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In the process of finding unknown parameters $\mu$(mean) and $\sigma^2$(variance) for the maximum likelihood function for Gaussian distribution, we arrive at the following equation.

$\begin{align*} &\ln\left [ \bf p\left ( x | \mu,\sigma ^2 \right ) \right ] = -\frac{N}{2}\ln (2 \pi) - \frac{N}{2}\ln(\sigma ^2) - \frac{1}{2\cdot \sigma ^2} \left \{ \sum_{n=1}^{N}(x_n - \mu)^2 \right \} \\ &\text{where } { \bf x } = \left ( x_1,x_2,x_3 \dots x_N \right )^T \text{ is a data set of observation. } \\ \end{align*}$

Now by differentiating and maximizing with respect to $\mu$ we get

\begin{align*} {\bf \mu_{ml}} = \frac{1}{N}\sum_{n=1}^{N}x_n \\ \end{align*}

Now how to evaluate Expected value of $\bf\mu_{ml}$ ? I am not able to apply summation rule for expected value. I tried $E\left ( \bf \mu_{ml} \right ) = \sum_{??}^{??} \left ( \mu_{ml} \right ) \cdot \left [ p \left ( {\bf x} | \mu_{ml},\sigma^2 \right ) \right ]$.

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    $\begingroup$ The equivalent to summation for continuous distributions is integration. $\endgroup$
    – Björn
    Jul 29, 2017 at 6:14
  • $\begingroup$ Thanks ! But what is the domain of integration? $\endgroup$
    – Debashish
    Jul 29, 2017 at 6:37
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    $\begingroup$ What is the domain of the random variable? $\endgroup$
    – Björn
    Jul 29, 2017 at 6:51
  • $\begingroup$ OK. That means it is same as the domain of the random variable $x$. It is not explicitly mentioned in the text about the domain of $x$. However the answer to $E(\mu_{ml})$ is given as $\mu$. Could you please help a little bit how they derived? $\endgroup$
    – Debashish
    Jul 29, 2017 at 6:57

1 Answer 1

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$ \begin{align*} &\text{Using} \; \\ &E\left [ X+Y \right ] =E[X]+E[Y] \\ &E[\mu_{\text{ml}}] = \frac{1}{N}\sum_{n = 1}^{N}E(x_n) \end{align*} $

Since each $x_i$ is a Random Variable picked uniformly from a Gaussian distribution, the expected value of each is just the mean of the distribution. That is $E[x_i] = \mu$ for each $x_i$.

Thus $E[\mu_{ml}] = \frac{1}{N}\cdot N \cdot \mu = \mu$

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