How do I find this probability for the given joint distribution?

Question

$f(x,y) = x+y, 0<x<1 $ and $ 0<y<1$

I am required to find $P(X+Y>\frac{1}{2})$.

I calculated the ranges for x and y,keeping y dependent on x as follows: $\frac{1}{2} - x <y< 1 $ and $ \frac{1}{2}<x<1$. Then I performed the following integration — $$\int_\frac{1}{2}^1 \int_{\frac{1}{2} - x}^1 (x+y)dydx$$ to get $\frac{17}{24}$. But that is wrong according to the answer key. Where have I gone wrong?

Answer 1

Note 1: When $x>1/2$ and $y=1/2-x$ then $y<0$, which conflicts with the supported interval, so your lower bound of the inner integral is clearly wrong. When $1/2<x<1$ and $0<y<1$ then $x+y>1/2$ is satisfied.

Note 2: When $0<x<1/2$ and $1/2-x<y<1$ then the constraint of $x+y>1/2$ is satisfied, so you are missing an entire interval.

$${\mathsf P(X+Y>1/2)}~=~{\int_0^{1/2}\int_{1/2-x}^1 (x+y)~\mathrm d y~\mathrm d x}+{\int_{1/2}^1\int_0^1 (x+y)~\mathrm d y~\mathrm d x}$$

Answer 2

The reason why your answer is wrong is because the region of integration is wrong. In particular, you have assumed that $x+y > 1/2$ implies $x \in [1/2,1]$ when this is not the case; e.g., $x = y = 1/3$ satisfies $x+y > 1/2$ but $x \not\in [1/2, 1]$.

In order to compute the correct region, consider the line $x+y = 1/2$ in the Cartesian coordinate plane. This line passes through the points $(x,y) = (1/2, 0)$ and $(0, 1/2)$. So when you sketch the complementary region $x + y \le 1/2$, it is the triangle with vertices $$(0,0), (1/2, 0), (0, 1/2).$$ The desired region is therefore a pentagonal shape in the unit square $(x,y) \in [0,1] \times [0,1]$ with vertices $$(1/2,0), (1,0), (1,1), (0,1), (0,1/2).$$

Since the integration over this more complicated pentagon shape is not as straightforward as the integration over the complementary triangular region, it is easier to compute $$\Pr[X + Y \le 1/2] = \int_{x=0}^{1/2} \int_{y=0}^{1/2 - x} f_{X,Y}(x,y) \, dy \, dx$$ and then use the fact that $$\Pr[X + Y > 1/2] = 1 - \Pr[ X + Y \le 1/2].$$ This I leave as an exercise.

If you insist on computing the integral directly, it is necessary to split the region of integration into two parts; e.g., $$\Pr[X + Y > 1/2] = \int_{x=0}^{1/2} \int_{y=1/2-x}^1 f_{X,Y}(x,y) \, dy \, dx + \int_{x=1/2}^1 \int_{y=0}^1 f_{X,Y}(x,y) \, dy \, dx.$$ I invite you to perform the computation this way as well, and see if the results agree with the above approach.