Given the continuous random variable $X$ with cumulative distribution function $F_{X}$, find $E[F_{X}(X)]$.

Attempt at solution:

I understand that the expected value, $E[X]$, of a random variable, $X$, is $\int^{+\infty}_{-\infty} x f_{X}(x)\operatorname{d}x$, where $f_{X}$ is the probability density function.

However, I'm a little thrown off by the wording of the question. Is $E[F_{X}(X)]$ the same thing as $E[X]$?

  • $\begingroup$ That is strange wording to me as well. I could read it as saying "what is the expected value of the cumulative distribution function itself", in which case you'd compute $F_X(x)$ and then take its average value with respect to $f_X(x)$. But it's far from clear. $\endgroup$ Jul 22, 2014 at 5:01
  • $\begingroup$ Ok, thank you for the input. I will attempt to see if I can contact the instructor to try to get a better grip on what he was asking. $\endgroup$
    – Swamp G
    Jul 22, 2014 at 5:08
  • 1
    $\begingroup$ The question should read: "Given the continuous random variable $X$ with cumulative distribution function $F_X$, find $E[F_X(X)]$". $\endgroup$
    – Did
    Jul 22, 2014 at 5:53

2 Answers 2


You are correct that for a continuous random variable, $X$, with probability density function $f_X$, the expected value is: $$\mathbb{E}[X]=\int_{-\infty}^{+\infty} x\cdot f_X(x)\operatorname{d}x$$ This can be extended. Let $g$ be a function of the continuous random variable $X$. The expected value of this function is: $$\mathbb{E}[g(X)]=\int_{-\infty}^{+\infty} g(x)\cdot f_X(x)\operatorname{d}x$$ So, for example, $\mathbb{E}[X^2]=\int\limits_{-\infty}^{+\infty} x^2\cdot f_X(x)\operatorname{d}x$.

Thus the expected value of the cumulative distribution function, $F_X$, is: $$\mathbb{E}[F_X(X)]=\int_{-\infty}^{+\infty} F_X(x)\cdot f_X(x)\operatorname{d}x$$ However, by definition the probability density function is the derivative of the cumulative distribution function. $f_X(x) =\frac{\operatorname{d} F_X(x)}{\operatorname{d}x }$ $$\begin{align}\therefore \mathbb{E}[F_X(X)] & =\int_{0}^{1} F_X(x)\operatorname{d}F_X(x) \\ & = \left[\tfrac 1 2F_X(x)^2\right]_{F_X(x)=0}^{F_X(x)=1} \\ & = \tfrac 12 \end{align}$$


Letting $Y=F_{X}(X)$, the key is find the distribution of $Y$ it turns out $Y\sim Uni(0,1)$. To show a quasi proof of this I will look at case $F_{X}(X)$ is invertible (which is case for many continuos distribution) so we know that distribution are characterized by their CDFs so we have,

$$F_{Y}(y)=P(Y\leq y)=P(F_{X}(X)\leq y)$$ $$=P(F^{-1}_{X}(F_{X}(X))\leq F^{-1}_{X}(y))=P(X\leq F^{-1}_{X}(y))=F_X(F^{-1}_{X}(y))=y \textrm{ for 0<y<1}$$

where this is CDF of $Uni(0,1)$

From above the we see that $E(Y)=E(F_{X}(X))=\frac{1}{2}$

  • 3
    $\begingroup$ Also just a cool fact, the above fact gives us that $F^{-1}(Y)=X$ where $Y\sim Uni(0,1)$ thus this means that any distribution that has an inverse that can be found than as long as we can simulate form $Uni(0,1)$ we can simulate from any of those distributions too $\endgroup$
    – Kamster
    Jul 24, 2014 at 4:02

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