429 lines
22 KiB
TeX
429 lines
22 KiB
TeX
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\chapter{Maximum likelihood estimation}
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\label{maximumlikelihoodchapter}
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\exercisechapter{Maximum likelihood estimation}
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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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\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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\section{Maximum likelihood}
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Let $p(x|\theta)$ (to be read as ``probability(density) of $x$ given
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$\theta$.'') the probability (density) distribution of data value $x$
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given parameter values $\theta$. This could be the normal distribution
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\begin{equation}
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\label{normpdfmean}
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p(x|\mu, \sigma) = \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
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\end{equation}
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defined by the mean $\mu$ and the standard deviation $\sigma$ as
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parameters $\theta$. If the $n$ observations $x_1, x_2, \ldots, x_n$
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are independent of each other and originate from the same probability
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density distribution (they are \enterm[i.i.d.|see{independent and
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identically distributed}]{i.i.d.}, \enterm{independent and
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identically distributed}), then the conditional probability
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$p(x_1,x_2, \ldots, x_n|\theta)$ of observing the particular data
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values $x_1, x_2, \ldots, x_n$ given some specific parameter values
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$\theta$ of the probability density is given by the product of the
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probability densities of each data value:
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\begin{equation}
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\label{probdata}
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p(x_1,x_2, \ldots, x_n|\theta) = p(x_1|\theta) \cdot p(x_2|\theta)
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\ldots, p(x_n|\theta) = \prod_{i=1}^n p(x_i|\theta) \; .
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\end{equation}
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Vice versa, the \entermde{Likelihood}{likelihood} of the parameters $\theta$
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given the observed data $x_1, x_2, \ldots, x_n$ is
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\begin{equation}
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\label{likelihood}
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{\cal L}(\theta|x_1,x_2, \ldots, x_n) = p(x_1,x_2, \ldots, x_n|\theta) \; .
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\end{equation}
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Note, that the likelihood ${\cal L}$ is not a probability in the
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classic sense since it does not integrate to unity ($\int {\cal
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L}(\theta|x_1,x_2, \ldots, x_n) \, d\theta \ne 1$). For given
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observations $x_1, x_2, \ldots, x_n$, the likelihood
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\eqref{likelihood} is a function of the parameters $\theta$. This
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function has a global maximum for some specific parameter values. At
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this maximum the probability \eqref{probdata} to observe the measured
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data values is the largest.
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Maximum likelihood estimators just find the parameter values
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\begin{equation}
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\theta_{mle} = \text{argmax}_{\theta} {\cal L}(\theta|x_1,x_2, \ldots, x_n)
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\end{equation}
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that maximize the likelihood \eqref{likelihood}.
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$\text{argmax}_xf(x)$ is the value of the argument $x$ for which the
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function $f(x)$ assumes its global maximum. Thus, we search for the
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parameter values $\theta$ at which the likelihood ${\cal L}(\theta)$
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reaches its maximum. For these parameter values the measured data most
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likely originated from the corresponding distribution.
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The position of a function's maximum does not change when the values
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of the function are transformed by a strictly monotonously rising
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function such as the logarithm. For numerical reasons and reasons that
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we discuss below, we instead search for the maximum of the logarithm
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of the likelihood
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(\entermde[likelihood!log-]{Likelihood!Log-}{log-likelihood})
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\begin{eqnarray}
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\theta_{mle} & = & \text{argmax}_{\theta}\; {\cal L}(\theta|x_1,x_2, \ldots, x_n) \nonumber \\
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& = & \text{argmax}_{\theta}\; \log {\cal L}(\theta|x_1,x_2, \ldots, x_n) \nonumber \\
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& = & \text{argmax}_{\theta}\; \log \prod_{i=1}^n p(x_i|\theta) \nonumber \\
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& = & \text{argmax}_{\theta}\; \sum_{i=1}^n \log p(x_i|\theta) \label{loglikelihood}
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\end{eqnarray}
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which is the sum of the logarithms of the probabilites of each
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observation. Let's illustrate the concept of maximum likelihood
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estimation on the arithmetic mean.
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\subsection{Arithmetic mean}
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Suppose that the measurements $x_1, x_2, \ldots, x_n$ originate from a
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normal distribution \eqref{normpdfmean} and we do not know the
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population mean $\mu$ of the normal distribution
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(\figrefb{mlemeanfig}). In this setting $\mu$ is the only parameter
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$\theta$. Which value of $\mu$ maximizes the likelihood of the data?
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\begin{figure}[t]
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\includegraphics[width=1\textwidth]{mlemean}
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\titlecaption{\label{mlemeanfig} Maximum likelihood estimation of
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the mean.}{Top: The measured data (blue dots) together with three
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normal distributions differing in their means (arrows) from which
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the data could have originated from. Bottom left: the likelihood
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as a function of the parameter $\mu$. For the data it is maximal
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at a value of $\mu = 2$. Bottom right: the log-likelihood. Taking
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the logarithm does not change the position of the maximum.}
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\end{figure}
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With the normal distribution \eqref{normpdfmean} and applying
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logarithmic identities, the log-likelihood \eqref{loglikelihood} reads
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\begin{eqnarray}
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\log {\cal L}(\mu|x_1,x_2, \ldots, x_n)
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& = & \sum_{i=1}^n \log \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x_i-\mu)^2}{2\sigma^2}} \nonumber \\
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& = & \sum_{i=1}^n - \log \sqrt{2\pi \sigma^2} -\frac{(x_i-\mu)^2}{2\sigma^2} \; .
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\end{eqnarray}
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Since the logarithm is the inverse function of the exponential
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($\log(e^x)=x$), taking the logarithm removes the exponential from the
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normal distribution. This is the second reason why it is useful to
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maximize the log-likelihood. To calculate the maximum of the
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log-likelihood, we need to take the derivative with respect to $\mu$
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and set it to zero:
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\begin{eqnarray}
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\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 \\
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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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\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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arithmetic mean maximizes the likelihood that the data originate from
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a normal distribution centered at the arithmetic mean
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(\figref{mlemeanfig}). Equivalently, the standard deviation computed
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from the data, maximizes the likelihood that the data were generated
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from a normal distribution with this standard deviation.
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\begin{exercise}{mlemean.m}{mlemean.out}
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Draw $n=50$ random numbers from a normal distribution with a mean of
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$\ne 0$ and a standard deviation of $\ne 1$.
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Plot the likelihood (the product of the probabilities) and the
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log-likelihood (given by the sum of the logarithms of the
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probabilities) for the mean as parameter. Compare the position of
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the maxima with the mean calculated from the data.
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\end{exercise}
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Comparing the values of the likelihood with the ones of the
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log-likelihood shown in \figref{mlemeanfig}, shows the numerical
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reason for taking the logarithm of the likelihood. The likelihood
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values can get very small, because we multiply many, potentially small
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probability densities with each other. The likelihood quickly gets
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smaller than the samlles number a floating point number of a computer
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can represent. Try it by increasing the number of data values in the
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exercise. Taking the logarithm avoids this problem. The log-likelihood
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assumes well behaving numbers that can be handled well by the
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computer.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Fitting probability distributions}
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Consider normally distributed data with unknown mean and standard
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deviation. From the considerations above we just have seen that a
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Gaussian distribution with mean at the arithmetic mean and standard
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deviation equal to the standard deviation computed from the data is
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the Gaussian that fits the data best in a maximum likelihood sense,
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i.e. the likelihood that the data were generated from this
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distribution is the largest. Fitting a Gaussian distribution to data
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is very simple: just compute the two parameter of the Gaussian
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distribution $\mu$ and $\sigma$ as the arithmetic mean and a standard
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deviation, respectively, directly from the data.
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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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\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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\begin{figure}[t]
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\includegraphics[width=1\textwidth]{mlepdf}
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\titlecaption{\label{mlepdffig} Maximum likelihood estimation of a
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probability density.}{Left: the 100 data points drawn from a 2nd
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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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\end{figure}
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Instead we should stay with maximum-likelihood estimation. Exactly in
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the same way as we estimated the mean value of a Gaussian distribution
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above, we can numerically fit the parameter of any type of
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distribution directly from the data by means of maximizing the
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likelihood. We simply search for the parameter values of the desired
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probability density function that maximize the log-likelihood. In
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general this is a non-linear optimization problem that is solved with
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numerical methods such as the gradient descent \matlabfun{mle()}.
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\begin{exercise}{mlegammafit.m}{mlegammafit.out}
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Generate a sample of gamma-distributed random numbers and apply the
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maximum likelihood method to estimate the parameters of the gamma
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function from the data.
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\end{exercise}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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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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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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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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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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\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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\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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\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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$\theta$ 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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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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\begin{figure}[t]
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\includegraphics[width=1\textwidth]{mlepropline}
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\titlecaption{\label{mleproplinefig} Maximum likelihood estimation
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of the slope of a line through the origin.}{The data (blue and
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left histogram) originate from a straight line $y=mx$ trough the origin
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(red). The maximum-likelihood estimation of the slope $m$ of the
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regression line (orange), \eqnref{mleslope}, is close to the true
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one. The residuals, the data minus the estimated line (right), reveal
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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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\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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\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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\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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\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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necessary for fitting the slope of a straight line, because the slope
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can be directly computed via \eqnref{mleslope}. More generally, this
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is the case also for fitting the coefficients of linearly combined
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basis functions as for example the slope $m$ and the y-intercept $b$
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of a straight line
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\[ y = m \cdot x +b \]
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or the coefficients $a_k$ of a polynomial
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\[ y = \sum_{k=0}^N a_k x^k = a_o + a_1x + a_2x^2 + a_3x^4 + \ldots \]
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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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\[ y = c \cdot e^{\lambda x} \quad , \quad c, \lambda \in \reZ \; . \]
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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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\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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\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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\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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\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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\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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\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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\begin{equation}
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\theta = \frac{1}{n} \sum_{i=1}^n x_i y_i = \text{cov}(x,y) = r_{x,y}
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\end{equation}
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For standardized data the slope of the regression line is the same as the
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correlation coefficient!
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Neural coding}
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In sensory systems certain aspects of the environment are encoded in
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the neuronal activity of populations of neurons. One example of such a
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population code is the tuning of neurons in the primary visual cortex
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(V1) to the orientation of an edge or bar in the visual
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stimulus. Different neurons respond best to different edge
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orientations. Traditionally, such a tuning is measured by analyzing
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the neuronal response strength (e.g. the firing rate) as a function of
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the orientation of a black bar and is illustrated and summarized
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with the so called \enterm{tuning-curve} (\determ{Abstimmkurve},
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figure~\ref{mlecodingfig}, top).
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\begin{figure}[tp]
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\includegraphics[width=1\textwidth]{mlecoding}
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\titlecaption{\label{mlecodingfig} Maximum likelihood estimation of
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a stimulus parameter from neuronal activity.}{Top: Tuning curve of
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an individual neuron as a function of the stimulus orientation (a
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dark bar in front of a white background). The stimulus that evokes
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the strongest activity in that neuron is the bar with the vertical
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orientation (arrow, $\phi_i=90$\,\degree). The red area indicates
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the variability of the neuronal activity $p(r|\phi)$ around the
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tuning curve. Center: In a population of neurons, each neuron may
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have a different tuning curve (colors). A specific stimulus (the
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vertical bar) activates the individual neurons of the population
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in a specific way (dots). Bottom: The log-likelihood of the
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activity pattern will be maximized close to the real stimulus
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orientation.}
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\end{figure}
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The brain, however, is confronted with the inverse problem: given a
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certain activity pattern in the neuronal population, what is the
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stimulus (here the orientation of an edge)? In the sense of maximum
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likelihood, a possible answer to this question would be: the stimulus
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for which the particular activity pattern is most likely given the
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tuning of the neurons.
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Let's stay with the example of the orientation tuning in V1. The
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tuning $\Omega_i(\phi)$ of the neurons $i$ to the preferred edge
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orientation $\phi_i$ can be well described using a van-Mises function
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(the Gaussian function on a cyclic x-axis) (\figref{mlecodingfig}):
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\begin{equation}
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\Omega_i(\phi) = c \cdot e^{\cos(2(\phi-\phi_i))} \quad , \quad c \in \reZ
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\end{equation}
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If we approximate the neuronal activity by a normal distribution
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around the tuning curve with a standard deviation $\sigma=\Omega/4$,
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which is proportional to $\Omega$, then the probability $p_i(r|\phi)$
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of the $i$-th neuron showing the activity $r$ given a certain
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orientation $\phi$ of an edge is given by
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\begin{equation}
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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} \; .
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\end{equation}
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The log-likelihood of the edge orientation $\phi$ given the
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activity pattern in the population $r_1$, $r_2$, ... $r_n$ is thus
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\begin{equation}
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{\cal L}(\phi|r_1, r_2, \ldots, r_n) = \sum_{i=1}^n \log p_i(r_i|\phi)
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\end{equation}
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The angle $\phi$ that maximizes this likelihood is then an estimate of
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the orientation of the edge.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\printsolutions
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