1. ## Convergence in Distribution

I am wondering if someone can give me some help on this problem.

Let $X_1, X_2,$ ... be i.i.d. and $X_{(n)} = max_{1 \le i \le n}X_i$. If $X_i ~ Beta (1, \beta)$, find a value $v$ so that
$n^v(1 - X_{(n)})$ converges in distribution.

I think it has to do with order statistics but I could be comletely off the right track.

2. Originally Posted by chabmgph
I am wondering if someone can give me some help on this problem.

Let $X_1, X_2,$ ... be i.i.d. and $X_{(n)} = max_{1 \le i \le n}X_i$. If $X_i ~ Beta (1, \beta)$, find a value $v$ so that
$n^v(1 - X_{(n)})$ converges in distribution.

I think it has to do with order statistics but I could be comletely off the right track.

Start by getting the distribution of $X_{(n)}$.

3. Ok. Let see if I can at least get this right...

$f_{X_i}=(1-x)^{\beta-1}$ since $X_i \sim Beta(1, \beta)$

Then $F_{X_i}=\int^x_0 (1-t)^{\beta-1} dt= \frac{1}{\beta}(1-(1-x)^{\beta})$
And since $X_n$ is the max, the p.d.f for it is
$f(x_n)=n*f(x_n)*F^{n-1}(x_n)$
$=n*(1-x_n)^{\beta-1}*\frac{1}{\beta^{n-1}}(1-(1-x_n)^{\beta})^{n-1}$

Am I in the right track?

Thanks again!

4. That's not a valid density.
You're leaving out the constant terms in your Beta.
And, yes the max is the largest order statistic.

5. I see where I messed up now...'cause I can't add...

So instead, $f_{X_i}=\beta(1-x)^{\beta-1}$

Then it will be $F_{X_i}=\beta \int^x_0 (1-t)^{\beta-1} dt= 1-(1-x)^{\beta}$
And p.d.f will be
$f(x_n)=n*f(x_n)*F^{n-1}(x_n)$
$=n\beta(1-x_n)^{\beta-1}(1-(1-x_n)^{\beta})^{n-1}$

Does that look better?

6. Instead of doing all of that, why don't you get the CDF of the max directly?

$P(X_{(n)}\le a)=P(X_1\le a,...., X_n\le a)=F^n(a)$

Then let $Y_n=n^v(1-X_{(n)})$ and find its CDF.

7. Originally Posted by matheagle
Instead of doing all of that, why don't you get the CDF of the max directly?

$P(X_{(n)}\le a)=P(X_1\le a,...., X_n\le a)=F^n(a)$
Oh, right, since they are i.i.d, I guess I could have done that. >"<

So the c.d.f is $F(x_{(n)})=[1-(1-x_{(n)})^{\beta}]^n$.

Edit:And I just saw the last line of your hint, let me try to see if I can get it out. Puzzle time! lol

8. So let $Y_n=n^v(1-X_{(n)})$

Then $F(Y_n)=P(Y_n\le y)$
$=P(n^v(1-X_{(n)})\le y)$
$=P(X_{(n)}\ge 1-\frac{y}{n^v})$
$=1- P(X_{(n)}\le 1-\frac{y}{n^v})$
$=1-F(1-\frac{y}{n^v})$
$=1-[1-(1-(1-\frac{y}{n^v}))^{\beta}]^n$
$=1-[1-(\frac{y}{n^v})^{\beta}]^n$

So now I have to prove that this converges in distribution.
So as $n$ approaches infinity, doesn't $(\frac{y}{n^v})^{\beta}$ approaches zero?

9. USE calc 2

$\bigg(1-{a\over x}\biggr)^x \to$?????

And you left out a power of n.

10. Originally Posted by matheagle
USE calc 2

$\bigg(1-{a\over x}\biggr)^x \to$?????

And you left out a power of n.
$e^a$!!!

So $v=\frac{1}{\beta}$ must be true?

Then $F(Y_n) \to 1-e^{y^{\beta}}$

And wow on the publication.

By the way, you are only 5 hours from me. lol

11. No problem, any time

12. Thank you again for your help!

13. Nope...........that limit is $e^{-a}$
you need a negative in the exponent

Originally Posted by chabmgph
$e^{-a}$!!!

So $v=\frac{1}{\beta}$ must be true?

Then $F(Y_n) \to 1-e^{-y^{\beta}}$

And wow on the publication.

By the way, you are only 5 hours from me. lol

14. Can't believe it..... I messed it up again.

Thanks again for catching that!

15. I noticed it the other night.
I was hoping you or someone else would point it out.
The exponent must be negative if you want F(y) to approach 1 as y goes to infinity.
And there is a power of n still missing.

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