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I'm trying to solve Problem 1.13 in An intermediate Course in Probability, 2nd Ed. (Gut, 2009):

Let $X$ and $Y$ have a joint density function given by

$$ f_{X,Y} (x, y) = \begin{cases} 1, & \text{for } 0 \le x \le 2, \text{max}(0, x − 1) \le y \le \text{min}(1, x),\\ 0, & \text{otherwise.} \end{cases} $$ Determine the marginal density functions and the joint and marginal distribution functions.

Earlier in the chapter, Gut shows us how to find these marginal density functions $f_X(x)$ and $f_Y(y)$:

$$ f_X(x) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy\\ f_Y(y) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dx $$

So in this case we have

$$ \begin{split} f_X(x) &= \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy\\ &= \int_{\text{max}(0,x-1)}^{\text{min}(1,x)}1dy\\ \end{split} $$

This is where I'm stuck. How do I solve this integral?

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1 Answer 1

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$\min(1,x)=1$ if $x\geqslant 1$ and $x$ if $x\leqslant 1$, $\max(0,x-1)=0$ if $x\leqslant 1$ and $x-1$ if $x\geqslant 1$. Calculate the integral when $x\geqslant 1$ first and then when $x\leqslant 1$. You can also say that $\int_a^b{1dy}=b-a$ for all $(a,b)\in\mathbb{R}^2$

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  • $\begingroup$ Using this, I found f_X(x) = 2. As far as I can tell, this is correct, but the answer I'm supposed to find is f_X(x) = x for 0 < x < 1, 2 − x for 1 < x < 2. I'll have to rethink what I'm supposed to do. $\endgroup$
    – Mossmyr
    Sep 10, 2019 at 18:01

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