# Given the Joint PDF find the value of $c$

Given the following Joint PDF

$$\begin{equation*} f(x,y) = \left\{ \begin{array}{ll} c & \quad -1< x \leq 1 ; \lvert x \rvert

$$1.$$ Find the value of $$c$$.

My attempt:

Since Given function is PDF, then by the the property of PDF we have:

$$\int_{-1}^1\int_{-y}^{y}c*dxdy=1$$

$$c\int_{-1}^1(2y)dy=1$$

$$c(1^2-(-1)^2)=1$$

$$c*0 = 1$$

$$0=1$$

Where am I making the mistake? Please guide me. Why value of $$c$$ is not being found?

• What is the actual support? You are integrating over $-1<y<1, -y < x < y$ (half of which is a negative area). Also $-1<x<1, \lvert x\rvert <y$ is invalid for a uniform distribution.. Commented Apr 23, 2021 at 8:10
• The question contains an error, because the region for which $-1 < x \le 1$ and $|x| < y$ has infinite area. Commented Apr 23, 2021 at 8:11

Probabily the region is $$\{(x,y) \in \mathbb{R}^2:-1
Assuming this, $$f_X(x)=c$$ is uniform over the $$(x,y)$$ support that is a triangle with area 1 thus it is self evident that $$c=1$$. If the correct exercise is different, the procedure is the same....do a drawing of the support area, calculate it and consider as $$f(x)$$ its reciprocal