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

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It is clearer to work via the cumulative density function, $F(x)=P(X

For our new variable $\lambda X$, we seek $P(\lambda X

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