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Suppose $X_1, X_2, . . . , X_6$ is a random sample of a normal random variable with mean µ and variance $σ^2$ .

Determine c such that c[$(X_1 − X_2)^2$ + $(X_3 − X_4)^2$ + $(X_5 − X_6)^2$] is an unbiased estimator of $σ^2$.

I've attempted to get the second moments of the $(X_i − X_j)$'s and then to use these and solve for c

So far I managed to find $E[X_i − X_j]$ = µ-µ = 0 and $Var[X_i − X_j]$ = $Var[X_i]$ + $Var[X_j]$ - 2$Cov[X_i,X_j]$ = 2$σ^2$- 2$Cov[X_i,X_j]$

Thus $E[(X_i − X_j)^2]$ = 2$σ^2$- 2$Cov[X_i,X_j]$

or

$E[(X_i − X_j)^2]$ = 2$σ^2$+2$µ^2$ - $E[X_iX_j]$

Does anyone have any hints on how to progress from here, there's no mention in the question that the sample is i.i.d

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  • $\begingroup$ It's most likely the case the person who posed the question is assuming independence. Without it, you could vary one of the covariances (say $\mbox{Cov}(X_{1},X_{2})$, while leaving the rest independent and thus vary the estimate of $\sigma^{2}$. $\endgroup$ – Brian Borchers Feb 17 '18 at 20:25
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    $\begingroup$ It says 'random sample', so you can reasonably assume the $X_i$ are independent. $\endgroup$ – BruceET Feb 17 '18 at 21:49
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Let's compute $$\mathbb E[(X_1 − X_2)^2] = \mathbb E[X_1^2] + \mathbb E[X_2^2] - \mathbb 2E[X_1 X_2] = \mathbb E[X_1^2] + \mathbb E[X_2^2] - 2\mathbb E[X_1]\mathbb E[X_2] = 2(\sigma^2 + \mu^2) -2\mu^2 = 2\sigma^2.$$ We thus get $$c\cdot \mathbb E[(X_1 − X_2)^2 + (X_3 − X_4)^2 + (X_5 − X_6)^2)] = 3c\mathbb E[(X_1 − X_2)^2] = 6c\sigma^2 .$$

We need $$6c\sigma^2 = \sigma^2 \implies c= \frac 1 6.$$

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