Let $X_1, ... X_n$ be i.i.d random variables that have an exponential distribution with parameter $\theta$. So we know that $\sum X_n \sim \Gamma(n, \theta)$.

This makes sense by working backward. Because if this is true, then $X_i$ should be $\sim \Gamma(1,\theta)$ . which in turn the correct answer. (because $\Gamma(1,\theta)$ and $\text{Exp}(\theta)$ are equivalent)

I tried to do the same thing for 2 parameter exponential distribution. That means if

$$f(x\mid\theta , a) = \frac{1}{\theta}e^{-(x-a)/\theta} ,x>a ,\theta>0$$. I wanted to know the distribution of sum. I saw in a book that $\sum X_n\sim \Gamma(n, \theta + a)$.

Then I tried to do the same thing by working backward. Then $X_i$ should be $\sim \Gamma(1 , a+ \theta)$. That means $f(x\mid\theta,a) $ and $\Gamma(1 , a+ \theta)$ should be equivalent.

But it is not. So what did I do incorrectly here? can anyone help me to figure it out?


1 Answer 1


Your book is wrong. The easy way to see this is that the support of $gamma(n, \theta + a)$ is $(0, \infty)$ while $\sum X_i \geq na$.

Instead, the sum should have the same distribution as a $na + Y$ where $Y \sim gamma(n, \theta)$. This essentially follows from the first case as each of your variables has the distribution of $a + Z$ with $Z \sim exp(\theta)$.


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