Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This appeared as a throwaway statement in a proof - that a strictly monotonic (increasing) transformation of a continuously distributed random variable (I am assuming that this means that the distribution function is continuous, not that the random variable is absolutely continuous) is also a continuously distributed random variable.

So the setup $X:\left(\Omega, \mathscr{F}, \mathbb{P}\right) \longmapsto \left(\mathbb{R}, \mathscr{B}(\mathbb{R}), \mathbb{P}_X\right)$ and $h: \mathbb{R} \longmapsto \mathbb{R}$ and $h(X) = Y$, where clearly $h$ needs to be measurable. So the claim is that if $h$ is monotonic, then $Y$ is a continuously distributed random variable.

A proof or a reference to a textbook would be appreciated.

share|cite|improve this question
What if $h$ is constant? – scineram Nov 2 '11 at 8:55
That is a good point; my fault, since I meant strict monotonicity. – Praetoria Nov 2 '11 at 8:58
Doesn't "$h$ is monotonic" imply $h$ is measurable (so that doesn't need to be a separate hypothesis)? – Michael Hardy Nov 2 '11 at 10:55
Yes, that was my (redundant) addition to the throwaway statement. – Praetoria Nov 2 '11 at 11:01

You are interested in what you call continuously distributed random variables, more commonly called random variables with atomless distributions. These are the random variables $X$ such that $\mathrm P(X=x)=0$ for every $x$.

Assume the function $h$ is (strictly) increasing and the distribution of $X$ is atomless. Then $h$ is measurable hence $Y=h(X)$ is a random variable and the task is to show that $\mathrm P(Y=y)=0$ for every $y$.

Fix some $y$, then $[Y=y]=[h(X)=y]=[X\in B]$, where $B=h^{-1}(y)=\{x\mid h(x)=y\}$. Let us study the set $B$. If $x'$ and $x''$ are both in $B$, then $h(x')=y=h(x'')$ hence, due to the strict monotonicity of $h$, both cases $x'\lt x''$ and $x''\lt x'$ are impossible. Thus, $B=\varnothing$ or $B$ is a singleton. In both cases, $\mathrm P(X\in B)=0$, hence $\mathrm P(Y=y)=0$ and you are done.

share|cite|improve this answer

Here I first assume $h$ is strictly increasing. The cumulative distribution function of $Y$ is $$ F_Y(y)=\Pr(Y \le y) = \Pr(h(X) \le y) = \Pr(X \le h^{-1}(y)) = F_X(h^{-1}(y)), $$ provided $y$ is not in an interval that is not in the image of $h$ because of a jump discontinuity.

The inverse of a strictly monotone function is continuous, but will be undefined on intervals corresponding to jump discontinuities of $h$. Since the probability that $Y$ is in any of those intervals is $0$, the c.d.f. $F_Y$ is constant on those intervals.

So on some intervals $F_Y$ is continuous because it's a composition of continuous functions. On some it's continuous because it's constant.

At the boundaries it's continuous for another reason: Say $h$ has a discontinuity at $x_0$. Then $y_0=\lim\limits_{x\to x_0-}h(x)$ exists and $y_1=\lim\limits_{x\to x_0+}h(x)$ exists. Then $\lim\limits_{y\to y_1+}h^{-1}(y) = x_0 = \lim\limits_{y\to y_0-}h^{-1}(y)$. For $y\in[y_0,y_1]$, the value of $F_Y$ will be constant. So $F_Y$ is continuous from the left at $y_0$, constant on $[y_0,y_1]$, and constinuous from the right at $y_1$, and the limits at $y_0$ and $y_1$ are equal to each other and to the constant value that $F_Y$ has on that interval.

For decreasing $h$, reverse the inequalities as needed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.