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This is a more general question about the transformation of a random variable. Say X is given as a certain distribution, and Y=g(X). If it asks to compute the pdf of Y, I am having trouble to determine the region of pdf of Y. (sorry I know this is confusing, let me have one example).

example: say Z is exponential (parameter 1/2), Y=exp(Z).

Then the cdf of Y=P(Z< lnz), pdf_Y =d/dz[ cdf_Y]=z^(-3/2). My question is what region does this answer defined?

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1 Answer 1

The random variable $Z$ of your question has density function $e^{-z/2}$ for $z\gt 0$ and $0$ for $z\lt 0$.

By your argument, we have $$\Pr(Y\le y)=\Pr(e^Z\le y).$$ If $y\le 1$, then $\Pr(e^Z\le y)=0$, since $Z\le 0$ with probability $0$. Thus the density function of $Y$ is $0$ for $y\le 1$.

If $y\gt 1$, then $$\Pr(e^Z\le y)=\Pr(Z\le \ln y)=1-e^{-(\ln y)/2}.$$ Differentiate. For $y\gt 1$, the density of $Y$ is $\frac{1}{2y}e^{-(\ln y)/2}$, which is $\frac{1}{2}y^{-3/2}$.

Thus, as is often the case in probability theory, the density function of $Y$ is defined by different expressions on different parts of its domain.

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