Question.
Joint probability function of continuous probability X, Y is here :
$f_{X,Y}(x,y) = k(|x|-|y|) \ \ \ \ \ \ \ \ \ \ (-1< y< x< 2)$
Then what is k?
I mean how can I differentiate absolute value range??
Please help me with a solution.
2026-04-04 03:52:50.1775274770
Joint pdf of X and Y with absolute value
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Joint pdf integrates out to 1, so you should have $$ 1 = \iint_{\mathbb{R}^2} f_{X,Y} (x,y) dxdy $$ Substituting you get $$ 1 = k \int_{-1}^2 \int_{-1}^x (|x| - |y|) dydx $$ can you finish this?