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Given the following Joint PDF

\begin{equation*} f(x,y) = \left\{ \begin{array}{ll} c & \quad -1< x \leq 1 ; \lvert x \rvert <y \\ 0 & otherwise \end{array} \right. \end{equation*}

$1.$ 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?

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  • $\begingroup$ 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.. $\endgroup$ Commented Apr 23, 2021 at 8:10
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    $\begingroup$ The question contains an error, because the region for which $-1 < x \le 1$ and $|x| < y$ has infinite area. $\endgroup$
    – heropup
    Commented Apr 23, 2021 at 8:11

1 Answer 1

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Probabily the region is $\{(x,y) \in \mathbb{R}^2:-1<x<1;|x|<y<1\}$

It is understood that this assumption is only one of the infinity possibilities...I assumed this only to explain to the OP how the exercise could work...

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

enter image description here

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