# Math Help - convergence in distribution

1. ## convergence in distribution

let X1,X2, . . ., Xn be independent random variables all with uniform distribution on [0,1]. Let 0<=a<=b
show P(a<=(X1*X2*...*Xn)^(1/Sqr[n]))<=b) tends to a limit as n tends to infinity and find an expression for it?

i have some problems with my solution and am not sure it is right ..... thanks for any help

Let Y=(X1*X2*...*Xn)^(1/Sqr[n]))

c.d.f FY(y) = P({Y<=y})= P({(X1*X2*...*Xn)^(1/Sqr[n]))<=y})
=P({X1^(n/Sqr[n])<=y}) = P({X1^(Sqr[n])<=y})
=P({X1<=Sqr[y] to the (Sqr[n])th root})

=
0 if (Sqr[y] to the (Sqr[n])th root)<=0
Sqr[y] to the (Sqr[n])th root if 0<= Sqr[y] to the (Sqr[n])th root <=1
1 if 1<= Sqr[y] to the (Sqr[n])th root

but not sure if this is right and then when considering looking at this function as n tends to infinity i got that it tends to 1 for all y eeeeek ?????

thanks for any help

2. These rvs are not the same.

So the statement $X_1X_2\cdots X_n\le a$

is not the same as $X_1^n\le a$

3. ok i see that now thankssss

to solve this i want to find c.d.f of Y=(X1*X2*...*Xn)^(1/Sqr(n))

Fy(y) = P(Y<=y) = P((X1*X2*...*Xn)^(1/Sqr(n))<=y)
= P(X1*X2*...*Xn<= Sqr[y] to the (1/Sqr(n))th root)

would i have to do an n length intergral of the joint density function to solve this??? is this right or is there an something better i can do ????

4. Have you thought of looking at $\log Y$? I can smell Central Limit theorem here

5. Log Y is correct, but I'm not sure you even need the CLT.
The logarithm turns thisthe product into a sum and the log of a uniform is just an exponential.
The sum of iid exponentials is a gamma...

I hate, I can't stand, I abhor, I dislike, I disdain, I detest, I loathe, I despise and I am repelled by cows
http://comics.com/pearls_before_swine/2009-12-29/
http://comics.com/pearls_before_swine/2009-12-30/

6. Sweet so i have

i have Fy(y) = P(Y<=y) = P((X1*X2*...*Xn)^(1/Sqr(n))<=y))
= P(log((X1*X2*...*Xn)^(1/Sqr(n))<=log(y))
= P((1/Sqr(n)log(X1*X2*...*Xn)<=log(y))
= P(log(X1*X2*...*Xn)<=Sqr(n)*log(y))
= P(log(X1)+log(X2)+...+log(Xn)<=log(y))

and the m.d.f

Flog(X1)(x) =
= exp[x] if x<0
= 1 if 0<=0

where do i go from here, what i have tryed dosnt come out very nice . . .

thanks

use [tex] in front and [/math ] no space, at the end of each line
and you should hit the Thanks link.

8. Hello,

Originally Posted by silk23
Sweet so i have

i have Fy(y) = P(Y<=y) = P((X1*X2*...*Xn)^(1/Sqr(n))<=y))
= P(log((X1*X2*...*Xn)^(1/Sqr(n))<=log(y))
= P((1/Sqr(n)log(X1*X2*...*Xn)<=log(y))
= P(log(X1*X2*...*Xn)<=Sqr(n)*log(y))
= P(log(X1)+log(X2)+...+log(Xn)<=log(y))

and the m.d.f

Flog(X1)(x) =
= exp[x] if x<0
= 1 if 0<=0

where do i go from here, what i have tryed dosnt come out very nice . . .

thanks
The Xi's are independent. So are the -log(Xi), which follow an exponential distribution, according to the cdf (slight transformation)
And as oldeagle says, the sum of iid exponential distributions is a Gamma distribution (see here : Exponential distribution - Wikipedia, the free encyclopedia)

The problem with this method is that it's impossible to get a defined formula for the cdf.

So the best solution seems to be approximating this by the Central Limit Theorem, since you're interested in the limit !
For this, you just need to know that $\frac{X_1+\dots+X_n-nE[X_1]}{\sqrt{var[X_1]}\cdot\sqrt{n}}$ converges in distribution to a normal distribution (0,1)
So the cdf converges too.