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Suppose that each measurement $X_i$ can be seen as $X_i = μ + U_i$, where the $U_i$ are i.i.d. $N(0, σ^2)$ random variables.
Consider the estimator $T=\frac{1}{20}\sum_{i=1}^{20}U_i^2$ for $σ^2$.
Calculate the MSE(T).
Hint: use that $E[U_i^4]=3σ^4$.

Solve:

$MSE(T)=Var(T)+(E[T]-\sigma^2)^2$

$E[T]=\frac{1}{20}\cdot 20 \cdot E[U_i^2]=E[U_i^2]=\sigma^2$

$Var(T)=\frac{1}{20^2} \cdot 20 \cdot Var(U_i^2)= \frac{Var(U_i^2)}{20}$

$Var(U_i^2)=E[U_i^4]-(E[U_i^2])^2=3σ^4-\sigma^4=2σ^4$

So $Var(T)=\frac{σ^4}{10}$ and $MSE(T)=\frac{σ^4}{10}+(\sigma^2-\sigma^2)^2=\frac{σ^4}{10}$

Is this the correct way to solve it? My only doubt is when I compute $E[T]$, I don't know if I have to assume it equal to zero since the distribution is $N=(0,\sigma^2)$ or maybe just assume that $T$ is un unbiased estimator for $\sigma^2$ (what I did)?

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    $\begingroup$ Same question was asked sometime back (now deleted) and surprisingly with the same formatting as far as I remember. $\endgroup$ – StubbornAtom Jan 27 at 16:19
  • $\begingroup$ Yes, it was a friend of mine (we are doing the same course and we are struggling with the same question) and since it was posted in a wrong moment (just a few views and no answer) we decided to repost it with my account and see if someone can help us! :) $\endgroup$ – Mark Jacon Jan 27 at 16:22
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    $\begingroup$ We don't draw attention to our questions by deleting previous copies and posting the same thing again (within a few hours in this case). Why not wait for someone to answer? $\endgroup$ – StubbornAtom Jan 27 at 16:28
  • $\begingroup$ Ah ok sorry, now I know it! $\endgroup$ – Mark Jacon Jan 27 at 16:29
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    $\begingroup$ @MarkJacon Do you have two accounts? $\endgroup$ – callculus Jan 27 at 16:32
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The calculation is correct. It is a fact that $T$ is an unbiased estimator of $\sigma^2$ and not something you need to assume.

The situation often seen where one has to divide by $n-1$ to have an unbiased estimator for the variance is when the data is first used to estimate the unknown population mean. However, that is not the case here.

Here it is given that $U_i$ has mean zero. Namely, $U_i = \sigma Z_i$ where $Z_i$ are iid standard Normal such that each $Z_i^2$ is a Chi-square distribution with $1$ degree of freedom and $\sum_{i=1}^n Z_i$ is a Chi-sq\ with $n$ degree of freedom (instead of the reduced $n-1$).

The mean of a standard Chi-sq is $1$ as in $E[Z_i] = 1$. Therefore, $T = \frac1n \sum U_i^2$ has mean $$E[T] = \frac1n \sum_{i=1}^n\left( \sigma^2 E[Z_i^2] \right) = \frac1n \sigma^2 \sum_{i=1}^n 1 = \sigma^2$$

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  • $\begingroup$ If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you. $\endgroup$ – Lee David Chung Lin Feb 7 at 8:16

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