For $X_1, X_2, ... X_n$ which are Weibull(3,$\alpha$), I am trying to find the distribution of $Y=\sum_{i=0}^n X_i^3$

I looked up the MGF to be $\sum_{n=0}^\inf \frac{t^n\alpha^n}{n!} \Gamma(1+\frac{n}{3})$ so I let $t=3$ and multiply it out, but I'm having trouble seeing what the distribution of this is, I think mostly due to the summation. Since the gamma function is present I assume it's either beta or gamma distributions but can't be sure.

And this is just a step in a homework problem having to do with hypothesis testing, not the problem itself, so an actual full answer would be much appreciated if that is alright since the problem is testing me on something else, not how to find the distribution.

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    $\begingroup$ You should specify what your parameters mean (e.g. what's the pdf or cdf in terms of your parameterization) -- is 3 the shape parameter? Is $\alpha$ the scale? $\endgroup$ – Glen_b -Reinstate Monica Apr 14 '15 at 1:40

Here's a big hint - do it in two steps

  1. work out the distribution of $X_i^3$. (This can be done by quite elementary means)

    [Let $W_i=X_i^3$. $P(W_i\leq w) = P(X_i^3 \leq w) = P(X_i \leq w^{1/3}) =\; ...$]

  2. Then worry about the distribution of the sum.

Both are quite straightforward.

(This isn't the only way to tackle this problem, but it's a very easy way.)


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