi;c,\phi_i)$ of a specific neuron $i$ as a function of the
+ orientation $\phi$ of a stimulus, a dark bar in front of a white
+ background. The preferred stimulus $\phi_i$ of that neuron, the
+ one that evokes the strongest firing rate response, is a bar with
+ vertical orientation (arrow, $\phi_i=90$\,\degree). The width of
+ the red area indicates the variability of the neuronal activity
+ $\sigma_i$ around the tuning curve. Center: In a population of
+ neurons, each neuron may have a different tuning curve (colors). A
+ specific stimulus activates the individual neurons of the
+ population in a specific way (dots). Bottom: The log-likelihood of
+ the activity pattern has a maximum close to the real stimulus
orientation.}
\end{figure}
The brain, however, is confronted with the inverse problem: given a
certain activity pattern in the neuronal population, what is the
-stimulus (here the orientation of an edge)? In the sense of maximum
-likelihood, a possible answer to this question would be: the stimulus
-for which the particular activity pattern is most likely given the
-tuning of the neurons.
+stimulus? In our example, what is the orientation of the bar? In the
+sense of maximum likelihood, a possible answer to this question would
+be: the stimulus for which the particular activity pattern is most
+likely given the tuning of the neurons and the noise (standard
+deviation) of the responses.
Let's stay with the example of the orientation tuning in V1. The
-tuning $\Omega_i(\phi)$ of the neurons $i$ to the preferred edge
-orientation $\phi_i$ can be well described using a van-Mises function
-(the Gaussian function on a cyclic x-axis) (\figref{mlecodingfig}):
+tuning of the firing rate $r_i(\phi)$ of neuron $i$ to the preferred
+bar orientation $\phi_i$ can be well described using a van-Mises
+function (the Gaussian function on a cyclic x-axis)
+(\figref{mlecodingfig}):
\begin{equation}
- \Omega_i(\phi) = c \cdot e^{\cos(2(\phi-\phi_i))} \quad , \quad c \in \reZ
+ \label{bartuningcurve}
+ r_i(\phi; c, \phi_i) = c \cdot e^{\cos(2(\phi-\phi_i))} \quad , \quad c > 0
\end{equation}
+Note the factor two in the cosine, because the response of the neuron
+is the same for a bar turned by 180\,\degree.
+
If we approximate the neuronal activity by a normal distribution
-around the tuning curve with a standard deviation $\sigma=\Omega/4$,
-which is proportional to $\Omega$, then the probability $p_i(r|\phi)$
-of the $i$-th neuron showing the activity $r$ given a certain
-orientation $\phi$ of an edge is given by
+around the tuning curve with a standard deviation $\sigma_i$, then the
+probability $p_i(r|\phi)$ of the $i$-th neuron having the observed
+activity $r$, given a certain orientation $\phi$ of a bar is
\begin{equation}
- p_i(r|\phi) = \frac{1}{\sqrt{2\pi}\Omega_i(\phi)/4} e^{-\frac{1}{2}\left(\frac{r-\Omega_i(\phi)}{\Omega_i(\phi)/4}\right)^2} \; .
+ p_i(r|\phi) = \frac{1}{\sqrt{2\pi\sigma_i^2}} e^{-\frac{1}{2}\left(\frac{r-r_i(\phi; c, \phi_i)}{\sigma_i}\right)^2} \; .
\end{equation}
-The log-likelihood of the edge orientation $\phi$ given the
+The log-likelihood of the bar 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)
\end{equation}
-The angle $\phi$ that maximizes this likelihood is then an estimate of
-the orientation of the edge.
+The angle $\phi_{mle}$ that maximizes this likelihood is an estimate
+of the true orientation of the bar (\figref{mlecodingfig}).
+
+The noisiness of the neuron's responses as quantified by $\sigma_i$
+usually is a function of the neuron's mean firing rate $r_i$,
+\eqnref{bartuningcurve}: $\sigma_i = \sigma_i(r_i)$. This dependence
+has a major impact of the maximum likelihood estimation. Usually, the
+stronger the response of a neuron, the higher its firing rate, the
+lower the noise. In this case, strong responses will have a stronger
+influence on the position of the maximum of the log-likelihood.
+
+Whether neural systems really implement maximum likelihood estimators
+is another question. There are many ways how a stimulus property can
+be read out from the activity of a population of neurons. The simplest
+one being a ``winner takes all'' rule. The preferred stimulus
+parameter of the neuron with the strongest response is the
+estimate. Another possibility is to compute a population vector. The
+estimated stimulus parameter is the sum of the preferred stimulus
+parameters of all neurons in the population weighted by the activity
+of the neurons. In case of angular stimulus parameters, like the
+orientation of the bar in our example, a vector pointing in the
+direction of the angle is used instead of the angle to incorporate the
+cyclic nature of the parameter.
+
+Using maximum likelihood estimators for analyzing neural population
+activity gives us an upper bound of how well a stimulus parameter is
+encoded in the activity of the neurons. The brain would not be able to
+do better.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
diff --git a/likelihood/lecture/mlecoding.py b/likelihood/lecture/mlecoding.py
index 3edd5a5..8be385f 100644
--- a/likelihood/lecture/mlecoding.py
+++ b/likelihood/lecture/mlecoding.py
@@ -5,11 +5,9 @@ import matplotlib.gridspec as gridspec
from plotstyle import *
rng = np.random.RandomState(4637281)
-lmarg=0.1
-rmarg=0.1
fig = plt.figure(figsize=cm_size(figure_width, 2.8*figure_height))
-spec = gridspec.GridSpec(nrows=4, ncols=1, height_ratios=[4, 4, 1, 3], hspace=0.2,
+spec = gridspec.GridSpec(nrows=4, ncols=1, height_ratios=[4, 5, 1, 3], hspace=0.2,
**adjust_fs(fig, left=4.0))
ax = fig.add_subplot(spec[0, 0])
ax.set_xlim(0.0, np.pi)
@@ -17,7 +15,7 @@ ax.set_xticks(np.arange(0.125*np.pi, 1.*np.pi, 0.125*np.pi))
ax.set_xticklabels([])
ax.set_ylim(0.0, 3.5)
ax.yaxis.set_major_locator(plt.NullLocator())
-ax.text(-0.2, 0.5*3.5, 'Activity', rotation='vertical', va='center')
+ax.text(-0.2, 0.5*3.5, 'Firing rate', rotation='vertical', va='center')
ax.annotate('Tuning curve',
xy=(0.42*np.pi, 2.5), xycoords='data',
xytext=(0.3*np.pi, 3.2), textcoords='data', ha='right',
@@ -32,27 +30,34 @@ ax.text(0.52*np.pi, 0.7, 'preferred\norientation')
xx = np.arange(0.0, 2.0*np.pi, 0.01)
pp = 0.5*np.pi
yy = np.exp(np.cos(2.0*(xx+pp)))
-ax.fill_between(xx, yy+0.25*yy, yy-0.25*yy, **fsBa)
+ss = 1.0/(0.75+2.0*yy)
+ax.fill_between(xx, yy+ss, yy-ss, **fsBa)
ax.plot(xx, yy, **lsB)
ax = fig.add_subplot(spec[1, 0])
ax.set_xlim(0.0, np.pi)
ax.set_xticks(np.arange(0.125*np.pi, 1.*np.pi, 0.125*np.pi))
ax.set_xticklabels([])
-ax.set_ylim(0.0, 3.0)
+ax.set_ylim(-0.1, 3.5)
ax.yaxis.set_major_locator(plt.NullLocator())
-ax.text(-0.2, 0.5*3.5, 'Activity', rotation='vertical', va='center')
+ax.text(-0.2, 0.5*3.5, 'Firing rate', rotation='vertical', va='center')
xx = np.arange(0.0, 1.0*np.pi, 0.01)
prefphases = np.arange(0.125*np.pi, 1.*np.pi, 0.125*np.pi)
responses = []
-xresponse = 0.475*np.pi
+sigmas = []
+xresponse = 0.41*np.pi
+ax.annotate('Orientation of bar',
+ xy=(xresponse, -0.1), xycoords='data',
+ xytext=(xresponse, 3.1), textcoords='data', ha='left', zorder=-10,
+ arrowprops=dict(arrowstyle="->", relpos=(0.0,0.0)) )
for pp, ls, ps in zip(prefphases, [lsE, lsC, lsD, lsB, lsD, lsC, lsE],
[psE, psC, psD, psB, psD, psC, psE]) :
yy = np.exp(np.cos(2.0*(xx+pp)))
- #ax.plot(xx, yy, color=cm.autumn(2.0*np.abs(pp/np.pi-0.5), 1))
ax.plot(xx, yy, **ls)
y = np.exp(np.cos(2.0*(xresponse+pp)))
- responses.append(y + rng.randn()*0.25*y)
+ s = 1.0/(0.75+2.0*y)
+ responses.append(y + rng.randn()*s)
+ sigmas.append(s)
ax.plot(xresponse, y, **ps)
responses = np.array(responses)
@@ -68,25 +73,23 @@ ax = fig.add_subplot(spec[3, 0])
ax.set_xlim(0.0, np.pi)
ax.set_xticks(np.arange(0.125*np.pi, 1.*np.pi, 0.125*np.pi))
ax.set_xticklabels([])
-ax.set_ylim(-1600, 0)
+ax.set_ylim(-210, 0)
ax.yaxis.set_major_locator(plt.NullLocator())
ax.set_xlabel('Orientation')
-ax.text(-0.2, -800, 'Log-Likelihood', rotation='vertical', va='center')
+ax.text(-0.2, -100, 'Log-Likelihood', rotation='vertical', va='center')
phases = np.linspace(0.0, 1.1*np.pi, 100)
probs = np.zeros((len(responses), len(phases)))
-for k, (pp, r) in enumerate(zip(prefphases, responses)) :
+for k, (pp, r, sigma) in enumerate(zip(prefphases, responses, sigmas)) :
y = np.exp(np.cos(2.0*(phases+pp)))
- sigma = 0.1*y
- probs[k,:] = np.exp(-0.5*((r-y)/sigma)**2.0)/np.sqrt(2.0*np.pi)/sigma
+ probs[k,:] = np.exp(-0.5*((r-y)/sigma)**2.0)/np.sqrt(2.0*np.pi*sigma**2)
loglikelihood = np.sum(np.log(probs), 0)
-maxl = np.max(loglikelihood)
maxp = phases[np.argmax(loglikelihood)]
ax.annotate('',
- xy=(maxp, -1600), xycoords='data',
- xytext=(maxp, -30), textcoords='data',
+ xy=(maxp, -210), xycoords='data',
+ xytext=(maxp, -10), textcoords='data',
arrowprops=dict(arrowstyle="->", relpos=(0.5,0.5),
connectionstyle="angle3,angleA=80,angleB=90") )
-ax.text(maxp+0.05, -1100, 'most likely\norientation\ngiven the responses')
-ax.plot(phases, loglikelihood, **lsA)
+ax.text(maxp+0.05, -150, 'most likely\norientation\ngiven the responses')
+ax.plot(phases, loglikelihood, clip_on=False, **lsA)
plt.savefig('mlecoding.pdf')