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


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!