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.

  • 3
    $\begingroup$ I suggest you draw a plot $\endgroup$ – Xi'an Mar 21 '19 at 21:41
  • $\begingroup$ And here is another way to solve this without any actual integration. $\endgroup$ – StubbornAtom Mar 21 '19 at 22:52
  • $\begingroup$ 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 $y=x$ $\endgroup$ – Sarina Mar 22 '19 at 2:28
  • $\begingroup$ Thank you StubbornAtom! I'm definitely looking that over as it looks much easier! $\endgroup$ – Sarina Mar 22 '19 at 2:29

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.

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

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  • $\begingroup$ 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)? $\endgroup$ – Sarina Mar 22 '19 at 2:27
  • $\begingroup$ It's one of the property of inequalities: $a < b \Leftrightarrow b > a$. $\endgroup$ – olooney Mar 22 '19 at 3:26
  • $\begingroup$ Haha, I'm so sorry. Thank you so much, this is so helpful!! $\endgroup$ – Sarina Mar 22 '19 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 $y<x$.

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  • $\begingroup$ Hi gunes, thank you! So is the first integral actually correct (not the second one?) $\endgroup$ – Sarina Mar 21 '19 at 20:46
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    $\begingroup$ 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. $\endgroup$ – gunes Mar 21 '19 at 20:51

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


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

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


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

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