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I was checking answers for a question and the following arised:

$$ P(Y>25\mid Y\sim \mathrm{GAM}(15,1)) = P(\chi ^2>3.33\mid\chi^2\sim \chi^2(2)) $$ Now i get that chi is just a special case of gamma with $\theta = 2$ and $r=v/2$ but can i just divide everything by $7.5$? It seems like blasphemy.

2 Answers2

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HINT: If $X \sim \mathrm{Gamma}(\nu/2, 2)$, then $X$ is identical to $\chi^2(\nu)$, the chi-squared distribution with $ν$ degrees of freedom. Conversely, if $Y \sim \chi^2(\nu)$ and $k$ is a positive constant, then $kY \sim \Gamma(\nu/2, 2k)$.

alexjo
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You have $$ \Pr(X>x) = e^{-x/15}. $$ So $$ \Pr\left(\frac{X}{7.5}>\frac{25}{7.5}\right) = \Pr(X>25) = e^{-25/15} = e^{-(25/7.5)/2} $$ because $$ \frac{25}{15} = \frac{2.5/7.5}{2}. $$