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Suppose $X_i \stackrel{i.i.d}{\sim}$ Exp$(1/\theta)$ which implies $\sum_{i =1}^{n} X_i \sim$ Gamma $(n, 1/\theta)$.

But, then, the book that I am reading says that $(2/\theta)\sum_{i =1}^{n} X_i \sim$ Gamma$(n, 2)=$$\chi^2(2n)$. I don't understand how this is true. Can someone please explain this step? Thanks!

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    $\begingroup$ I'm not making sense of this: Do you mean $|X_i| \sim \mathsf{Exp}(1/\theta)?$ $\endgroup$
    – BruceET
    May 17, 2020 at 18:04
  • $\begingroup$ @BruceET $\left|X_i\right|$ is not mentioned anywhere in the book but I've added one more detail in my question that I had accidentally left out earlier. I know that the gamma distribution is a special case of the Chi-squared distribution but I am having trouble applying this logic here. $\endgroup$ May 17, 2020 at 18:14
  • $\begingroup$ Still something wrong. See Wikipedia on Laplace distribution (also known as double exponential).Also on Gamma under relationships with other distributions. $\endgroup$
    – BruceET
    May 17, 2020 at 19:26
  • $\begingroup$ My answer ignores Laplace dist'n. It's a mess as is and needs to be fixed. // My answ Illustrates that sum of $n$ indep exponential RVs with same rate is gamma with shape parameter $n$ and that rate. (Follows immedately by looking at MGFs.) Also, relationship btw chi-sq RV with particular DF and gamma parameters. (Simple linear transformation.) $\endgroup$
    – BruceET
    May 17, 2020 at 19:38
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    $\begingroup$ @BruceET I've edited my question. Hopefully, it's not nonsensical anymore. Also, with this edited question, your answer fits perfectly. Thanks! $\endgroup$ May 17, 2020 at 21:44

1 Answer 1

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I'll let you use moment generating functions or other methods to prove that $X_i \stackrel{iid}{\sim}\mathsf{Exp}(\mathrm{rate}=1/\theta)$ implies $T = \sum_{i=1}^n X_i \sim \mathsf{Gamma}(\mathrm{shape}=n, \mathrm{rate} = 1/\theta).$

For $n=5, \theta=10, \lambda = 1/\theta = 0.1,$ the following simulation in R verifies that this 'works' for a specific case. Recall that $E(T) = n\theta = n/\lambda = 50.$

n = 5; lam = 0.1
set.seed(517)
t = replicate(10^5, sum(rexp(n, lam)))
mean(t)
[1] 49.99174   # aprx 50

hist(t, prob=T, br=50, col="skyblue2", main="GAMMA(shape=5, rate=.1)")
 curve(dgamma(x, n, lam), add=T, col="red", lwd=2)

enter image description here

Also, continuing from above you can show that $Y = 2T/\theta = 2\lambda T \sim \mathsf{Chisq}(\nu=2n).$ Recall that $E(Y) = \nu = 10.$

y = 2*lam*t
mean(y)
[1] 9.998347  # aprx 10

hist(y, prob=T, br=50, col="skyblue2", main="CHISQ(10)")
curve(dchisq(x, 2*n), add=T, col="red", lwd=2)

enter image description here

More generally, $\mathsf{Chisq}(\nu) \equiv \mathsf{Gamma}(\mathrm{shape} = \nu/2, \mathrm{rate} = 1/2)$ $\equiv \mathsf{Gamma}(\mathrm{shape} = \nu/2, \mathrm{scale} = 2).$

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