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Let $X = (X_1, . . . , X_n)$ consist of independent and identically Normal $N(0, θ)$ random variables, with mean $0$ and variance $θ \gt 0$.

The Moment Estimator for $\theta$ is given by $\hat \theta (X)= \frac{1}{n} \sum^n_{i=1}X^2_i$

I would like to compute the mean and variance of the ME. So, by using the help of this question :

$\Bbb E[\frac{1}{n}\sum^n_{i=1}X^2_i]$ $= \Bbb E[\frac{1}{n}] \Bbb E[\sum^n_{i=1}X^2_i]$$ = \frac{1}{n} \frac{n}{2} 2 \theta = \theta$ Since the gamma distribution has mean $k \theta$

And then the variance:

$Var[\frac{1}{n} \sum^n_{i=1}X^2_i]$ = $\frac{1}{n^2} var(\sum^n_{i=1}X^2_i)= \frac{n}{2n^2}4 \theta^2 = \frac{2 \theta^2}{n}$ Since the variance of the gamma distribution is $k \theta^2$. Is this correct?

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    @M.Austin, the variance is $\theta$, not the deviation. – gunes Feb 02 '20 at 19:33
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    Sorry, misread the post. Please disregard my prior comment. It is wrong. Since the variance of your Xs is $\theta$ (with mean = 0), the value $E(X_i^2) = \theta$ for all $i$. So the expected value is $\theta$ as you have written. I am not sure where you get the 2 and 1/2, but I agree with the answer to your first part. – ma83 Feb 02 '20 at 19:34

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They're correct. For the first one, you could also use $E[X_i^2]=\theta$ and obtain the result w/o going into gamma distribution. For the second one, an alternate solution would be using Chi-squared distribution, by defining $Z_i=X_i/\sqrt\theta$, yielding the same result since: $$\operatorname{var}\left(\frac{1}{n}\sum X_i^2\right)=\frac{\theta^2}{n^2}\sum_{i=1}^n \overbrace{\operatorname{var}(Z_i^2)}^2=\frac{2\theta^2}{n}$$

gunes
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