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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.

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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

2 Answers 2

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.

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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.

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