Let $X,Z$ be random variables with probability density functions $p_X,p_Z$. Suppose $Z=f(X)$, where $f$ is continuous and differentiable. How is $p_Z$ related to $p_X$? It's tempting to say $p_Z(z) = p_X(f^{-1}(z))$, but I think that is not correct: I think it might be

$$p_Z(z) = {p_X(f^{-1}(z)) \over f'(f^{-1}(z))},$$

where $f'$ is the derivative of $f$, but I am not sure whether I've got that right. What is the correct rule?


The correct rule is

$$p_Z(z) = \sum_{x \in f^{-1}(z)} {p_X(x) \over f'(x)},$$

as there can be multiple possible values $x$ that satisfy $f(x)=z$.

One way to derive this, in the case of a monotonic $f$ (so there is only a single $x$ such that $f(x)=z$), is to consider the cdfs $f_X,f_Z$. Then we have

$$F_Z(z) = \Pr[Z \le z] = \Pr[f(X) \le z] = \Pr[X \le f^{-1}(x)] = F_X(f^{-1}(z))$$


$$\begin{align*} p_Z(z) &= {d \over dz} F_Z(z) = {d \over dz} (F_X(f^{-1}(z)))\\ &= ({d \over dz} F_X)(f^{-1}(z)) \cdot {d \over dz} f^{-1}(z)\\ &= p_X(f^{-1}(z)) \cdot 1/f'(f^{-1}(z)). \end{align*}$$

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