# Why is sample mean minus location parameter of shifted exponential gamma distributed?

My book says the following

Suppose $$X_{i} \sim$$ iid $$Exp(1,\eta)$$

Where $$Exp(\theta,\eta)$$ is the shifted exponential ie has density

$$\frac{1}{\theta}e^\frac{-(x-\eta)}{\theta}$$ for $$x \ge \eta$$ and zero otherwise.

Then my book states

$$\bar X-\eta$$ $$\sim$$ $$Gamma(\frac{1}{n},n)$$

But I do not see why this is true.

I tried using mgf but it is hard to work with sample mean in mgf.

So why is it true?

• What expression for the the gamma pdf are you using? – Sycorax Apr 8 at 5:14
• k Shape- $\theta$ Scale – Quality Apr 8 at 5:17

If we expand your final expression, we have:$$\bar{X}-\eta=\frac{1}{n}\sum ( X_i - \eta )$$
$$X_i-\eta$$ is classic exponential RV with parameter $$\theta = 1$$. By denition Sum of $$n$$ iid exp RVs (with parameter $$\theta$$) is gamma distributed, i.e. $$G(n,\theta)$$, so, the sum $$\sum (X_i-\eta)\sim G(n, 1)$$. And gamma Rv has scaling property, i.e. if $$Y$$ is Gamma with $$(k,\theta)$$, then $$cY\sim G(k,c\theta)$$. Applying here $$\bar{X}-\eta\sim G(n,1/n)$$. I think the parameter order is swapped in your book, I’m following wiki convention, where you can also find sum and scale properties.