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

This makes sense by working backward. Because if this is true  ,then $X_i$ should be $\sim \text{Gamma}(1,\theta)$ . which in turn the correct answer. (because $\text{Gamma}(1,\theta)$ 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 $\sum X_n$ $\sim \text{Gamma}(n, \theta + a)$.

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

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

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?

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

This makes sense by working backward. Because if this is true ,then $X_i$ should be ~ $gamma (1,\theta)$ . which in turn the correct answer. (because $gamma (1,\theta)$ and $exp(\theta)$ are equivalent)

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

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

Then I tried to do the same thing by working backward. Then $X_i$ should be ~ $gamma (1 , a+ \theta)$. That means $f(X/\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 ?

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

This makes sense by working backward. Because if this is true ,then $X_i$ should be $\sim \text{Gamma}(1,\theta)$ . which in turn the correct answer. (because $\text{Gamma}(1,\theta)$ 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 $\sum X_n$ $\sim \text{Gamma}(n, \theta + a)$.

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

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