Recent content by stevenham

I think we can do it some shorter way. I looked at some previous examples and Y=-Ʃln(1-Xi) has a Gamma distribution with α=2n and β=1/(θ+1) However I'm stumped on how to get there. I know how to find the probability distribution of W=-ln(1-X), but I don't know what to do from there...

We're basically doing the Central limit theorem and confidence intervals. I think we just have to do it the long way. So I first solved for H=-ln(1-X) and I got 1-e-(θ+1)h I think that's an exponential with mean 1/θ+1 I'm assuming the next step would be to find the probability...

Homework Statement Let X1, X2,...Xn be a random sample from the distribution with probability density function fX (x;θ) = (θ+1)(1-x)θ, 0<x<1 θ>-1 a) What is the probability distribution of Y= -\sum ln(1-Xi from i=1 -> n b) Suggest a (1-α)100% confidence interval for θ based on Y=...

Do I have to calculate the CDF? After calculating the CDF and doing the limit as n goes to infinity, I get 1. Did I calculate the MGF incorrectly? Should I have done E(e^ti) instead?

Homework Statement Suppose P(Xn = i) =[SIZE="5"] \frac{n+i}{3n+6}, for i=1,2,3. Find the limiting distribution of Xn Homework Equations The Attempt at a Solution I first found the MGF by Expectation(etx) which resulted in [SIZE="5"]etx(\frac{n+1}{3n+6} + \frac{n+2}{3n+6} +...

I'm trying to figure out what the summation notation of (1-x1)^θ * (1-x2)^θ * ...(1-xn)^θ would be for summation I know I need to convert this to a summation notation in order to solve my problem, but I can't figure out how to convert it. Any help will be appreciated.