These rvs are not the same.
So the statement
is not the same as
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
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 ????
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/
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
Hello,
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 converges in distribution to a normal distribution (0,1)
So the cdf converges too.
This is way better to read than many other threads.please use tex, this is hard to read
use [tex] in front and [/math ] no space, at the end of each line
and you should hit the Thanks link.
And he knows the Thanks link (he thanked 2 posts), but it's up to him to decide if a post deserves one...