Given $N$ points ($x_k$, k from $1$ to $N$) generated from a normal distribution (1-dimensional case) with known mean $\mu$, the Maximum Likelihood estimation of the variance is $\frac{1}{N}\sum_{k=1}^{\infty} (x_k-\mu)^2$.
How is it justified that this estimation is biased for finite N?
thanks, Nikos
On
The "bias for finite N" is for a different estimator where $\mu$ is replaced by $\overline{x} = \frac{x_1 + \cdots + x_N}{N}$. If the "true" value of $\mu$ is given the estimator in the question is unbiased.
This was the subject of an earlier question,
Intuitive Explanation of Bessel's Correction
and in my answer there it is explained why the calculation with $\mu$ is unbiased.
Filling in some details left out of the comments by @Michael Hardy, @zyx, and myself, suppose $\vec{X} = (X_1, \ldots, X_N)$ where the $X_i, 1 \leq i \leq N,$ are independent $N(\mu, v)$ random variables with known mean $\mu$ and unknown variance $v$. The joint density function is $$f_{\vec{X}}(\vec{x}) = (2\pi v)^{-N/2}\exp(-a/v) ~\text{where}~ \vec{x} = (x_1, \ldots, x_N)~ \text{and}~ a = \frac{1}{2} \sum_{i=1}^N (x_i - \mu)^2.$$ If $\vec{X}$ is observed to have value $\vec{x}$, the likelihood function is $L(v) = (2\pi v)^{-N/2}\exp(-a/v)$ and it is easy to show that $L(v)$ attains maximum value at $$v = \frac{2a}{N} = \frac{1}{N}\sum_{i=1}^N (x_i - \mu)^2$$ and so the maximum likelihood estimator (MLE) of $v$ is $\frac{1}{N}\sum_{i=1}^N (x_i - \mu)^2$. We have $$ E\left [ \frac{1}{N}\sum_{i=1}^N (X_i - \mu)^2 \right ] = \frac{1}{N}\sum_{i=1}^N E[(X_i - \mu)^2] = \frac{1}{N}\sum_{i=1}^N v = v, $$ and thus, contrary to any alleged claims in an unspecified textbook in the possession of Nikos, the MLE for $v$ is unbiased in this instance.
What if $\mu$ is also unknown? The likelihood function is now $$ L(\mu, v) = (2\pi v)^{-N/2}\exp\left [ -\frac{1}{2v}\sum_{i=1}^N (x_i - \mu)^2\right ] $$ and has a global maximum at $$ (\mu, v) = \left (\frac{1}{N}\sum_{i=1}^N x_i, \frac{1}{N}\sum_{i=1}^N \left (x_i -\frac{1}{N}\sum_{i=1}^N x_i \right)^2 \right ) = \left ( \bar{x}, \frac{1}{N}\sum_{i=1}^N (x_i -\bar{x})^2 \right ).$$ The MLE for $v$ is thus $\frac{1}{N}\sum_{i=1}^N (x_i -\bar{x})^2$ and is biased since $$ E\left [ \frac{1}{N}\sum_{i=1}^N \left (X_i -\frac{1}{N}\sum_{i=1}^N X_i \right)^2 \right ] = \left(\frac{N-1}{N}\right)v $$ but, as noted by Nikos's textbook, the MLE for $v$ is asymptotically unbiased in the limit as $N \to \infty$. On the other hand, it should be obvious from the above description that $\frac{1}{N-1}\sum_{i=1}^N (x_i -\bar{x})^2$ is an unbiased estimator for $v$ for all $N \geq 2$.