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Given the following graph:

enter image description here

$$f_{X,Y}(x,y) = \begin{cases} a &, x,y \in OABCDE \\ 0 &, \text{otherwise}\end{cases}$$

My solution: $$f_{X,Y}(x,y) = \begin{cases} a &, 0 \leq x \leq 2, 0 \leq y \leq 2 \\ 0 &, \text{otherwise}\end{cases}$$ Is this correct?

Andrei0408
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1 Answers1

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Nope, You have described a square. $OABCDE$ is not a square.

$$f_{X,Y}(x,y) = \begin{cases} a& ,0\le x \le 2 , \max(0, x-1) \le y \le \max(1, x) \\ 0 & \text{,otherwise}\end{cases}$$

Furthermore, we can find what is $a$.

First find the area of $OABCDE$, $a$ is its reciprocal.

Siong Thye Goh
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  • To find a, could I also use the formula $\int_0^{2} \int_{max(0, x-1)}^{max(1,x)} a dydx = 1$? If so, how would I proceed to find the exact domain of the second integral? – Andrei0408 Jun 26 '21 at 10:06
  • Yup, you can use that, just that I think it's faster if you just view it as a geometry problem, you can shift some triangle around to form a rectangle to get the area quickly. – Siong Thye Goh Jun 26 '21 at 10:08
  • Thank you! I got a = 1/2 using the area. I have one more question: If I want to find the marginal PDF of X, on what domain to I integrate to find it? I mean I can't integrate on $max(0, x-1)$ and $max(1,x)$. – Andrei0408 Jun 26 '21 at 10:40