How to find bounds of integration for this question?

Question

I am currently studying joint pdfs and am struggling with finding the bounds of integrations for this problem. Consider a pair of random variables X, Y with constant joint density on the triangle with vertices at $(0, 0)$, $(3, 0)$, and $(0, 3)$. Find $P(X +Y > 2)$.

The fact that it says constant density tells me that this is continuous uniform with area $0.5 \cdot 3 \cdot 3 = 4.5$ so $f(x) = 2/9$

Since $P(X +Y > 2)$ I thought that this would mean that the bounds of integration would be $\displaystyle\int_2^3\int_{2-x}^3\cfrac{2}{9}dydx = \cfrac{7}{9}$

But my book says the answer is $\cfrac{5}{9}$

What am I misunderstanding here? What are the proper bounds of integration and why?

Answer 1

If you are using integral, then notice it is the shaded area you need to integrate over. So easier is to integrate over the unshaded part and subtract from $1$.

So the integral should be

$P(X+Y \lt 2) = \displaystyle\int_0^2\int_{0}^{2-y} \cfrac{2}{9} \ dx \ dy = \cfrac{4}{9}$

So $P(X+Y \gt 2) = \displaystyle 1 - \frac{4}{9} = \frac{5}{9}$.

Answer 2

The endpoints of the integral are wrong in a few places: it should be $$ \int_0^3\int_{\max\{2-x,0\}}^{3-x}\cfrac{2}{9} \, dy \, dx = \frac59. $$

• When you put the lower limit of the first integral equal to $2$, that prohibited $X$ from being less than $2$;
• Putting $2-x$ as the lower limit of the second integral is an error because $2-x$ can be negative, and $f(x,y)=0$ for $y<0$;
• Having $3$ as the upper limit of the second integral allowed pairs $(x,y)$ with $x+y>3$, contrary to the initial domain of the random variables.

Answer 3

A direct calculation can be made by observing that the entire domain is an isosceles right triangle with side length $3$ so the area is $\frac{9}{2}$. The condition is exclusion of a similar right triangle with side length $2$ and area$\frac{4}{2}$ leaving the area for $P(X+Y\gt 2)$ as $\frac{5}{2}$ and a probability of $\frac{5}{9}$.