Probability density function (PDF) problem

Question

Problem

$X$ and $Y$ random variables, the common probability density function of $X$ and $Y$ is given as follows: $$f(x,y)= \begin{cases} ke^{-x-y},&\textrm{when } x\geqslant y\geqslant 0\\ 0\;,&\textrm{otherwise } \end{cases} $$

a) Find the constant $k$

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Proposed Solution

$$\int_{y=0}^{\infty}\int_{x=0}^{\infty} ke^{-x-y} dxdy = 1$$

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Is any of my work correct? Any feedback is much appreciated, and if you think I should add more details to my calculations, please point it out and I will edit my work accordingly.

Thank you for your time.

Answer 1

a) $\int_0^{\infty} \int_y^{\infty} k\cdot e^{-x-y} \, dx \, dy$

inner integral: $\int_y^{\infty} k\cdot e^{-x-y} \, dx=k\cdot e^{-y}\int_y^{\infty} e^{-x} \, dx$

$$=k\cdot e^{-y}\cdot \left(-e^{-x}\bigg|_y^{\infty}\right)=k\cdot e^{-y}\cdot(-0-(-e^{-y}))=k\cdot e^{-2y}$$

Now the outer integral: $\int_0^{\infty} k\cdot e^{-2y} \, dy=k\cdot \int_0^{\infty} e^{-2y} \, dy$

$$=k\cdot \left(-\frac12 \cdot e^{-2y}\bigg|_0^{\infty}\right)=k\cdot(-0-(-\frac12))=\frac{k}2=1$$

$\Rightarrow k=...$

Answer 2

Reprising this post

After understanding that $k=2$, your joint density is the following

$$f_{XY}(x,y)=2e^{-x}e^{-y}\mathbb{1}_{[0;+\infty)}(x)\mathbb{1}_{[0;x]}(y)$$

or equivalently

$$f_{XY}(x,y)=2e^{-x}e^{-y}\mathbb{1}_{[0;+\infty)}(y)\mathbb{1}_{[y;+\infty)}(x)$$

now to get the marginals you have only to integrate the opposite variable "copying" the integral extremes in the $f(x,y)$ formula

Thus

$$f_X(x)=\int_0^x [2e^{-x}e^{-y} ]dy=2e^{-x}(1-e^{-x})\mathbb{1}_{[0;+\infty)}(x)$$

$$f_y(y)=\int_y^{+\infty} [2e^{-x}e^{-y} ]dx=2e^{-2y}\mathbb{1}_{[0;+\infty)}(y)$$

$Y\sim Exp(2)$