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Let $X_1,..., X_n$ be a random sample. Then one version of the sample variance formula is

$$S^2 = \frac{1}{2n(n-1)} \Sigma_{i=1}^n \Sigma_{j=1}^n (X_i - X_j)^2$$

Then suppose $n = 4$ and $E(X_i)=0$, I know $S^2 = \frac{1}{24} \Sigma_{i=1}^n \Sigma_{j=1}^n (X_i - X_j)^2$

But one reference said $E(S^2) = 24 E(X_i)^2$

I do not see how to derive the above.

Jonathen
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  • what is the reference ? – WeakLearner Nov 14 '22 at 03:30
  • The question is originally from Casella's Statistical inference Exercise 5.8 (b). There is an online solution available. Maybe the solution has typo or is wrong. It happens sometimes. – Jonathen Nov 14 '22 at 03:32
  • $E(S^2) = 24 E(X_i)^2$ does not make sense. Since it is related to the second central moment, it should not be affected by the location of the $X_i$. – Henry Nov 14 '22 at 08:19
  • @Henry So what should it be in terms of the second central moment? In this simpler case, we assume E(X)=0 – Jonathen Nov 14 '22 at 13:00

1 Answers1

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If $i \not = j$ then $\mathbb E\left[(X_i-X_j)^2\right] = 2 \sigma^2$ so $E\left[S^2\right] = \mathbb E\left[\frac{1}{2n(n-1)} \sum\limits_{i=1}^n \sum\limits_{j=1}^n (X_i - X_j)^2\right] = \sigma^2$, which is why this is a suggested measure of the sample variance.

$\mathbb E\left[X_i^2\right] = \sigma^2 + \left(E\left[X_i\right]\right)^2$, so if $E\left[X_i\right] = 0$ then you would have $\mathbb E\left[X_i^2\right] =\mathbb E\left[\frac1n \sum\limits_{i=1}^n X_i^2\right]= \sigma^2= \mathbb E\left[S^2\right]$.

But $\frac1n \sum\limits_{i=1}^n X_i^2$ is not the sample variance: it is a measure of the second moment about $0$ rather than the second moment about the sample mean.

Henry
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  • but here we assume sample mean is zero. So it is also a measure of the second moment about the sample mean, – Jonathen Nov 14 '22 at 13:40