How can I use Joint Distribution Function to Find the Probability for 2 random variables in a certain interval?

Question

We are given this density function :

$$ f(x,y) = \begin{cases} \frac{xy}{96}, & 0<x<4,& 1<y<5 \\ 0, & \text{otherwise} \end{cases}$$

and are supposed to find $P(X\ge3, Y\le2)$

rather than using the double integral how can we calculate this probability with the distribution function?

I calculated the joint dist func : $$F(x,y)=\frac{x^2(y^2-1)}{384}$$

then i tried : $P(X\ge3, Y\le2) = F(4, 2)-F(3,1)$

but this gives $\frac{48}{384}$ and the correct answer is $\frac{21}{384}$

By definition: $F(x,y)=P(X \leq x, Y \leq y)$, so it should be $F(4,2)-F(3,\color{red}{2})$. — Anurag A, Jul 01 '21 at 21:39
@AnuragA for Y<=2 shouldn't we include the complete interval just like we did with X? — g0x0, Jul 01 '21 at 21:46
First draw the rectangular region $[0,4] \times [1,5]$. Then $F(4,2)$ represents the probability over the region $[0,4] \times [1,2]$ and $F(3,2)$ will represent the probability over $[0,3] \times [1,2]$. Now you subtract one from the other. — Anurag A, Jul 01 '21 at 21:47

score 0 · Accepted Answer · answered Jul 01 '21 at 21:48

0

You want the green area of probability in this

The green and purple areas of probability together represent $F(4,2)$

while the purple area of probability is $F(3,2)$ not $F(3,1)$

so take the difference

answered Jul 01 '21 at 21:48

Henry

1 Answers1