When X is r.v with exponential distribution then X-a is r.v with Exponential distribution too? (a is a constant)

Question

When r.v. $X$ ~ Expo($\lambda$) then r.v. $(X-a)$ ~ Expo($\lambda$) ?

Let $Y = X-a$.

CDF of $Y$ : $P(Y < y) = P(X-a < y) = P(X < a+y) = 1-e^{\lambda(a+y)}$

It doesn't seem like $Y$ is not r.v with Expo$(\lambda)$ but i want to make sure whether my logic is correct.

And i wanna know why given $X$ ~ Expo($\lambda$), $E(X|X>a) = a + E(X-a|X>a) = a + E(X)$

I really would be appreciated if you enlighten me, Thanks!

Answer 1

The exponential distribution's support is $[0,\infty)$, whereas $Y$ has support $[-a,\infty)$. Thus, if $a\ne0$ $Y$ cannot be an exponential distribution. It does have the same shape as one, though.

We calculate $E(X\mid X>a)$ as $$\frac{\int_a^\infty x\lambda e^{-\lambda x}\,dx}{\int_a^\infty \lambda e^{-\lambda x}\,dx}=\frac{(1+a\lambda)e^{-a\lambda}}{\lambda e^{-a\lambda}}=a+\frac1\lambda=a+E(X)$$