# Finding probability $P(X+Y < 1)$ with CDF

Suppose I have a Cumulative Distribution Function like this:

$$F(x,y)=\frac { (x\cdot y)^{ 2 } }{ 4 }$$

where $0<x<2$ and $0<y<1$.

And I want to find the probability of $P(X+Y<1)$.

Since $x<1-y$ and $y<1-x$, I plug these back into the CDF to get this:

$$F(1-y,1-x)=\frac { ((1-y)\cdot (1-x))^{ 2 } }{ 4 }$$

Because of the constraint where $0<x<2$ and $0<y<1$, I integrate according to the range of values:

$$\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 2 }{ \frac { ((1-y)\cdot (1-x))^{ 2 } }{ 4 } dxdy } } =\frac { 1 }{ 18 }$$

This answer, however, is incorrect. My intuition for doing this is that because the two variables are somewhat dependent on each other to maintain the inequality of less than $1$, I want to "sum"(or integrate) all the probabilities within the possible range of values of $x$ and $y$ that satisfy the inequality. Somehow, the answer, which is $\frac{1}{24}$, doesn't seem to agree with my intuition.

What have I done wrong?

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Your intuition would be correct if this were the PDF, but it's the CDF. –  yohBS Mar 11 '12 at 8:51

We have the cumulative distribution function (CDF)

$$F_{X,Y}(x,y)=\int_0^y\int_0^x f_{X,Y}(u,v)dudv=\frac{(xy)^2}{4}.$$

Differentiate with respect to both $x$ and $y$ to obtain the probability density function (PDF)

$$f_{X,Y}(x,y)=\frac{d^2}{dxdy}\frac{(xy)^2}{4}=xy.$$

Finally, how do we parametrize the region given by $x+y<1$ inside the rectangle? Well, $x$ must be nonnegative so $y$ can be anything from $0$ to $1$, and simultaneously $x$ must be between $0$ and $1-y$;

$$P(X+Y<1)=\int_0^1\int_0^{1-y} xy \; dx dy =\int_0^1 \frac{(1-y)^2}{2}ydy=\frac{1}{24}.$$

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what if we ask to parametrice x+y>=1? Thanks –  user57191 Jan 14 '13 at 1:47
@user57191 $P[x+y\geq 1] = 1 - P[x+y <1]$ –  David Simmons Jul 17 '14 at 11:11