If X has an exponential distribution with expectation 1/λ, find the probability density function for λX. What is the distribution of λX?

Question

I am not sure if the PDF of λX is λ^2(e^(−λx))? Then what is the distribution?

Answer 1

It is clearer to work via the cumulative density function, $F(x)=P(X<x)=1-e^{-\lambda x}$.

For our new variable $\lambda X$, we seek $P(\lambda X<x)=P(X<x/ \lambda) = F(x/ \lambda )= 1-e^{-x}$. Thus the new PDF is $e^{-x}$, ie. $\lambda X$ has an exponential distribution with parameter 1.

This shouldn't surprise us. The mean of the new distribution is $\lambda \times 1/ \lambda=1 $ and similarly for the variance.