A Question about Expected value

[Artusartos]

Homework Statement

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.

Homework Equations

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?

Thanks in advance

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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?

Thanks in advance

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 \}\\<br /> = F_X(y)^n for iid $X_i$?

 

[Artusartos]

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 \}\\<br /> = F_X(y)^n for iid $X_i$?

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...

 

[Ray Vickson]

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...

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

 

[Artusartos]

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

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

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...