Let $X_1$, $X_2$ have a joint pdf and support set $S$. Suppose random variables $Y_1$, $Y_2$ are given by $Y_1=u_1(X_1,X_2)$, $Y_2=u_2(X_1,X_2)$ where the functions define one to one transformation that maps $S$ onto $T$ (support of $(Y_1,Y_2)$ ).

Let $f$ be joint pdf of $(X_1,X_2)$ and $g$ be joint pdf of $(Y_1,Y_2)$. Then can we say that if $f(x_1,x_2) > 0$ where $(x_1,x_2)$ is in the support of $(X_1,X_2)$ then $g(u_1(x_1,x_2), u_2(x_1,x_2)) > 0$ ? How can I prove this?


1 Answer 1


The conclusion is false, even in benign cases where the mapping is differentiable almost everywhere.

One way to guarantee its truth is for the map $(u_1,u_2):\mathbb{R}^2\to\mathbb{R}^2$ to be continuously differentiable.

Counterexamples exist even in one dimension (you may let $u_2$ be the identity if you insist on using two dimensions), so let's focus on $u=u_1$. Let $S=(-1,1) \subset \mathbb{R}^1$, let $X$ have a uniform distribution on $S$, and set $$u(x)=x^{1/3}=y.$$

Graph of $u$

This graph of $u$ demonstrates it is continuous, one-to-one, and differentiable everywhere but $0$.


  • $u$ is one-to-one onto the interval $(-1,1) = T$ because it has an inverse $x = y^3$.

  • The CDF of $X$ is $$F_X(x) = (1+x)/2$$ for $-1 \lt x \lt 1$. Therefore $X$ has a PDF $f_X$ and it is nonzero when $-1 \lt x \lt 1$, where it is constantly equal to $$1/2 = F_X^\prime(x) = f_X(x).$$

  • Thus, the CDF of $Y$ when $-1 \lt y \lt 1$ is $$F_Y(y) = \Pr(Y \le y) = \Pr(X^{1/3} \le y) = \Pr(X \le y^3) = (1+y^3)/2.$$

  • Consequently $Y$ has a PDF $f_Y$ given by the derivative of $F_Y$, equal to $$3y^2/2 = F_Y^\prime(y) = f_Y(y)$$ when $-1 \lt y \lt 1$.

The foregoing establishes that $X$, $Y$, and $u$ satisfy all the conditions in the question. However, although $f_X(0)=1/2$, $$f_Y(u(0)) = f_Y(0) = 3(0^2)/2 = 0.$$

There is the counterexample.

If you would like to see $f_Y$, here are the results of a quick simulation (followed by R code that will reproduce it). Now the problem is visually obvious: $u$ "stretches" all neighborhoods of $0$ so thinly that right at $0$ all probability vanishes. For this to happen, the derivative of $u$ in those neighborhoods had to become arbitrarily large.


Colors correspond to values of $X$. All histograms show the same amount of each color. The effect of $u$ is to squeeze the reds towards either end and stretch the cyan away from $0$. The amount of stretching is controlled by a parameter $h$: $u$ is defined through its inverse as $u^{-1}(y)=(1-h)y^3+h y$, so that the original $u$ corresponds to $h=0$ (at the right). This figure varies $h$ to help us see the gradual migration of probability away from $0$ as $X$ is slowly transformed into $Y$ at the far right.

x <- runif(1e5, -1, 1)
y <- abs(x)^(1/3) * sign(x)
hist(x, freq=FALSE, breaks=50)
hist(y, freq=FALSE, breaks=50)
  • $\begingroup$ Nice. I made the following mistake in a my attempt to answer: I wrongly stated that is all neighborhoods of $y$ have a positive measure, then the density is positive in $y$. I don’t know what I had in mind. $\endgroup$
    – Elvis
    May 26, 2016 at 15:26

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