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Let $Y$ be an exponential random variable with parameter $\frac12$. Let $X=e^{-Y/2}$. Determine the pdf of $X$.

$$f(t)=\frac{d}{dt}P(X\le t)=\frac{d}{dt}P(e^{-Y/2}\le t)\\=\frac{d}{dt}P(Y\ge-2\ln t)=\frac{d}{dt}(1-P(Y<-2\ln t))=-\frac12e^{\ln t}=-\frac{t}2$$

How come I end up with a negative value? What's wrong?

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    NB: This is `probability` not `probability-theory`.2012-04-19

2 Answers 2