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I want to calculate the marginal $f_X(x)$ of $f_{X, Y}(x, y) = 2 e^{-(x+y)} \mathbb 1_A (x,y)$ where $A = \{(x, y): 0 \leq y \leq x \}$.

It is clear that I have to integrate $f_{X, Y}$, but how do I handle this "$\mathbb 1_A (x, y)$" at the end? Sorry for the stupid question, but since this seems very basic to me, it's important that I understand how to compute these things.

StMan
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  • this recent discussion can be also useful https://math.stackexchange.com/questions/4128208/compute-mathbbex-y2-of-two-dependent-random-variables?noredirect=1#comment8539057_4128208 – tommik May 07 '21 at 08:24

2 Answers2

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Using the definition you get

$$f_X(x)=\int_{-\infty}^{\infty}f_{XY}(x,y)dy$$

that in your case is

$$f_X(x)=2e^{-x}\int_{0}^{x}e^{-y}dy=2e^{-x}(1-e^{-x})$$

the other marginal is

$$f_Y(y)=2e^{-y}\int_y^{\infty}e^{-x}dx=2e^{-2y}$$

that is $Y\sim Exp(2)$

tommik
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All you need to do, is to apply the set $A$ in your integration. In this case, $$ f_X(x)=\int_{y\ :\ (x,y)\in A} 2e^{-x-y}dy=\int_0^x 2e^{-x-y}dy. $$ If you wished to find $f_Y(y)$, it should be $$ f_Y(y)=\int_{x\ :\ (x,y)\in A} 2e^{-x-y}dx=\int_y^\infty 2e^{-x-y}dy. $$

Mostafa Ayaz
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