I am stucking on this problem from days, I need help, if anyone could help me. I have $Y=\log(1+X)$ with $X$ that follows an exponential distribution, I would understand which is the $Y$ distribution, is $Y$ also exponential?
Thank you in advance
I am stucking on this problem from days, I need help, if anyone could help me. I have $Y=\log(1+X)$ with $X$ that follows an exponential distribution, I would understand which is the $Y$ distribution, is $Y$ also exponential?
Thank you in advance
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No. If $X\sim\operatorname{Exp}(\lambda)$, the CDF of $Y$ is$$P(Y\le y)=P(X\le\exp y-1)=1-\exp(-\lambda(\exp y-1))$$for $y\ge0$. This is a Gompertz distribution with $\eta=\lambda,\,b=1$.