Prove the probability-integral transformation, i.e., if $F_X$ is continuous, then $F_X(x)\overset{d}{=}\mathsf{Unif}(0,1)$, by finding the mgf of the random variable $Y=F_X(X)$ where $X$ is absolutely continuous and has cdf $F_X$.

This is easy to show by noting that

$$\mathsf P\left(F_X(X)\geq x\right)= \mathsf P\left(X\geq F_X^{-1}(x)\right) = 1-F_X\left(F_X^{-1}(x)\right)=1-x\tag{1}$$

but I'm having trouble showing this by mgf. Since the mgf of a $\mathsf{Unif}(0,1)$ random variable is given by $\frac{e^t-1}{t}$ then we need to show that

$$M_Y(t)=\int_{-\infty}^{\infty} e^{tY} f_Y(y)dy = \frac{e^t-1}{t}$$

The only way I can think about showing this is by noting that

$$\begin{align*} M_Y(t) &=\int_{-\infty}^{\infty} e^{ty} f_Y(y)dy\\\\ &= \int_{-\infty}^{\infty} e^{ty} dF_Y(y)\\\\ &= \int_{0}^{1} e^{ty} dy\\\\ &=\frac{e^t-1}{t} \end{align*}$$

but this requires the knowledge that $F_Y(y)=y$ which is already sufficient to showing that $Y\sim\mathsf{Unif}(0,1)$. Is there a way to show this without making use of (1)?


1 Answer 1


Your idea is the correct one, but you need to take a slightly different route to get to the end.

To be rigorous, we ought to be working with the characteristic function (cf) of $Y=F_X(X),$ which (unlike the mgf) always exists and determines the distribution. (If you don't like this, simply erase all occurrences of "$i$" in the following.)

Let's just plug stuff into the definition of the cf, which begins these equalities.

$$\begin{aligned} E\left[\exp(it\,Y)\right] &= E\left[\exp\left(it\,F_X(X)\right)\right] & \text{definition of } Y\\ &= \int_{\mathbb{R}} \exp\left(it\, F_X(x)\right)\,\mathrm{d}F_X(x)& \text{expectation formula} \\ &= \int_0^1 \exp(it\,y)\,\mathrm{d}y & \text{formal substitution }y=F_X(x)\\ &= \frac{e^{it}-1}{it}. \end{aligned}$$

This derivation only required knowing that

  1. $F_X$ is the cdf of a continuous variable (if you use the Lebesgue-Stieltjes integral) or at least differentiable (if you use the Lebesgue integral) and

  2. $F_X:\mathbb{R}\to[0,1]$ is a one-to-one function.

No assumptions about $F_Y$ were made (or harmed) in the derivation of this result.

  • $\begingroup$ Thanks! The only thing I don't understand is where $d F_X(x)$ comes from. Why is it not $d F_Y(y)$? $\endgroup$
    – Remy
    Commented Sep 8, 2020 at 20:57
  • 1
    $\begingroup$ By the LOTUS, the expectation of a function $g(X)$ is $$E[g(X)] = \int_{\mathbb{R}} g(x)\,\mathrm{d}F_X(x).$$Apply this to the function $g(x)=\exp(itF_X(x)).$ $\endgroup$
    – whuber
    Commented Sep 8, 2020 at 21:02
  • 2
    $\begingroup$ "No assumptions about $F_Y$ were made (or harmed) in the derivation of this result." Nice! $\endgroup$ Commented Sep 8, 2020 at 21:14

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