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**approx** Suppose that a random variable Y has a gamma distribution w. parameters $\displaystyle \alpha = 2$ and an unknown $\displaystyle \beta$. We have earlier proved that $\displaystyle 2Y/\beta$ has a chi-square distribution w. 4 df. Using $\displaystyle 2Y/\beta$ as a pivotal quantity. derive a 90% confidence interval for \beta.

I don't know if I'm on the right way. I create a new random variable $\displaystyle U = 2Y/\beta$ and look for $\displaystyle P(a\leq U\leq b) = 0.90$

=

$\displaystyle P(a\leq 2Y/\beta\leq b) = 0.90$

I then have a similar example in the book that tells me to set it up like this:

$\displaystyle P(U<a) = \int_{0}^{a}{} f(u) du = 0.05$

$\displaystyle P(U>b) = \int_b^\infty f(u) du = 0.05$

Must I derive the pdf of U manually or can I use that $\displaystyle 2Y/\beta$ is chi-square distributed and put in the pdf of a chi-square distribution?

That would be:

$\displaystyle (y)^{(\upsilon /2)-1}e^{-y/2}/2^{\upsilon/2}\Gamma (\upsilon /2)$ but with y=u and $\displaystyle \upsilon$ = 4

I've tried to do that but I don't get the right answer, which is $\displaystyle 2Y/9.49, 2Y/.711$.

Thanks for your help.