5
$\begingroup$

I'd like some help solving this problem about multivariate probability densities.

Let the random variables X and Y have the joint density f(x,y) = 1/y for 0 < x < y < 1 and 0 otherwise. Find P(X+Y > 0.5).

My problem is that any way I know of to set up the double integral, I get an integral of 1/y (or, switching to variables s and t, 1/t) with a lower bound of integration of 0, but such an integral doesn't converge. If I didn't know that the answer was 0.6534, I would assume that the lower bound of integration in the vertical direction should be x, but that produces a result that includes ln x, not a constant.

I emailed my professor asking for help, but her response was not helpful to me. Maybe it will be to you. She said, "you missed the condition for the density to be positive, so you have to integrate over a different area. And to make things simpler, calculate P(X+Y<0.5) and subtract from 1."

I understand why it's often easier to find a complementary probability and subtract that from 1, but here I don't see how it would help because I don't know how to make the lower bound of integration in the vertical direction anything other than 0 or x. I must be trying to integrate over the wrong region.

The region that I think is correct is the region of the unit square above the line x+y=0.5. That is, a trapezoid with vertices (0, 0.5), (0, 1), (1, 1), (1, 0), and (0.5, 0).

How would you define the region differently, or how would you integrate differently? Thanks so much for your help.

UPDATE: I figured out that I did indeed have the wrong region of integration. When my professor said I missed the condition for the density to be positive, she was implying that I needed to enforce the condition that Y > X. So I needed to add the line y=x to my graph and restrict the integration to the area above it (but still below y=1, obviously).

Thus, the region of integration became the quadrilateral with vertices (0, 1/2), (0, 1), (1, 1), and (1/4, 1/4). The only way I know how to do integration requires dividing this region into two sub-regions. To divide them horizontally, with dx on the outside and dy on the inside, these are the two double integrals (for the left and right sub-regions, respectively):

$ P(X+Y>0.5) = \int_0^.25 \int_{0.5-x}^1 \frac{1}{y}\,dy\,dx + \int_{0.25}^1 \int_x^1 \frac{1}{y}\,dy\,dx $

The left integral equals 0.25, and the right integral equals 0.75 + 0.25(ln 0.25), for a total of 1 + 0.25(ln 0.25) = 0.6534.

$\endgroup$
  • $\begingroup$ I divided the trapezoidal region into two regions: a top rectangular region from y=0.5 to y=1, and a bottom trapezoidal region from y=0 to y=0.5. The top rectangle gives me no problems. For the bottom trapezoidal region, I did the x-direction (switching to variable s) on the inside and the y-direction (variable t) on the outside. The inner integral was from 0.5 - y (the line x+y=0.5) to 1, and the integrand was 1/t ds. The outer integral was from 0 to 0.5, with an integrand of (1+2y)/2t dt. $\endgroup$ – John Petrie Oct 24 '15 at 21:13
  • $\begingroup$ Here, I tried to code it in LaTex. It looks almost just like this: $\int_0^{0.5}\int_{0.5-y}^{1} \frac{1}{t} \;ds\;dt$ $\endgroup$ – John Petrie Oct 24 '15 at 21:42
  • $\begingroup$ Draw a picture. Another illustrated example is at stats.stackexchange.com/a/95725. $\endgroup$ – whuber Oct 25 '15 at 15:58
6
$\begingroup$

The use of multiple integrals requires keeping track of all constraints. A neat approach works through indicators functions, using as many as needed. The probability $\mathbb{P}(X+Y > 0.5)$ can be represented as \begin{align*}\underbrace{\iint_{[0,1]^2}}_{\substack{\text{maximum range}\\\text{for $x$ and $y$}}} \mathbb{I}_{x+y>0.5}\,\mathbb{I}_{0<x<y<1}\,\frac{1}{y}\text{d}x\text{d}y&=\underbrace{\overbrace{\int_0^1\,\int_0^y}^{\substack{\text{outer limits:}\\\text{range of $y$}}}}_{\substack{\text{inner limits:}\\\text{conditional}\\\text{range of $x$}}} \mathbb{I}_{x+y>0.5}\,\frac{1}{y}\text{d}x\text{d}y\\ &=\int_0^1\,\int_0^y \mathbb{I}_{x>0.5-y}\,\frac{1}{y}\text{d}x\text{d}y\\ &=\int_0^1\,\int_{\max(0,0.5-y)}^y \underbrace{\mathbb{I}_{y>0.5-y}\,}_{\substack{\text{as $x<y$}\\\text{and $x>0.5-y$}}}\frac{1}{y}\text{d}x\text{d}y \end{align*} and \begin{align*}\int_0^1\,\int_{\max(0,0.5-y)}^y \mathbb{I}_{y>0.5-y}\,\frac{1}{y}\text{d}x\text{d}y&=\int_0^{0.5}\mathbb{I}_{y>0.25}\,\int_{0.5-y}^y \frac{1}{y}\text{d}x\text{d}y+\overbrace{\int_{0.5}^1\,\int_{0}^y}^{\substack{\text{no constraint on}\\\text{$y$ when $y>0.5$}}} \frac{1}{y}\text{d}x\text{d}y\\ &=\int_{0.25}^{0.5}\,\int_{0.5-y}^y \frac{1}{y}\text{d}x\text{d}y+\int_{0.5}^1\,\int_{0}^y \frac{1}{y}\text{d}x\text{d}y \end{align*} From there you should be able to conclude about the value of this probability, $1-\log(\sqrt{2})$.

$\endgroup$
  • $\begingroup$ Thank you. I've never heard of indicator functions before, but that roughly makes sense. $\endgroup$ – John Petrie Oct 25 '15 at 16:45
  • $\begingroup$ @xian, No, thank you. I did want to update my post indicating that I had the region of integration wrong and suggesting a more "basic" way to solve it (without indicator functions), but I wasn't sure if that was frowned upon if someone else has already given a correct and complete answer. $\endgroup$ – John Petrie Oct 27 '15 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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