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I've come across this question while revising for finals:

Let $X \sim \operatorname{\Gamma}(k,2)$, $Y_i \sim \operatorname{EXP}(2)$, $Z \sim \operatorname{N}(0,1)$, where $Y_1, \cdots, Y_k$ are independent.

I've already successfully showed that $\sum_{i=1}^k Y_i \sim X \sim \operatorname{\Gamma}(k,2)$ and that as $k \rightarrow \infty$, $\frac{X-2k}{2 \sqrt{k}} \xrightarrow{d} \operatorname{N}(0,1)$ (via the Central Limit Theorem and that $X$ and $\sum_{i=1}^k Y_i$ have the same distribution).

But I get stuck while calculating $2k \frac{Z^2}{\sum_{i=1}^k Y_i}$. I know that $Z^2 \sim \chi^2(1)$, so the expression could have a $t$-distribution or an $F$- distribution, but I can't see how to relate $\frac{2k}{\sum_{i=1}^k Y_i}$ to a $\chi^2$-distributed random variable.

Any help is greatly appreciated.

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  • $\begingroup$ Note that $\chi^2$ distribution is a special case of gamma distribution. Can you check if this is indeed the special case? $\endgroup$ – BGM Apr 19 '18 at 16:24
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Observe that $$ \sum_{i=1}^k Y_i\sim\text{Gamma}(k,2) $$ But a $\text{Gamma}(k,2)$ is also equivalently a chi square distribution with $2k$ degrees of freedom whence $$ \frac{Z^2}{(\sum_{i=1}^k Y_i)/2k}\sim F(1,2k) $$ assuming that $Z, Y_1,\dotsc, Y_k$ are independent.

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  • $\begingroup$ Thanks! Could you perhaps elaborate a bit more on where this result (gamma(k, 2) is equivalent to chi^2(2k)) comes from? $\endgroup$ – Alex Lostado Apr 19 '18 at 17:19

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