# A Question about Expected value

1. Feb 10, 2013

### Artusartos

1. The problem statement, all variables and given/known data

Suppose $X_1, ... , X_n$ are iid with pdf $f(x,\theta)=2x/(\theta^2)$, $0<x\leq\theta$, zero elsewhere. Note this is a nonregular case. Find:

a)The mle $\hat{\theta}$ for $\theta$.
b)The constant c so that $E(c\hat{\theta})=\theta$.
c) The mle for the median of the distribution.

2. Relevant equations

3. The attempt at a solution

a) I got $\hat{\theta} = max\{X_1, ... ,X_2\}=Y_n$
b) I was stuck here...

I need to find the pdf for $Y_n$ using the order statistics formula, right?

So this is what I got...

$f_n(y_n)=\frac{n!}{(n-1)!(n-n)!}[F(y_n)]^{n-1}[1-F(y_n)]^{n-n}f(y_n)$

= $n(F(y_n))^{n-1}f(y_n)= n(\frac{-2x}{\theta})^{n-1}(\frac{2x}{\theta^2})$

= $n(\frac{-2x}{\theta})^{n}(\frac{\theta}{-2x})(\frac{2x}{\theta^2})$

= $n(\frac{-1}{\theta})(\frac{-2x}{\theta})^n$

So now I want to find the expected value, but I'm not sure what the boundaries need to be for the integral...so can anybody help me?

2. Feb 10, 2013

### Ray Vickson

The formula for $f_n(y)$ must not have x in it! Do you mean
$$f_n(y) = n \left( \frac{-1}{\theta} \right) \left(\frac{-2x}{\theta}\right)^n?$$
I hope not, because for even n this whole expression is < 0, so you have a negative probability density.

Why not just use $$P\{ \max(X_1,X_2, \ldots, X_n) \leq y \} = P\{ X_1 \leq y, X_2 \leq y, \ldots, X_n \leq y \}\\ = F_X(y)^n$$ for iid $X_i$?

3. Feb 10, 2013

### Artusartos

Thank you. So...

$F_X(y)=\frac{y^2}{\theta^2}$, right? So...

$P\{ \max(X_1,X_2, \ldots, X_n) \leq y \} = P\{ X_1 \leq y, X_2 \leq y, \ldots, X_n \leq y \} = P(X_1 \leq y)P(X_2 \leq y)...P(X_n \leq y) = (\frac{y^2}{\theta^2})^n$, right?

But I'm still a bit confused about the boundaries for the intergral for the expected value...

4. Feb 10, 2013

### Ray Vickson

Of course you need $0 \leq y \leq \theta$ in the above. For $y > \theta$ we have $F_Y(y) = 1$.

5. Feb 11, 2013

### Artusartos

Thanks a lot.

When I use the technique that you suggested, I get...

$P\{ \max(X_1,X_2, \ldots, X_n) \leq y \} = P\{ X_1 \leq y, X_2 \leq y, \ldots, X_n \leq y \} = P(X_1 \leq y)P(X_2 \leq y)...P(X_n \leq y) = (\frac{y^2}{\theta^2})^n$

When I use the order statistics (I did it again, because the one I did before was wrong), I get...

$f_n(y_n)=\frac{n!}{(n-1)!(n-n)!}[F(y_n)]^{n-1}[1-F(y_n)]^{n-n}f(y_n)$

= $n(F(y_n))^{n-1}f(y_n)= n(\frac{{y_n}^2}{\theta^2})^n(\frac{\theta^2}{y_n})(\frac{2y_n}{\theta^2}) =(\frac{2n}{y_n})(\frac{{y_n}^2}{\theta^2})^n$

This question is also solved in this link (question 2):

http://www.math.harvard.edu/~phorn/362/362assn6-solns.pdf [Broken]

All three of these answers are different. I keep repeating this to see if I did something wrong, but I just keep getting the same answers. It's really frustrating...

Last edited by a moderator: May 6, 2017