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I am trying to show that the estimator $\frac{n(k-1)S}{(nk-2)}$ is an inconsistent estimator of $\sigma^2$. I must show that the estimator will converge in probability to $\frac{(k-1)\sigma^2}{k}$.

Note that $S = \sum_{i=1}^{n} \sum_{j=1}^{k}$ $\frac{(X_{ij} - \overline{X}_{i})^{2}}{(n(k-1))}$ where $X_{i1},...,X_{ik}$ are iid $\mathcal N(\theta_{i}, \sigma^{2})$.

Edit: For the sake of context, I should note that this is part of a larger problem, and I have reduced the problem to this.

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  • $\begingroup$ Please write explicitly the expression for $\bar X_i$. What does it average? Also, I guess the variables are assumed independent over $i$? $\endgroup$ – Alecos Papadopoulos Feb 19 '17 at 17:14
  • $\begingroup$ Also, confirm that $k$ is fixed and it is only $n$ that goes to infinity? $\endgroup$ – Alecos Papadopoulos Feb 19 '17 at 17:17
  • $\begingroup$ k is fixed. Only n is going to infinity. $ \overline X_{i}$ averages $X_{i1}$ to $X_{ik}$ Everything here is independent. $\endgroup$ – shmiggens Feb 19 '17 at 17:37
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    $\begingroup$ Do you already know that $S$ is a consistent estimator? If so, then you only need to establish that $n(k-1)/(nk-2)$ does not have a limit of $1$ as $n\to\infty$. $\endgroup$ – whuber♦ Feb 19 '17 at 18:16
  • $\begingroup$ I think that is the crux of the problem. I don't know if we can conclude that S is a consistent estimator. $\endgroup$ – shmiggens Feb 19 '17 at 18:21
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I think the key observation is to note that

$$\sum_{j=1}^{k} (X_{i,j} - \bar{X}_i)^2 = \sigma^2 \sum_{j=1}^{k} \Big(\frac{X_{i,j} - \bar{X}_i}{\sigma}\Big)^2 \sim \sigma^2\chi^2(k-1)$$

where $\chi^2(k-1)$ denotes a Chi squared distribution with $k-1$ degrees of freedom.

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  • $\begingroup$ What should I do with that? $\endgroup$ – shmiggens Feb 19 '17 at 18:14
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    $\begingroup$ @shmiggens Try to apply Law of Large Numbers knowing that inner sum follow a Chi squared distribution $\endgroup$ – Łukasz Grad Feb 19 '17 at 18:22
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I believe I have figured out the solution, and it was staring me in the face the whole time. Lukasz Grad got me thinking that I needed to apply the Strong Law of Large Numbers.

Note that $S = \sum_{i=1}^{n} \sum_{j=1}^{k}$ $\frac{(X_{ij} - \overline{X}_{i})^{2}}{(n(k-1))} = \sum_{i=1}^{n}$ $ \frac{S_{i}^{2}}n$, where $S^{2}_{i} = $$\sum_{j=1}^{k}$ $\frac{(X_{ij} - \overline{X}_{i})^{2}}{k-1}$

We know that $E[S^{2}_{i}] = \sigma^{2}$. Then by the Strong Law of Large numbers, S converges to $\sigma^{2}$ almost surely. The solution then follows.

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