Example to a Maximum Likelihood estimation

Let us take N number of y_i measured datapoint, that are indepedent but were produced by the same process which can be described by a normal distribution $$ \wp \left ( y_{i}\mid\mu ,\sigma \right) =\frac{1}{\sqrt{2\pi \sigma ^{2}}} e^{-\frac{\left ( y_{i}-\mu \right )^{2}}{2\sigma^{2}}}$ with $$ \mu$ expected value, and $$ \sigma$ variance.

The joint probability density function of the independent y_i events is the product of the individual pdf's:
$$ \wp \left ( \mathbf{y}\mid\mu ,\sigma \right) =\prod_{i=1}^{N}\frac{1}{\sqrt{2\pi \sigma ^{2}}} e^{-\frac{\left ( y_{i}-\mu \right )^{2}}{2\sigma^{2}}}=\left [\frac{1}{\sqrt{2\pi \sigma ^{2}}} \right ]^{N} e^{-\sum_{i=1}^{N}\frac{\left ( y_{i}-\mu \right )^{2}}{2\sigma^{2}}}$

The log-likelihood function reads:
$$ \ln \wp \left ( \mathbf{y}\mid\mu ,\sigma \right) =\frac{N}{2}\ln \left ( \frac{1}{2\pi \sigma ^{2}} \right )-\sum_{i=1}^{N}\frac{\left ( y_{i}-\mu \right )^{2}}{2\sigma^{2}}$

We now look for the maximum by differentiating and setting the result equal to zero:
$$ \partial _{\mu } \ln \wp \left ( \mathbf{y}\mid\mu ,\sigma \right) =-\sum_{i=1}^{N}\frac{\left ( y_{i}-\mu \right )}{\sigma^{2}}=0$
from that our estimation for $$ \mu$ is:
$$ \widehat{\mu}=\sum_{i=1}^{N}\frac{ y_{i} }{N}$
the average of the measured data. The partial derivatives according to $$ \sigma$ :
$$ \partial _{\sigma }\ln \wp \left ( \mathbf{y}\mid\mu ,\sigma \right) =-\frac{N}{\sigma}-\sum_{i=1}^{N}\frac{\left ( y_{i}-\mu \right )^{2}}{\sigma^{3}}=0$
from that:
$$ \widehat{\sigma}^{2}=\sum_{i=1}^{N}\frac{\left ( y_{i}-\mu \right )^{2}}{N}$

For completeness we mention that the formula for the variance we still have an unknown $$ \mu$ , and we may get tempted to estimate it by $$\widehat{\mu}$ . As it is well known this would result in a biased estimate, that can be corrected by a factor of N/(N-1).

Now we proceed with estimation theory and the Bayesianestimators.

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