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Let $X_1, X_2, ..., X_n$ be a random distribution such that the mean $\mu = 0$ and the variance $\sigma^2$ is unknown.

I'm finding a constant $c$ such that $$U(X) = c \sum^{n-1}_{i = 1}(X_{i+1} - X_i)^2$$ is an unbiased estimator for $\sigma^2$.

So far I've reached $$E[U(X)] = c \sum^{n-1}_{i=1}E[(X_{i+1} - X_i)^2] $$ where $$E[(X_{i + 1} - X_i)^2] $$ $$= E[(X_{i+1})^2 + (X_i)^2 - 2X_{i + 1}X_i] $$ $$ = E[(X_{i+1})^2] + E[(X_i)^2] - 2E[X_{i+1}]E[X_i]$$

which is where I reach a dead end and am not sure how to continue.

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Assuming the $X_i$ are independent and identically distributed as $X$, then for all $i$, $\mathbb E[ X_i ] = \mathbb E[X] = \mu$ and $\mathbb E [X_i^2] - \mathbb E[X_i]^2 = \mathbb E [X^2] - \mathbb E[X]^2 = \sigma^2$.

Picking up where you left off and using the above, $$ \mathbb E [ X_{i+1}^2 ] + \mathbb E [ X_{i}^2 ] - 2 \mathbb E [ X_{i+1} ] \mathbb E [ X_i ] = 2 \mathbb E[X^2] - 2 \mathbb E[X]^2 = 2 \sigma^2. $$

Can you take it from here?

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  • $\begingroup$ I'm just trying to work out how you went from the term $E[(X_{i + 1})^2] + E[(X_i)^2]$ to the term $E[X_{i + 1}^2] + E[X_{i + 1}^2]$ $\endgroup$
    – Goshawk
    Apr 7 '16 at 14:04
  • $\begingroup$ That was a typo -- it should match what you have. I just fixed it! The point is that, assuming the $X_i$ are iid (did the problem say this?), the subscripts don't matter, and you can replace $X_i$ with $X$, where $X$ is the identical distribution $\endgroup$ Apr 7 '16 at 14:07
  • $\begingroup$ Sorry, yes the $X_i$ are iid. Thanks for the help! I've arrived at C = $\frac{1}{2n - 2}$. Hopefully this is correct. $\endgroup$
    – Goshawk
    Apr 8 '16 at 10:39
  • $\begingroup$ Where was the fact that the mean is 0 used? It seems odd that this information is given but not used. $\endgroup$
    – Goshawk
    Apr 8 '16 at 15:26
  • $\begingroup$ It could have been intended that you use the equation $\sigma^2 = \mathbb E [ (X - \mu)^2 ]$ rather than $\sigma^2 = \mathbb E [X^2] - \mathbb E [X]^2$, and if you use the first equation, knowing $\mu = 0$ would simplify calculations. Howevever, we don't actually need to know $\mu = 0$ as far as I can tell, and we didn't use that fact. $\endgroup$ Apr 8 '16 at 15:38

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