Could anyone help me with the following proof, and give me any lead how to prove or refute this statement?
$X\sim \text{Exp}(\lambda)$. For each constant $a\ne 0$ and constant $b$, the random variable $Y = aX + b$ is distributed exponentially.
Thank You!
If $P(X \le x) = 1-e^{-\lambda x}$ for $x \ge 0$ and $P(X \le x) =0$ for $x \le 0$
and if $Y = aX+b$ with $a \gt 0$ so $P(Y \le y) = P(aX+b \le y) = P(X \le \frac{y}{a} -\frac{b}{a})$
then $P(Y \le y) = 1-e^{\frac{b\lambda}{a}}e^{- \frac{\lambda}{a}y}$ for $y \ge b$ and $P(Y \le y) =0$ for $y \le b$
If $b=0$, this simplifies to $P(Y \le y) = 1-e^{- \frac{\lambda}{a}y}$ for $y \ge 0$ and so is a exponential distribution with rate $\frac{\lambda}{a}$
Personally I would not call the distribution for $Y$ an exponential distribution if $b\not=0$. I might instead call it a shifted exponential distribution, or something similar. And if $a <0$ then I might call it a reversed exponential distribution, or something similar.