Just to clarify, Fk is the cumulative distribution function. So for k=1 to n, F1, F2, F3 etc are separate cumulative distribution functions
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 ?
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
Yes sorry that's right, given that Fk is strictly increasing (quite important assumption), follows a uniform distribution over (0,1)
And we know (we can check it with cdf) that follows an exponential distribution with parameter 1. And its expectation is 1.
So the expectation of is 0.
And here, we have ... so, can you apply the CLT here ?
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
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)