[likelihood] improved text
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@ -20,12 +20,11 @@
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\section{TODO}
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\begin{itemize}
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\item Fitting psychometric functions:
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\item Fitting psychometric functions, logistic regression:
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Variable $x_i$, responses $r_i$ either 0 or 1.
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$p(x_i, \theta)$ is Weibull or Boltzmann function.
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$p(x_i, \theta)$ is Weibull or Boltzmann or logistic function.
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Likelihood is $L = \prod p(x_i, \theta)^{r_i} (1-p(x_i, \theta))^{1-r_i}$.
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Use fminsearch for fitting.
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\item GLM model fitting?
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Use fminsearch for fitting?
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\end{itemize}
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\end{document}
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@ -7,15 +7,15 @@
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The core task of statistics is to infer from measured data some
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parameters describing the data. These parameters can be simply a mean,
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a standard deviation, or any other parameter needed to describe the
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distribution the data a re originating from, a correlation
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coefficient, or some parameters of a function describing a particular
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dependence between the data. The brain faces exactly the same
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problem. Given the activity pattern of some neurons (the data) it
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needs to infer some aspects (parameters) of the environment and the
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internal state of the body in order to generate some useful
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behavior. One possible approach to estimate parameters from data are
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\enterm[maximum likelihood estimator]{maximum likelihood estimators}
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(\enterm[mle|see{maximum likelihood estimator}]{mle},
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distribution the data are originating from, a correlation coefficient,
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or some parameters of a function describing a particular dependence
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between the data. The brain faces exactly the same problem. Given the
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activity pattern of some neurons (the data) it needs to infer some
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aspects (parameters) of the environment and the internal state of the
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body in order to generate some useful behavior. One possible approach
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to estimate parameters from data are \enterm[maximum likelihood
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estimator]{maximum likelihood estimators} (\enterm[mle|see{maximum
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likelihood estimator}]{mle},
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\determ{Maximum-Likelihood-Sch\"atzer}). They choose the parameters
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such that they maximize the likelihood of the specific data values to
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originate from a specific distribution.
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@ -124,7 +124,7 @@ and set it to zero:
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\Leftrightarrow \quad \sum_{i=1}^n - \frac{2(x_i-\mu)}{2\sigma^2} & = & 0 \nonumber \\
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\Leftrightarrow \quad \sum_{i=1}^n x_i - \sum_{i=1}^n \mu & = & 0 \nonumber \\
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\Leftrightarrow \quad n \mu & = & \sum_{i=1}^n x_i \nonumber \\
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\Leftrightarrow \quad \mu & = & \frac{1}{n} \sum_{i=1}^n x_i \;\; = \;\; \bar x
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\Leftrightarrow \quad \mu & = & \frac{1}{n} \sum_{i=1}^n x_i \;\; = \;\; \bar x \label{arithmeticmean}
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\end{eqnarray}
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Thus, the maximum likelihood estimator of the population mean of
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normally distributed data is the arithmetic mean. That is, the
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@ -173,28 +173,28 @@ For non-Gaussian distributions, for example a
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\entermde[distribution!Gamma-]{Verteilung!Gamma-}{Gamma-distribution}
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\begin{equation}
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\label{gammapdf}
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p(x|\alpha,\beta) \sim x^{\alpha-1}e^{-\beta x} \; ,
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p(x|\alpha,\beta) \sim x^{\alpha-1}e^{-\beta x}
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\end{equation}
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however, such simple analytical expressions for the parameters of the
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distribution do not exist. This is the case, for example, for the
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shape parameter $\alpha$ of the Gamma-distribution. How do we fit such
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a distribution to some data? That is, how should we compute the
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values of the parameters of the distribution, given the data?
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A first guess could be to fit the probability density function by
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minimization of the squared difference to a histogram of the measured
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data in the same way as we fit a a function to some data. For several
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reasons this is, however, not the method of choice: (i) Probability
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densities can only be positive which leads, for small values in
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particular, to asymmetric distributions of the estimated histogram
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around the true density. (ii) The values of a histogram estimating the
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density are not independent because the integral over a density is
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unity. The two basic assumptions of normally distributed and
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independent samples, which are a prerequisite for making the
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minimization of the squared difference to a maximum likelihood
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estimation (see next section), are violated. (iii) The estimation of
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the probability density by means of a histogram strongly depends on
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the chosen bin size \figref{mlepdffig}).
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with a shape parameter $\alpha$ and a rate parameter $\beta$
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(\figrefb{mlepdffig}), in general no such simple analytical
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expressions for estimating the parameters directly from the data do
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not exist. How do we fit such a distribution to the data? That is,
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how should we compute the values of the parameters of the
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distribution, given the data?
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A first guess could be to fit the probability density function by a
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\enterm{least squares} fit to a normalized histogram of the measured data in
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the same way as we fit a function to some data. For several reasons
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this is, however, not the method of choice: (i) Probability densities
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can only be positive which leads, in particular for small values, to
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asymmetric distributions of the estimated histogram around the true
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density. (ii) The values of a histogram estimating the density are not
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independent because the integral over a density is unity. The two
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basic assumptions of normally distributed and independent samples,
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which are a prerequisite for making the minimization of the squared
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difference to a maximum likelihood estimation (see next section), are
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violated. (iii) The estimation of the probability density by means of
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a histogram strongly depends on the chosen bin size.
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\begin{figure}[t]
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\includegraphics[width=1\textwidth]{mlepdf}
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@ -203,9 +203,9 @@ the chosen bin size \figref{mlepdffig}).
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order Gamma-distribution. The maximum likelihood estimation of the
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probability density function is shown in orange, the true pdf is
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shown in red. Right: normalized histogram of the data together
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with the real (red) and the fitted probability density
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functions. The fit was done by minimizing the squared difference
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to the histogram.}
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with the true probability density (red) and the probability
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density function obtained by a least squares fit to the
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histogram.}
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\end{figure}
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Instead we should stay with maximum-likelihood estimation. Exactly in
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@ -228,13 +228,13 @@ numerical methods such as the gradient descent \matlabfun{mle()}.
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\section{Curve fitting}
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When fitting a function of the form $f(x;\theta)$ to data pairs
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$(x_i|y_i)$ one tries to adapt the parameter $\theta$ such that the
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$(x_i|y_i)$ one tries to adapt the parameters $\theta$ such that the
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function best describes the data. In
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chapter~\ref{gradientdescentchapter} we simply assumed that ``best''
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means minimizing the squared distance between the data and the
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function. With maximum likelihood we search for the parameter value
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$\theta$ for which the likelihood that the data were drawn from the
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corresponding function is maximal.
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function. With maximum likelihood we search for those parameter
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values $\theta$ that maximize the likelihood of the data to be drawn
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from the corresponding function.
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If we assume that the $y_i$ values are normally distributed around the
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function values $f(x_i;\theta)$ with a standard deviation $\sigma_i$,
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@ -242,35 +242,49 @@ the log-likelihood is
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\begin{eqnarray}
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\log {\cal L}(\theta|(x_1,y_1,\sigma_1), \ldots, (x_n,y_n,\sigma_n))
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& = & \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 \\
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& = & \sum_{i=1}^n - \log \sqrt{2\pi \sigma_i^2} -\frac{(y_i-f(x_i;\theta))^2}{2\sigma_i^2} \\
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& = & \sum_{i=1}^n - \log \sqrt{2\pi \sigma_i^2} -\frac{(y_i-f(x_i;\theta))^2}{2\sigma_i^2}
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\end{eqnarray}
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The only difference to the previous example is that the averages in
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the equations above are now given as the function values
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$f(x_i;\theta)$. The parameter $\theta$ should be the one that
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maximizes the log-likelihood. The first part of the sum is independent
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of $\theta$ and can thus be ignored when computing the the maximum:
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The only difference to the previous example of the arithmetic mean is
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that the means $\mu$ in the equations above are given by the function
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values $f(x_i;\theta)$. The parameters $\theta$ should maximize the
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log-likelihood. The first term in the sum is independent of $\theta$
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and can be ignored when computing the the maximum. From the second
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term we pull out the constant factor $-\frac{1}{2}$:
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\begin{eqnarray}
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& = & - \frac{1}{2} \sum_{i=1}^n \left( \frac{y_i-f(x_i;\theta)}{\sigma_i} \right)^2
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\end{eqnarray}
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We can further simplify by inverting the sign and then search for the
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minimum. Also the factor $1/2$ can be ignored since it does not affect
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the position of the minimum:
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minimum. Also the factor $\frac{1}{2}$ can be ignored since it does
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not affect the position of the minimum:
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\begin{equation}
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\label{chisqmin}
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\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
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\end{equation}
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The sum of the squared differences normalized by the standard
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deviation is also called $\chi^2$ (chi squared). The parameter
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The sum of the squared differences between the $y$-data values and the
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function values, normalized by the standard deviation of the data
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around the function, is called $\chi^2$ (chi squared). The parameter
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$\theta$ which minimizes the squared differences is thus the one that
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maximizes the likelihood of the data to actually originate from the
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given function. Therefore, minimizing $\chi^2$ is a maximum likelihood
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estimation.
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given function.
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Whether minimizing $\chi^2$ or the \enterm{mean squared error}
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\eqref{meansquarederror} introduced in
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chapter~\ref{gradientdescentchapter} does not matter. The latter is
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the mean and $\chi^2$ is the sum of the squared differences. They
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simply differ by the constant factor $n$, the number of data pairs,
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which does not affect the position of the minimum. $\chi^2$ is more
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general in that it allows for different standard deviations for each
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data pair. If they are all the same ($\sigma_i = \sigma$), the common
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standard deviation can be pulled out of the sum and also does not
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influence the position of the minimum. Both \enterm{least squares} and
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minimizing $\chi^2$ are maximum likelihood estimators of the
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parameters $\theta$ of a function.
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From the mathematical considerations above we can see that the
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minimization of the squared difference is a maximum-likelihood
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estimation only if the data are normally distributed around the
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function. In case of other distributions, the log-likelihood
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\eqnref{loglikelihood} needs to be adapted accordingly.
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\eqref{loglikelihood} needs to be adapted accordingly.
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\begin{figure}[t]
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\includegraphics[width=1\textwidth]{mlepropline}
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@ -283,26 +297,32 @@ function. In case of other distributions, the log-likelihood
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the normal distribution of the data around the line (right histogram).}
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\end{figure}
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Let's go on and calculate the minimum \eqref{chisqmin} of $\chi^2$
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analytically for a simple function.
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\subsection{Example: simple proportionality}
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The function of a line with slope $\theta$ through the origin is
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\[ f(x) = \theta x \; . \]
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The $\chi^2$-sum is thus
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\[ \chi^2 = \sum_{i=1}^n \left( \frac{y_i-\theta x_i}{\sigma_i} \right)^2 \; . \]
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To estimate the minimum we again take the first derivative with
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respect to $\theta$ and equate it to zero:
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\begin{eqnarray}
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\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 \\
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& = & \sum_{i=1}^n \frac{\text{d}}{\text{d}\theta} \left( \frac{y_i-\theta x_i}{\sigma_i} \right)^2 \nonumber \\
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& = & -2 \sum_{i=1}^n \frac{x_i}{\sigma_i} \left( \frac{y_i-\theta x_i}{\sigma_i} \right) \nonumber \\
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& = & -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 \\
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\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
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\end{eqnarray}
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\subsection{Straight line trough the origin}
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The function of a straight line with slope $m$ through the origin
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is
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\[ f(x) = m x \; . \]
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With this specific function, $\chi^2$ reads
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\[ \chi^2 = \sum_{i=1}^n \left( \frac{y_i-m x_i}{\sigma_i} \right)^2
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\; . \] To calculate the minimum we take the first derivative with
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respect to $m$ and equate it to zero:
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\begin{eqnarray}
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\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}
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\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 \\
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& = & -2 \sum_{i=1}^n \frac{x_i}{\sigma_i} \left( \frac{y_i-m x_i}{\sigma_i} \right) \nonumber \\
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& = & -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 \\
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\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 \\
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\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}
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\end{eqnarray}
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This is an analytical expression for the estimation of the slope
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$\theta$ of the regression line (\figref{mleproplinefig}).
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This is an analytical expression for the estimation of the slope $m$
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of the regression line (\figref{mleproplinefig}). We do not need to
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apply numerical methods like the gradient descent to find the slope
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that minimizes the squared differences. Instead, we can compute the
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slope directly from the data by means of \eqnref{mleslope}, very much
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like we calculate the mean of some data by means of the arithmetic
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mean \eqref{arithmeticmean}.
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\subsection{Linear and non-linear fits}
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A gradient descent, as we have done in the previous chapter, is not
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@ -317,47 +337,68 @@ or the coefficients $a_k$ of a polynomial
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\matlabfun{polyfit()}.
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Parameters that are non-linearly combined can not be calculated
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analytically. Consider for example the rate $\lambda$ of the
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exponential decay
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analytically. Consider, for example, the factor $c$ and the rate
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$\lambda$ of the exponential decay
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\[ y = c \cdot e^{\lambda x} \quad , \quad c, \lambda \in \reZ \; . \]
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Such cases require numerical solutions for the optimization of the
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cost function, e.g. the gradient descent \matlabfun{lsqcurvefit()}.
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\subsection{Relation between slope and correlation coefficient}
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Let us have a closer look on \eqnref{mleslope}. If the standard
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deviation of the data $\sigma_i$ is the same for each data point,
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i.e. $\sigma_i = \sigma_j \; \forall \; i, j$, the standard deviation drops
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out of \eqnref{mleslope} and we get
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Let us have a closer look on \eqnref{mleslope} for the slope of a line
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through the origin. If the standard deviation of the data $\sigma_i$
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is the same for each data point, i.e. $\sigma_i = \sigma_j \; \forall
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\; i, j$, the standard deviation drops out and \eqnref{mleslope}
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simplifies to
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\begin{equation}
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\label{whitemleslope}
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\theta = \frac{\sum_{i=1}^n x_i y_i}{\sum_{i=1}^n x_i^2}
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m = \frac{\sum_{i=1}^n x_i y_i}{\sum_{i=1}^n x_i^2}
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\end{equation}
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To see what this expression is, we need to standardize the data. We
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make the data mean free and normalize them to their standard
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deviation, i.e. $x \mapsto (x - \bar x)/\sigma_x$. The resulting
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numbers are also called \entermde[z-values]{z-Wert}{$z$-values} or
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$z$-scores and they have the property $\bar x = 0$ and $\sigma_x =
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1$. $z$-scores are often used in Biology to make quantities that
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differ in their units comparable. For standardized data the variance
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To see what the nominator and the denominator of this expression
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describe, we need to subtract from the data their mean value. We make
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the data mean free, i.e. $x \mapsto x - \bar x$ and $y \mapsto y -
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\bar y$ . For mean-free data the variance
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\begin{equation}
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\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
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\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
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\end{equation}
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is given by the mean squared data and equals one.
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The covariance between $x$ and $y$ also simplifies to
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is given by the mean squared data. In the same way, the covariance
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between $x$ and $y$ simplifies to
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\begin{equation}
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\text{cov}(x, y) = \frac{1}{n} \sum_{i=1}^n (x_i - \bar x)(y_i -
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\bar y) =\frac{1}{n} \sum_{i=1}^n x_i y_i
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\bar y) =\frac{1}{n} \sum_{i=1}^n x_i y_i \; ,
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\end{equation}
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the averaged product between pairs of $x$ and $y$ values. Expanding
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the fraction in \eqnref{whitemleslope} by $\frac{1}{n}$ we get
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\begin{equation}
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\label{meanfreeslope}
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m = \frac{\frac{1}{n}\sum_{i=1}^n x_i y_i}{\frac{1}{n}\sum_{i=1}^n x_i^2}
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= \frac{\text{cov}(x, y)}{\sigma_x^2}
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\end{equation}
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the averaged product between pairs of $x$ and $y$ values. Recall that
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the correlation coefficient $r_{x,y}$,
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\eqnref{correlationcoefficient}, is the covariance normalized by the
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product of the standard deviations of $x$ and $y$,
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respectively. Therefore, in case the standard deviations equal one,
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the correlation coefficient is identical to the covariance.
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Consequently, for standardized data the slope of the regression line
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\eqnref{whitemleslope} simplifies to
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Recall that the correlation coefficient $r_{x,y}$ is the covariance
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normalized by the product of the standard deviations of $x$ and $y$:
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\begin{equation}
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\label{meanfreecorrcoef}
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r_{x,y} = \frac{\text{cov}(x, y)}{\sigma_x\sigma_y}
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\end{equation}
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If furthermore the standard deviations of $x$ and $y$ are the same,
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i.e. $\sigma_x = \sigma_y$, the slope of a line trough the origin is
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identical to the correlation coefficient.
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This relation between slope and correlation coefficient in particular
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holds for standardized data that have been made mean free and have
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been normalized by their standard deviation, i.e. $x \mapsto (x - \bar
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x)/\sigma_x$ and $y \mapsto (y - \bar x)/\sigma_y$. The resulting
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numbers are called \entermde[z-value]{z-Wert}{$z$-values} or
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\enterm[z-score]{$z$-scores}. Their mean equals zero and their
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standard deviation one. $z$-scores are often used to make quantities
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that differ in their units comparable. For standardized data the
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denominators for both the slope \eqref{meanfreeslope} and the
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correlation coefficient \eqref{meanfreecorrcoef} equal one. For
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standardized data, both the slope of the regression line and the
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corelation coefficient reduce to the covariance between the $x$ and
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$y$ 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.
|
||||
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
|
||||
@ -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')
|
||||
|
||||
Reference in New Issue
Block a user