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While doing some statistics exercises I found a question that I don't know how to solve. The question is as follows:

Let $X_{1},...X_{n} \stackrel{iid}{\sim} N(0,\sigma^2)$, where $\sigma^2$ $\in R^{+}$ is unknown and density function:

$f_{X_{i}} (x)= \frac{1}{\sqrt{2 \pi \sigma^2}} e^-\frac{x^2}{2 \sigma^2}$, $x \in R$

a) find an unbiased estimator for $\sigma^2$ based only on the first observation $X_{1}$ and determine its variance.

My thoughts: I know that the first observation (i.e. the smallest) is computed as $X_{1} = n (1-F(x))^{n-1} f(x)$, but I need to compute the integral of the above density function, but is there an easier way? Can you give me some hints?

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    $\begingroup$ Are you sure that $X_1$ refers to the smallest observation? If you say $X_1,\dots,X_n$ are iid, that implies that $X_1$ is not always the smallest. The notation for the smallest observation is something like $X_{(1)}$. $\endgroup$ – Mike Earnest Sep 4 '18 at 20:14
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    $\begingroup$ E(X1^2) = sigma^2 + mu^2 ,here mean = 0 , hence E(X1^2) = sigma^2 , hence X1^2 is unbiased estimator for population variance. you can find variance of X1^2 easily $\endgroup$ – cyberboy Sep 4 '18 at 20:23
  • $\begingroup$ You're right, the notation is different. But then what's the point of the exercise? What is the difference between considering the first observation and any other one? $\endgroup$ – Zhang_anlan Sep 5 '18 at 1:41
  • $\begingroup$ Perhaps the point of the exercise is to show there is more than one unbiased estimator. Later you can show the different estimators can have different variances $\endgroup$ – Henry Sep 5 '18 at 8:06
  • $\begingroup$ The point is that you can use only one observation, it does not matter that it is the first one. $\endgroup$ – Mike Earnest Sep 5 '18 at 15:07
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Consider the estimator $X_1^2$. It is unbiased since $$\mathsf{E}X_1^2=\sigma^2$$ Its variance is also $$\mathsf{Var}(X_1^2)=\mathsf{E}X_1^4-\mathsf{E}^2X_1^2=3\sigma^4-\sigma^4=2\sigma^4$$

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    $\begingroup$ $X_1$ notation doesn't mean the smallest. It implies only one (i.e. the first) observation. $\endgroup$ – msm Sep 4 '18 at 20:33

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