Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$f_{X,Y}(x,y) = \frac{2}{3} (x+2y)$ for $0 < x < 1, 0 < y < 1$; find $P(X\gt Y)$.

I got 1/9 by evaluating $$\int_0^1\int_0^{x-1} \frac{2}{3}(x+2y) dy dx$$

share|cite|improve this question
Your upper bound on $y$ in the (inner) integral is discrete but should be continuous. – bgins Mar 12 '12 at 10:16
In the inner integral, $x$ is a fixed number in the interval $(0,1)$ and so the upper limit $x-1$ of your inner integral is smaller than the lower limit. Now, if $y \in (x-1, 0)$, then $f_{X,Y}(x,y) = 0$ as per your definition, and so the integral should have been $$\int_0^1\int_0^{x-1} f_{X,Y}(x,y) \mathrm dx\mathrm dy = 0 \neq \int_0^1\int_0^{x-1} \frac{2}{3}(x+2y) \mathrm dx\mathrm dy$$. – Dilip Sarwate Mar 12 '12 at 11:58
up vote 1 down vote accepted

I will use the standard notation, with (lowercase) $f$ for the density function $f(x,y)=\frac23(x+2y)$. Then $$ P(X\gt Y) = \int_{0}^{1} \int_{0}^{x} f(x,y) \, dy \, dx \, . $$ If you were taking summations, in the discrete case, and $x,y$ had integer values, then your upper limit for $y$ would be $x-1$. But in the continuous case, the upper limit for $y$ must be $x$ (the integral only accumulates value over an interval, so you shouldn't worry about including the endpoint $y=x$).

That this is the density and not the CDF on $R=[0,1]^2$ is evident because $0\le f\le 2$ and $\int_{R}f(x,y)\,dxdy=1$ as shown below, where we simultaneously work out the answer. $$ \eqalign{ \int_{R}f(x,y)\,dxdy &= \int_{0}^{1} \int_{0}^{1} \frac23\left(x+2y\right) \, dy \, dx \\&= \frac23 \int_{0}^{1} \left[ xy+y^2 \right]_{0}^{1} \, dx \\&= \frac23 \int_{0}^{1} \left(x+1\right) \, dx \\&= \frac23 \left[ \frac12x^2+x \right]_{0}^{1} \, dx \\&= \frac23\cdot \frac32=1 } \qquad \qquad \eqalign{ P(X\gt Y) &= \int_{0}^{1} \int_{0}^{x} \frac23\left(x+2y\right) \, dy \, dx \\&= \frac23 \int_{0}^{1} \left[xy+y^2\right]_{0}^{x} \, dx \\&= \frac43 \int_{0}^{1} x^2 \, dx \\&= \frac43 \left[ \frac13 x^3 \right]_{0}^{1} \, dx \\&=\frac49 } $$

share|cite|improve this answer
It is a common convention in probability theory that $f$ is used for probability density functions and $F$ is used for cumulative probability distribution functions. I request that you change notation so as to avoid confusing generations of later readers of your answer. – Dilip Sarwate Mar 12 '12 at 11:51
I agree completely. – bgins Mar 12 '12 at 13:17
Thank you again :) – Josh holt Mar 12 '12 at 15:44

You want your bounds to represent your event, so to speak. The event x > y could be satisfied by any value of x, and any value of y such that y < x. Since y depends on x, I agree with the order of integration. But why do you have y ranging from 0 to x-1? That integral is not necessarily interpretable in the context of probability theory. How could you repair your upper bound on y to reflect the event x > y?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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