# 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!