Z=(-1/sqrt(n)) * sum from k=1 to n of [1+log(1-Fk)] where Fk is a cumulative distribution function which is continious and strictly increasing.

Show that as n->infinity, Z converges to a normal distribution with mean 0 and var 1

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- Aug 20th 2010, 08:19 PMGekkoCentral limit proof
Z=(-1/sqrt(n)) * sum from k=1 to n of [1+log(1-Fk)] where Fk is a cumulative distribution function which is continious and strictly increasing.

Show that as n->infinity, Z converges to a normal distribution with mean 0 and var 1 - Aug 23rd 2010, 05:47 AMGekko
Just to clarify, Fk is the cumulative distribution function. So for k=1 to n, F1, F2, F3 etc are separate cumulative distribution functions

- Aug 23rd 2010, 11:01 PMMoo
Hello,

Since Fk is a cdf, where is the variable ? You gotta have something like Fk(x) (Surprised) So could you clarify this ? - Aug 23rd 2010, 11:04 PMGekko
Yes sorry. It's Fk(Xk).

So this will take value from 0 to 1 and is uniform - Aug 23rd 2010, 11:22 PMMoo
Hm I don't understand. You mean it's the cdf taken at the value of a random variable, Xk ?

Isn't it that Fk is the cdf of Xk ?

And what kind of random variable is Xk ?

Why do you say Fk is uniform ? Would it mean that it's the cdf of a uniform distribution ? Over what interval ?

Is there the exact writing of the problem somewhere ? - Aug 23rd 2010, 11:47 PMGekko
Xk is an independent random variable and Fk is it's associated cumulative distribution function.

Sorry if I didn't make this clear. Missed this from the original question description.

So, Fk(Xk) will become a uniform distribution from 0 to 1 since it has equL likelihood of any vale from 0 to 1 given that Xk is independent and random. At least this is what I'm thinking now - Aug 24th 2010, 11:36 AMMoo
Yes sorry that's right, given that Fk is strictly increasing (quite important assumption), $\displaystyle U_k=F_k(X_k)$ follows a uniform distribution over (0,1)

And we know (we can check it with cdf) that $\displaystyle -\log(1-U_k)$ follows an exponential distribution with parameter 1. And its expectation is 1.

So the expectation of $\displaystyle -(1+\log(1-U_k))$ is 0.

And here, we have $\displaystyle -\frac{1}{\sqrt{n}}\sum_{k=1}^n 1+\log(1-U_k)=\frac{1}{\sqrt{n}}\sum_{k=1}^n -(1+\log(1-U_k))$... so, can you apply the CLT here ? :D - Aug 24th 2010, 03:18 PMGekko
This is good. Thanks.

I took the approach of calculating the moments of a log uniform by taking the standard approach but replacing the function, x, with ln(x) ie

1/(b-a) integral from a to b of ln(x)

Using this gave me -1 for the expected value and 1 for the variance.

I didnt think of using the quartile function and this looks to be a better approach. How do you calculate the variance though? Understand the mean is 1/parameter = 1 - Aug 28th 2010, 04:52 AMMoo
Find the variance of the exponential distribution of parameter 1 (that is 1 ;D) and since the random variables are independent, the variance of the sum is the sum of the variances !

The cdf is sometimes much more easier to use than calculating all the moments ^^ (use the inversion method)