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Let $X_1,...,X_n\sim\mathcal{N}(0,\sigma^2)$. I know that the sample variance $$\hat{\sigma}^2=\frac{1}{n-1}\sum_{i=0}^{n}X_{i}^2$$ of this data is the UMVUE for $\sigma^2$.

But what if we have only a single data point? $\hat{\sigma}^2$ is not defined because of the $n-1 $ factor in the denominator. I'm wondering whether there exists some other unbiased estimator for $\sigma^2$ in this case.

It might seem like a weird question, but I think I've constructed such an estimator for the standard deviation $\sigma$. We know that $E(|X_1|)=\sigma\sqrt{2/\pi}$, so rearranging gives $E(|X_1|\sqrt{\pi/2})=\sigma$.

Theo Walker
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  • UMVUE of $\sigma^2$ based on a sample of size $n(\ge 1)$ is $\frac1n\sum_{i=1}^n X_i^2$. The other unbiased estimator is fine. – StubbornAtom Jan 16 '21 at 05:37

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Dividing by $n-1$ instead of $n$ is called Bessel's correction. You should only be using Bessel's correction when you are also subtracting by the sample mean. If you already know the population mean and the sample mean is unnecessary then you shouldn't be using the correction (see the Wikipedia page).

In the case where you didn't know the population mean you'd need to subtract the sample mean which in the one data point case would be that single data point. You would then end up with $0/0$ which is undefined and that makes sense intuitively since one point can't tell you anything about the variability of the population.

Pedro Amaral
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