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

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

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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
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2 Answers

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. $$ \eqalign{ \int_{R}f(x,y)\,dxdy &= \int_{0}^{1} \int_{0}^{1} \frac23\left(x+2y\right) \, dy \, dx \\&= \frac23 \int_{0}^{1} \left[ xy+y^2 \right]_{0}^{1} \, dx \\&= \frac23 \int_{0}^{1} \left(x+1\right) \, dx \\&= \frac23 \left[ \frac12x^2+x \right]_{0}^{1} \, dx \\&= \frac23\cdot \frac32=1 } \qquad \qquad \eqalign{ P(X\gt Y) &= \int_{0}^{1} \int_{0}^{x} \frac23\left(x+2y\right) \, dy \, dx \\&= \frac23 \int_{0}^{1} \left[xy+y^2\right]_{0}^{x} \, dx \\&= \frac43 \int_{0}^{1} x^2 \, dx \\&= \frac43 \left[ \frac13 x^3 \right]_{0}^{1} \, dx \\&=\frac49 } $$

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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
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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?

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