Compute $P(Y<3X)$ using joint PDF

Question

I'm given a joint pdf

$f_{X,Y}(x,y)=2e^{-x-y}, 0<x<y, 0<y $

and asked to compute $P(Y<3X)$. To do this, I let $Y=3X$ (the boundary) and found that the region of integration is under this line.

To find $P(Y<3X)$, it seems to me that the integral for this region be written as

$P(Y<3X)=\int_0^\infty \int_0^{3x} 2e^{-x-y} dy dx $

However this isn't right, as the boundaries are actually

$P(Y<3X)=\int_0^\infty \int_x^{3x} 2e^{-x-y} dy dx $

Why is this? I'm having a hard time seeing how to construct these boundaries (especially the fact that the $y$ boundary goes from $x$ to $3x$ instead of $0$ to $3x$)

Thank you.

Answer 1

Before we even consider $P(Y < 3X)$, note that we have the condition $0<x<y$. Because this specifies both the upper and lower bounds for $x$, the natural way to write the the integral over the support of $f_{X,Y}$ is:

$$ P(\Omega) = \int_0^\infty \int_0^{\color{red}y} 2 e^{-x} e^{-y} dx \, dy $$

You can confirm for yourself as a preliminary exercise that this equals 1. Note that the inner integral is over $x$ with $y$ appearing as bound; therefore it will be convenient if we first solve the inequality for $x$:

$$ \begin{align} y & < 3x & \\[0.7em] 3x & > y & \\[0.7em] x & > y/3 \end{align} $$

We can use this to restrict the inner integral by adjusting the lower bound of the inner interval. The upper bound of $y$ is retained from above, the lower bound of $y/3$ is new and comes from the inequality which defines our event.

$$ \begin{align} P(Y < 3X) & = \int_0^\infty \int_{\color{red}{\tfrac{y}{3}}}^{\color{red}y} 2 e^{-x} e^{-y} dx \, dy \\[1em] & = \int_0^\infty \Bigg( -2 e^{-x} e^{-y} \Bigg|_{\tfrac{y}{3}}^{y} \Bigg) dy \\[1em] & = \int_0^\infty (-2e^{-y}e^{-y}) - (-2e^{-y/3}e^{-y}) dy \\[1em] & = \int_0^\infty -2e^{-2y} + 2 e^{-4y/3} dy \\[1em] & = \frac{-2 e^{-2y}}{-2} + \frac{ 2 e^{-4y/3} }{-4/3} \Bigg|_0^\infty \\[1em] & = e^{-2y} - \frac{3}{2} e^{-4y/3} \Bigg|_0^\infty \\[1em] & = (0 - 0) - (1 - \frac{3}{2}) \\[1em] & = \frac{1}{2} & \square \end{align} $$

Answer 2

The boundary for $y$ could actually go from $-\infty$ to $3x$, not just $0$ if the inside expression was written as $f(x,y)$. Then, you’d be forced to write $x$ to $3x$ in the inner integral limits when substituting $2e^{-x-y}$ in place of the joint since it is actually $0$ when $y<x$.

Answer 3

If the joint PDF of $X,Y$ is $$f_{X,Y}(x,y)=2e^{-x-y}, 0<x<y, 0<y $$

Then

$$ P(Y < \theta X) = \int_0^\infty \int_x^{\theta x} 2e^{-x-y} dy dx $$

The inner lower bound is $x$ because of the $0 < x < y$ constraint. A good Calc 2 exercise is to calculate that integral and see that it equals

$$\frac{\theta-1}{\theta+1}$$

So when $\theta = 3$, this equals $(3-1)/(3+1) = 1/2$