Variance of the Mean

Question

Hi I've a question that may seem very basic or may be breaking some math so sorry if i am, just couldn't find the answer online.

$Var(\mu) = E(\mu^2)-E(\mu)^2$

$Var(\mu) = \mu^2 - \mu^2$

$Var(\mu) = 0$

Thanks for any help or clarification.

Answer 1

You’re not finding the variance of a number ($\mu$) unless of course your distribution is constant. Variance is defined as the sum of the squares from the mean or $\sum\limits_{i\in S}(x_i-\mu)^2f(x)=E(x^2)-E(x)$ You are correct in saying that $E(\mu^2)$ is constant and $\mu^2,$ however it does not make any sense for you to calculate like that, you do not find expected value and variances of constants, even if your function was constant, $E(x^2)$ would still equal that constant and not the constant squared. The variance of a constant distribution would obviously be 0.