When we have a random variable $x$ with a probability density $p(x)$, and a function $y = f(x)$ that is differentiable and can be solved for $x = g(y)$, the change of variable formula leads us to a density for $y$ given by

$$ p(x) \, dx = p(x) \left| g'(y) \right| \, dy = p(x) \left| \frac{1}{f'(x)} \right| \, dy = p(x) \left| \frac{dx}{dy} \right| \, dy $$ where $\frac{dx}{dy}$ is called (to my knowledge even in the univariate case) the Jacobian of the transformation (as in Zill & Wright, p. 792). In general this would be a determinant of a Jacobian matrix $\mathbf{J}(\mathbf{g}(\mathbf{y}))$, obviously. But I never understood why does it enter in absolute value? I have read somewhere that it's because $f(x)$ could have a negative derivative whereas probabilities are confined to be positive, but that sounds more like a post-hoc justification than a mathematical result. Is there a way to derive this fact?


1 Answer 1


For a specific example, in addition to @whuber's advice, let $y=f(x)=-2x$, and $x=g(y)=-y/2$; and $x \in [0,1]$, i.e. the support. Then, $y$ would be in the range $[-2,0]$. Also, we have $g'(y)=-1/2, f'(x)=-2$.

Normally, you'd take the integral $\int p(g(y))\left|\frac{dx}{dy}\right|dy$ from $-2$ to $0$, while using the formula. However, it actually is from $0$ to $-2$, since $x$ and $y$ directions differ, i.e. $$\int_{0}^{-2} p(g(y))\frac{dx}{dy}dy=\int_{-2}^{0} p(g(y))\left(-\frac{dx}{dy}\right)dy=\int_{-2}^{0} p(g(y))\left|\frac{dx}{dy}\right|dy$$

The use of absolute value removes the need of considering the inverse directions (i.e. negative directions of $x$ and $y$ which is reflected by negative derivatives).

  • $\begingroup$ Thank you. That makes a lot of sense. $\endgroup$
    – Durden
    Jun 10, 2019 at 15:52

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