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I'm trying to show whether or not $\bar(Y^2)$ = $\\µ^2$ Or the mean of the sample squared) is a biased or unbiased estimator of the population mean squared.

I can prove that Ybar is an unbiased estimator of the population mean, but it's not clear how to prove the same for Ybar squared.

So I have something like:

$E[\bar(Y)^2]$ = $\frac{1}{N}E[\sum_{i=1}^{n}(Y_i)^2]$

I'm wondering where to go from here. I can swap the sum and the expected value components, but it doesn't seem to simplify anything.

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    $\begingroup$ What happens if $\mu$ is $0$? Can you envision any circumstances that would would result in $E\left[\bar{Y}^2\right]$, the expected value of a nonnegative quantity, having value $0$? $\endgroup$ – Dilip Sarwate Oct 20 '15 at 16:38
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    $\begingroup$ No, I can't. I used that sort of reasoning to figure out that the estimate is biased, but I'm not sure how to formally prove it outside of the example given in the answer to this question. $\endgroup$ – Parseltongue Oct 20 '15 at 16:50
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The problem is, that $E[Y^2] \neq \mu^2$ - take $Y$, such that $$P(Y=-1)=P(Y=1)=0.5,$$ clearly $EY=0$, but $E[Y^2]=1$.

From this example you can can see that $\frac{1}{N}\sum_{i=1}^{n}(Y_i)^2$ does not estimate $\mu^2$.

Edit: You've (nearly) proved in your edit that $\frac{1}{N}\sum_{i=1}^{n}(Y_i)^2$ is unbiased estimate of $E[Y^2]$. But then once again $E[Y^2] \neq (EY)^2.$

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    $\begingroup$ Ah, I see your point. I'm just wondering if there was a moral formal proof, starting with the summation definition of Ybar squared, to show that it doesn't simplify to mu squared. $\endgroup$ – Parseltongue Oct 20 '15 at 17:16
  • $\begingroup$ Yes, you have proved what the sum of $\frac{1}{N}\sum_{i=1}^{n}(Y_i)^2$ is and you now that $\mu^2$ differs from this sum, thats a formal proof. $\endgroup$ – iiivooo Oct 20 '15 at 17:26
  • $\begingroup$ I believe OP wanted to know what the expectation of Ybar squared actually ended up being, which is addressed by Dilip Sarwate's answer here. $\endgroup$ – Abraham Philip Sep 10 '17 at 21:13
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Suppose that $\bar{Y} = \frac 1n \sum_{i=1}^n Y_i$ where the $Y_i$ are independent random variables with the same mean $\mu$ and (positive) variance $\sigma^2$. The $Y_i$ need not be identically distributed but you can make this assumption if it makes you feel more comfortable.

Then, $\bar{Y}$ is a random variable with mean $\mu$ and variance $\frac{\sigma^2}{n}$ and so $\displaystyle E\left[\left(\bar{Y}\right)^2\right] = \mu^2 + \frac{\sigma^2}{n} > \mu^2.$

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