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
$E\left ( \bf \mu_{ml} \right ) = \sum_{??}^{??} \left ( \mu_{ml} \right ) \cdot \left [ p \left ( {\bf x} | \mu_{ml},\sigma^2 \right ) \right ]$.
Best Answer
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 $E[\mu_{ml}] = \frac{1}{N}\cdot N \cdot \mu = \mu$