Let $X$ be a random variable with distribution function $F_X$. Consider $$P=\int_0^\infty (1-F_X(x))e^{-x}dx.$$ Because $1-F_X(x)$ is the probability of $X>x$ and $e^{-x}$ is the pdf of an exponential random variable (with $\lambda=1$), this integral $P$ apparently equals the probability of $X>Y$ where $Y$ is an exponential random vairbale with $\lambda=1$.

Can anyone explain how we can conclude that $P$ is the probability of the event $X>Y$? I guess $X$ is defined over the same probability space as $X$?

  • 3
    $\begingroup$ Use the equation (correct only for independent variables $X$ and $Y$) $$\Pr(X\gt Y)=E_Y[E_X[\mathcal{I}(X\gt Y)]]$$ and evaluate the inner expectation using $F_X$ and the outer expectation by integrating against $f_Y.$ For the comparison $X\gt Y$ to make any sense at all, you must consider $(X,Y)$ to be defined on a common probability space. $\endgroup$
    – whuber
    Commented Feb 25, 2021 at 23:01
  • $\begingroup$ @whuber cool, thank you for the quick response. If you leave a short answer, I can accept and upvote it :) $\endgroup$
    – Alex
    Commented Feb 25, 2021 at 23:08
  • $\begingroup$ @Alex I have no idea how to sort this out nor I understand what I wrote, I just copied and pasted few lines form one another site. You may not reward me for copy and paste. Also I am not sure if this answers your question, but looks both X and Y are exponential random variables. $\endgroup$ Commented Feb 26, 2021 at 0:30

2 Answers 2


I am going to assume that $X$ is a non-negative continuous random variable so that $F_X(0) = 0$.

Let $Y \sim \text{Exp}(1)$ be independent of $X$ and note that $\mathbb{P}(Y < x) = 1-e^{-x}$. To get the result you want, define the functions $u(x) \equiv 1-F_X(x)$ and $v(x) \equiv -e^{-x}$ and use integration by parts to get:

$$\begin{align} \int \limits_0^\infty (1-F_X(x)) e^{-x} \ dx &= \int \limits_0^\infty u(x) \ v'(x) \ dx \\[6pt] &= \Bigg[ u(x) \ v(x) \Bigg]_{x=0}^{x \rightarrow \infty} - \int \limits_0^\infty u'(x) \ v(x) \ dx \\[6pt] &= \Bigg[ -(1-F_X(x)) e^{-x} \Bigg]_{x=0}^{x \rightarrow \infty} - \int \limits_0^\infty f_X(x) \ e^{-x} \ dx \\[6pt] &= \Bigg[ 0-(-1) \Bigg] - \int \limits_0^\infty f_X(x) \ e^{-x} \ dx \\[6pt] &= 1 - \int \limits_0^\infty f_X(x) \ e^{-x} \ dx \\[6pt] &= \int \limits_0^\infty f_X(x) \ (1-e^{-x}) \ dx \\[6pt] &= \int \limits_0^\infty f_X(x) \ \mathbb{P}(Y<x) \ dx \\[12pt] &= \mathbb{P}(Y<X). \\[6pt] \end{align}$$

(The final step follows from the law of total probability and the independence of $X$ and $Y$.)

  • $\begingroup$ Why do you assume $F_X(x) = 0$, and would you like to say $X$ is not exponential RV? $\endgroup$ Commented Feb 26, 2021 at 9:41
  • $\begingroup$ If $F_X$ is unknown distribution how can you possible calculate probabilities. $\endgroup$ Commented Feb 26, 2021 at 9:43
  • $\begingroup$ Sorry, that was a typo --- now edited to say that $F_X(0)=0$, which follows from assuming that $X$ is a non-negative continuous random variable. $\endgroup$
    – Ben
    Commented Feb 26, 2021 at 21:53

Can anyone explain how we can conclude that P is the probability of the event $\mathbb P(X>Y)$?

Since $1-F_{X}(x)$ also called survival function is $\int_{x}^{\infty} f_{X}(t) \mathrm{d} t = P(X \geq x)$ because $F_{X}(x) = \int_{0}^{x} f_{X}(t) \mathrm{d} t$ where $\int_{0}^{\infty} f_{X}(t) \mathrm{d} t =1$, where $f_{X}(x) = e^{-x}$ is our PDF.


$$ \int_{0}^{\infty}\left(1-F_{X}(x)\right) \mathrm{d} x=\int_{0}^{\infty} P(X \geq x) \mathrm{d} x=\int_{0}^{\infty} \int_{x}^{\infty} f_{X}(t) \mathrm{d} t \mathrm{~d} x $$ Then change the order of integration $$ =\int_{0}^{\infty} \int_{0}^{t} f_{X}(t) \mathrm{d} x \mathrm{~d} t=\int_{0}^{\infty}\left[x f_{X}(t)\right]_{0}^{t} \mathrm{~d} t=\int_{0}^{\infty} t f_{X}(t) \mathrm{d} t $$ Then substitute

$$ = \int_{0}^{\infty} x f_{X}(x) \mathrm{d} x = \mathbb E(X) $$

I copied it all.

I read before you may omit $F_{X}(x)$ and write just $F(x)$ same for the $f(x)$ when you know what RV you deal with.


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