Is f(X)=X+a exponentially distributed when X is exponential R.V and a>0?

Question

Suppose $X$ is an exponential R.V with parameter $\lambda$. Then, I am interested to find the distribution of $f(X)=X+a$, for $a>0$. To this end, I have computed the following: $$\Bbb E\left[f(X)\right]=a+1/\lambda$$ and $${\rm Var}\left(f(X)\right)=1/\lambda^2$$ Since ${\rm Var}\left(f(X)\right)\neq \left(\Bbb E\left[f(X)\right] \right)^2$ which is usually the case for an exponential r.v, I suspect that $f(X)$ is not an exponential r.v. Am I correct in my conjecture?

Answer 1

Let $Y:=X+a$, then \begin{align}F_{Y}(y)=P(Y\le y)=P(X+a\le y)=P(X\le y-a)=F_X(y-a)\end{align} Hence \begin{align}f_{Y}(y)&=\frac{d}{dy}F_Y(y)=\frac{d}{dy}F_X(y-a)=f_X(y-a)=λe^{-λ(y-a)}\end{align} for $a<y$. This is called the shifted exponential distribution with shift $a$. Intuitively your "exponential life" starts at time $a$ and not at time $0$.