finding probability of joint distribution.

Question

I am given a joint density function of two random variables $X$ and $Y$ which is as follows:

$$ f(x,y) = \left\{ \begin{array}{ll} \frac{1}{210}(2x+y) & \quad 2<x <6, 0<y<5 \\ 0 & \quad\text{otherwise} \end{array} \right. $$

Now I want to calculate the probability $P(X+Y < 4)$. Will I do it like this?:

$$ \int_{0}^{4}\int_{2}^{4} \frac{1}{210}(2x+y) dxdy $$

Answer 1

This is technically an exercise on multiple integrals, and in these cases you would realise that your job becomes easier if you can picture the desired region of integration.

Proceed step by step to focus on the domain of integration.

\begin{align} P(X+Y<4)&=E\left[\mathbf1_{X+Y<4}\right] \\&=\iint \mathbf1_{x+y<4}\,f(x,y)\,dx\,dy \\&=\frac{1}{210}\iint \mathbf1_{x+y<4}\,(2x+y)\mathbf1_{2<x<6,\,0<y<5}\,dx\,dy \\&=\frac{1}{210}\iint (2x+y)\,\mathbf1_{x+y<4,\,2<x<6,\,0<y<5}\,dx\,dy \end{align}

This is how the region $\{(x,y):x+y<4,\,2<x<6,\,0<y<5\}$ looks like:

Can you evaluate the double integral now, keeping any one of $x$ and $y$ fixed at a time depending on ease of computation ?

Answer 2

No, that integral gives $\mathbb{P}(2\leq X\leq 4,\, 0\leq Y \leq 4)$. For example, this includes $(4,4)$ which clearly does not satisfy $X + Y < 4$. You're integrating over a rectangular region, you should be integrating over a triangular region. Try

$$ \int_2^4\int_0^{4-x} \frac{1}{210}(2x+y)\,dy\,dx $$