Let $\{ X_i | i = 1, 2, . . . ,n \}$ be a sequence of independent and identically distributed (IID) random variables from a population and define $\mu \equiv \mathbb{E}(X)$ and $\sigma^2 \equiv \mathbb{V}(X)$. Suppose we think that the mean is $\mu = \mu_0$ for some number $\mu_0$ (but we may be wrong). Find the bias in the estimator:

$$\tilde{\sigma}^2 \equiv \frac{1}{n} \sum_{i=1}^n (X_i - \mu_0)^2,$$

as a function of $\mu$. When is this estimator unbiased for $\sigma^2$?

I know that when we find this estimator using $\bar{X}$, it is biased and we must divide it by $n-1$ instead of $n$ to get an unbiased estimator. But how does one do it for a number such as $\mu_0$?

  • $\begingroup$ Use the definition of bias. This looks like it should be marked as self-study. $\endgroup$
    – Glen_b
    Commented Nov 6, 2018 at 23:45

1 Answer 1


$\begin{align}E(\hat\sigma^2)&=E(\frac 1n(\sum_{i=1}^n (X_i-\mu_0)^2)\\ &=E(\frac 1n(\sum_{i=1}^n (X_i-\mu + \mu -\mu_0)^2) \\ &= E(\frac 1n(\sum_{i=1}^n (X_i-\mu)^2 + (\mu -\mu_0)^2 +2((X_i-\mu)(\mu -\mu_0))) \\ &= \frac 1n \left\{E(\sum_{i=1}^n (X_i-\mu)^2) + nE((\mu -\mu_0)^2) +2E(\sum_{i=1}^n(X_i-\mu)(\mu -\mu_0))\right \}\\ &= \sigma^2 + (\mu -\mu_0)^2 \end{align}$

So the bias is $(\mu -\mu_0)^2 $. When $\mu_0 = \mu$, the estimator is unbiased.

  • $\begingroup$ I think so. My only question is E((mu - mu0)^2) is being treated as a constant here. Is that right? Or should it be Var(mu0)? $\endgroup$ Commented Nov 6, 2018 at 18:19
  • $\begingroup$ @Moss Murderer: No, the answer here is correct. The use of Bessel's correction only applies when you measure the sample variance, taken around the sample mean instead of the true mean. $\endgroup$
    – Ben
    Commented Nov 6, 2018 at 23:35
  • $\begingroup$ Please try to avoid just doing people's homework for them. $\endgroup$
    – Glen_b
    Commented Nov 6, 2018 at 23:46
  • $\begingroup$ @Ben Yes you're right, I did not realise that $\mu$ denotes population mean. $\endgroup$ Commented Nov 7, 2018 at 0:22

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