5
$\begingroup$

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 $X$?

$\endgroup$
3
  • 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

3
$\begingroup$

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 $Y$.)

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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.

Also,

$$ \int_{0}^{\infty}\left(1-F_{X}(x)\right) \mathrm{d} x=\int_{0}^{\infty} P(X \geq x) \mathrm{d} x=\int_{0}^{\infty} \int_{x}^{\infty} f_{X}(t) \mathrm{d} t \mathrm{~d} x $$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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