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Let $X_1, X_2...X_n$ be iid with $f(x,\theta)=\dfrac{2x}{\theta^2}$ and $0<x\leq\theta$. Find $c$ such that $\mathbb{E}(c\hat{\theta})=\theta$ where $\hat{\theta}$ denotes MLE of $\theta$.

What I have tried: I found the MLE of $f(x;\theta)$ to be $\max\{X_1,X_2\cdots X_n\}$ (which aligns with the answer at the back) but now I am stuck at this question. The answer given is $\dfrac{2n+1}{2n}$.

I would have proceeded as: $$\begin{align*} \mathbb{E}(c\hat{\theta})&=\theta\\ \int_0^\theta c \dfrac{2x}{y^2} &=\theta \quad (y = \max\{x_1,x_2,\cdots,x_n\})\\ \dfrac{1}{y^2}\int_0^\theta c .{2x}{} &=\theta \end{align*}$$ But continuing this way gives me an answer far off from the one in the book (I don't have a term of n to begin with).

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  • $\begingroup$ You can't find the expected value of the MLE (hence, you can't correct the bias) without knowing the distribution of the MLE. It is related to order statistics as suggested below, but you don't need to be familiar with order statistics to solve this problem. If $Y=\textrm{max}\left\{X_1,X_2,\ldots,X_n\right\}$, what is the CDF of $Y$, $F_Y(y)=P(Y\leq y)$? Hint: If $Y\leq y$, what does that say about each of the $X_i$? $\endgroup$ – assumednormal Oct 30 '12 at 3:48
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    $\begingroup$ @Max Working on your hint $Y \leq y$ implies that each $X_i\leq y$ but how do I use this in my problem? I understand that my expectation is taken over $\hat{\theta}$ and thus, we must use its distribution. Now that brings up 2 questions in my head: 1. Is it different from the distribution of $X_i$? The answer seems to be no (although I am not completely sure why). 2. If it's different, then what is it? $\endgroup$ – Kplee Oct 30 '12 at 3:55
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    $\begingroup$ So,$$P(\max \{X_i\}<y)=P(X_1<y,X_2<y,...,X_n<y) = \prod_{i=1}^n P(X_i<y)$$ $$=(\int_0^y\dfrac{2x}{\theta^2})^n=\dfrac{y^{2n}}{\theta^{2n}}$$ $\endgroup$ – Kplee Oct 30 '12 at 4:08
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    $\begingroup$ $$f(y) = \dfrac{2n.y^{2n-1}}{\theta^{2n}}$$ $$E(c\hat{\theta}) = \int_0^\theta\dfrac{c\cdot2n.y^{2n}dy}{\theta^{2n}}=\dfrac{c\cdot 2n}{\theta^{2n}}\times\dfrac{\theta^{2n+1}}{2n+1}=\theta$$ $$c = \dfrac{2n+1}{2n}$$ Awesome! Thanks! I'll mark the existing answer as the solution. $\endgroup$ – Kplee Oct 30 '12 at 4:17
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    $\begingroup$ @Max Actually, an easier calculation would be $$E[\hat{\theta}] = E[\max X_i] = \int_0^{\theta} 1 - \left(\frac{y}{\theta}\right)^{2n}\,\mathrm dy = \theta\left(1-\frac{1}{2n+1}\right)$$ giving $c = \frac{2n+1}{2n}$ $\endgroup$ – Dilip Sarwate Oct 30 '12 at 4:25
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What's the distribution of the largest observation from a sample of size $n$ with distribution F?

You want a special case of the formula for $f_{X_k}(X)$ where $k=n$ (which formula should be obvious in any case):

http://en.wikipedia.org/wiki/Order_statistic#The_joint_distribution_of_the_order_statistics_of_an_absolutely_continuous_distribution

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    $\begingroup$ We haven't covered Order Statistics yet, any way to do without that? $\endgroup$ – Kplee Oct 30 '12 at 3:22
  • $\begingroup$ Yes, as the answer pretty much implies ("should be obvious in any case"), you can work it out yourself from first principles, using simple probability arguments. The probability that the largest value of a sample of $n$ is $\leq x$ is the probability that all $n$ observations are $\leq x$. Write that probability down (a power of a cdf at $x$), take derivates w.r.t $x$ to find the density. Then take expectations. to get $c$. (Though there are other ways to go about it, they're all going to involve some effort.) $\endgroup$ – Glen_b -Reinstate Monica Oct 30 '12 at 20:20

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