Find $P(X>Y)$ if $X,Y$ are independent exponential random variables

Question

Let $ X,Y $ be two independent exponential random variables with means $ 1 $ and $3$, respectively. Find $P(X>Y)$.

I know $F(X,Y)$ is $\iint 3e^{-x-3y}\,dx\,dy$. But I'm confused on the limits of $P(X>Y)$.

Answer 1

It helps to draw a picture of the plane, and draw the region containing points $(x,y)$ such that $x<y$.

In the end, either $\int_0^\infty \int_y^\infty \cdots \, dx \, dy$ or $\int_0^\infty \int_0^x \cdots \,dy \, dx$ will workwork.

Answer 2

We want $x\ge y$ for the integral, and $y$ should range from $0$ to $\infty$. So the integral is $$\int_{y=0}^{\infty}\int_{x=y}^{\infty}\frac{1}{3}e^{-x-\frac{1}{3}y}\, dx \, dy.$$

(I realised the PDF you wrote was incorrect because we are told the means. If an exponential random variable has mean $\beta$, then its PDF is $\frac{1}{\beta}e^{-\frac{x}{\beta}}$ (for $x \ge 0$).)

Answer 3

This answer is not completely okay. It assumes that $\lambda=3$ (this on base of the integral that was mentioned in the question) while actually $\lambda=\frac13$. The idea is the same, though and I have decided to leave it like this. See the answer of Minus One-Twelfth (who observed the mistake and edited) for a correct version.

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Let $\chi(x,y)$ take value $1$ if $x>y$ and take value $0$ otherwise.

Then here:

$$P(X>Y)=\mathbb E[\chi(X,Y)]=\int\int\chi(x,y)f_X(x)f_Y(y)dxdy=\int_0^{\infty}\int_y^{\infty}3e^{-x-3y}dxdy$$

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edit (I could not resist the comment)

$\begin{aligned}\int_{0}^{\infty}\int_{y}^{\infty}3e^{-x-3y}dxdy & =3\int_{0}^{\infty}e^{-3y}\int_{y}^{\infty}e^{-x}dxdy\\ & =3\int_{0}^{\infty}e^{-3y}\left[-e^{-x}\right]_{y}^{\infty}dy\\ & =3\int_{0}^{\infty}e^{-3y}e^{-y}dy\\ & =3\int_{0}^{\infty}e^{-4y}dy\\ & =3\left[-\frac{1}{4}e^{-4y}\right]_{0}^{\infty}\\ & =\frac{3}{4} \end{aligned} $

Answer 4

we have $$f_X(x) = e^{-t}, f_Y(y) = \frac{1}{3}e^{-\frac{y}{3}}$$ Hence $$P(Y \lt y) = F_Y(y) = \int_{-\infty}^{y} f_Y(y^\prime) dy^\prime = \int_{0}^{y} \frac{1}{3}e^{-\frac{y^\prime}{3}} dy\prime = 1 - e^{\frac{y}{3}}$$ So $$P(X>Y) = \int_{0}^{\infty} \int_{0}^{t} f_X(t) f_Y(y) dydt = \int_{0}^{\infty}f_X(t)P(Y \lt t)dt$$ $$= \int_{0}^{\infty}e^{-t}(1-e^{-t/3})dt = \int_{0}^{\infty}e^{-t}-e^{-4t/3}dt=1/4$$