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Here's the question again: what is the expected value of $Z^2$ when $Z\backsim \frac{n+1}{n}Y$ and $\displaystyle Y\backsim f(y\mid\theta )=\frac{n}{\theta}\left(\frac y \theta \right)^{n-1}$ from $0$ to $\theta$

I know the set up of the problem is to make it $$\int_0^\theta z^2f(z\mid \theta) \, dz $$ but the problem is how to figure out the density for z is. I also do not know what the variance is, so I cannot use that formula to figure that out.

Michael Hardy
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David Kang
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    The density of Z is not needed. // In the PDF of Y, is n=3? – Did Oct 31 '16 at 21:46
  • n is the number of iid random variables $X_i \backsim Unif[0,\theta]$ that we are trying to estimate an MLE for, so it is unknown and not relevant I don't think. (the MLE turns out to be biased, so the $\frac{n+1}{n}$ is to make it biased, and we are now looking for the variance for the unbiased multiple of the MLE – David Kang Oct 31 '16 at 21:50
  • @MichaelHardy you are right, I apologize. I misread my own handwriting. I have fixed it in the OP – David Kang Oct 31 '16 at 22:10

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$$ \operatorname{E} \left( \left( \frac {n+1} n Y \right)^2 \right) = \left( \frac {n+1} n \right)^2 \operatorname{E}(Y^2). $$ So you need to find $\operatorname{E}(Y^2)$.

\begin{align} \operatorname{E}(Y^2) & = \int_0^\theta y^2 \Big( \frac n \theta \left( \frac y \theta\right)^{n-1} \, dy\Big) = \theta^2 \int_0^\theta \left( \frac y \theta\right)^2 n\left( \frac y \theta \right)^{n-1} \frac{dy} \theta \\[10pt] & = \theta^2 \int_0^1 u^2\cdot nu^{n-1} \, du = \theta^2 \frac n {n+2}. \end{align}

PS: If you're ultimately looking for $\operatorname{var}\left( \dfrac{n+1} n Y \right)$, you will have \begin{align} & \operatorname{var}\left( \frac {n+1} n Y \right) = \left( \frac {n+1} n \right)^2 \operatorname{var}(Y) = \left( \frac {n+1} n \right)^2 \Big( \operatorname{E} (Y^2) - \left( \operatorname{E}(Y) \right)^2 \Big) \\[10pt] = {} & \left( \frac{n+1} n \right)^2 \left( \theta^2 \left( \frac n {n+2} \right) - \theta^2 \left(\frac n {n+1}\right)^2 \right) = \cdots \end{align} ok, so far --- what is just above is routine. The next part is trivial and thus you may think it doesn't need much attention, but the answer is somewhat enlightening: If I'm not mistaken, it comes down to this: $$ \frac {\theta^2}{n(n+2)}. $$

Michael Hardy
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  • That seems correct based on the information that I have provided, and the variance of Z, which I'm looking for, turns out to be $$\theta^2(\frac{(n+1)^2}{n(n+2)}-1)$$ which seems pretty reasonable. Thank you – David Kang Oct 31 '16 at 22:28
  • You'd better look at my postscript. As I said, "The next part is trivial and thus you may think it doesn't need much attention". Maybe that's just the mistake you're making at the end. – Michael Hardy Oct 31 '16 at 22:39
  • BTW, not only is the MLE biased, but the best unbiased estimator is inferior to (in the sense of having a larger mean squared error than) another estimator that is a constant multiple of $Y. \qquad$ – Michael Hardy Oct 31 '16 at 22:41
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    I am actually getting $\frac{\theta^2}{n(n+2)}$ and I'm not sure I follow the intuition of it – David Kang Oct 31 '16 at 23:24
  • I missed the $n$ somehow; I've fixed it. One thing to notice about that is that it's quadratic in $n$, whereas when a sample mean is used to estimate a population mean, the denominator in the mean squared error is linear in $n$. So estimators of sharp boundaries that no point in the population can cross are more accurate than estimators of means. $\qquad$ – Michael Hardy Oct 31 '16 at 23:40