Let $X$ and $Y$ be two continuous random variables with joint probability density function $$f_{x,y}(x,y)= \begin{cases} 1/2, & \left\lvert x\right \rvert+\left\lvert y\right \rvert\leq 1\\ 0, & \text{otherwise}\end{cases} $$

show that the marginal probability density function of $x$ is $$f_x(x)= \begin{cases}1-\left\lvert x\right \rvert , & \left\lvert x\right \rvert\leq1\\ 0, & \text{otherwise}\end{cases} $$

I know that it is give by the integral of $1/2$ dy but i just can't work out the bounds with the modulus signs involved.

  • $\begingroup$ Hint: $$-1+|x|\leqslant y\leqslant1-|x|$$ $\endgroup$ – Did Aug 15 '15 at 10:08
  • $\begingroup$ Thank you!! seems obvious now $\endgroup$ – jess Aug 15 '15 at 10:17

For $y\leq0$, we must have $-y\leq1-|x|$ i.e $|x|-1\leq y\leq0$ and also $|x|\leq1$ (this is necessary due to the condition $|x|+|y|\leq1$), and for $y\geq0$ we have $0\leq y\leq1-|x|$ and $|x|\leq1$. Then, $$f_x(x):=\int_{y\in\mathbb{R}}f_{x,y}(x,y)\mathrm{d}y$$ $$=\frac{1}{2}\left(\int_{|x|-1\leq y\leq0}\mathrm{d}y+\int_{0\leq y\leq1-|x|}\mathrm{d}y.\right)\mathrm{1}_{|x|\geq1}(x)=\left(1-|x|\right)\mathrm{1}_{|x|\geq1}(x).$$


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