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

I'm given a joint pdf

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

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.

• I suggest you draw a plot – Xi'an Mar 21 at 21:41
• And here is another way to solve this without any actual integration. – StubbornAtom Mar 21 at 22:52
• Thank you Xi'an! I drew a plot to come up with the bounds in the first equation, as it looked like the y-boundary went from $0$ to $3x$;; I'm not super sure why they started from $y=x$ – Sarina Mar 22 at 2:28
• Thank you StubbornAtom! I'm definitely looking that over as it looks much easier! – Sarina Mar 22 at 2:29

Before we even consider $$P(Y < 3X)$$, note that we have the condition $$0. Because this specifies both the upper and lower bounds for $$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}

• Thank you so much! This might be a dumb question, but how did you go from $x<3x$ to $3y>y$ (when solving the inequality for x)? – Sarina Mar 22 at 2:27
• It's one of the property of inequalities: $a < b \Leftrightarrow b > a$. – olooney Mar 22 at 3:26
• Haha, I'm so sorry. Thank you so much, this is so helpful!! – Sarina Mar 22 at 19:43

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.

• Hi gunes, thank you! So is the first integral actually correct (not the second one?) – Sarina Mar 21 at 20:46
• on the contrary, the second one is correct because joint pdf is $0$ when $y<x$. The first one’d only be correct if you write $f(x,y)$ to the inside of the integral, in which not only $0,3x$ but also $-\infty, 3x$ would be correct. – gunes Mar 21 at 20:51

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

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 $$(3-1)/(3+1) = 1/2$$

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