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Let $U \sim U(0,1)$ and $V\sim U(0,1)$ be two independent uniform random variables. Find the pdf of $U=X-Y$.

My attempt: First find the CDF of U. $$P(U\leq t) = P(X-Y \leq t) = \int_0^1 P(X\leq t+y) F_Y(y) = \int_0^1 F_X(t+y)F_Y(y)$$ Evaluating this integral, I get $F_U(t) = \frac{t}{2}+1/3$. To find the pdf, I differentiate this to obtain $f_U(t) = 1/2$ which I know is wrong but I have no idea where I went wrong.

Any help is appreciated. Thanks.

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    $\begingroup$ use a geometric sketch : it will be quite evident ! $\endgroup$
    – G Cab
    Sep 16, 2020 at 17:59

1 Answer 1

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Slow solution: first, find pdf of $Z=-Y$, which is quite straightforward in two steps: first, find CDF $F_{Z}(z) = P(Y\geq -z)$. The inequality changed because $Z=\varphi(Y)$ is a decreasing function, therefore you want $Y\geq \varphi^{-1}(z)$. Once you have got it, differentiate wrt $z$ to get $f_Z(z): Z \sim R[-1,0]$ is Now you have $$ U=X+Z $$ It is easier to find pdf of $U$ rather than CDF using convolution formula: $$ f_{U}(u) = \int f_{Z}(u-x)f_{X}(x)dx $$ Obvisouly $f_{Z}(u-x)$ is not $0$ only when $-1 <u-x<0$. Therefore you can split the support of the integral into 2 intervals: $$ \int_{0}^{x<u+1}f_X(x)dx \ \text { if } -1 <u<0\\ \int_{x>u}^{1}f_{X}(x)dx \ \text { if } 0<u<1 $$ Outside of these intervals it is $0$.

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  • $\begingroup$ Thanks a ton! How does this expression imply $1-|u|$ for $u \in[-1,1]$? $\endgroup$ Sep 16, 2020 at 21:27
  • $\begingroup$ Solve both integrals, and you get your expression (or $\{1+u, 1-u\}$) if you wish, which is the same thing $\endgroup$
    – Alex
    Sep 16, 2020 at 21:34
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    $\begingroup$ Great. Accepted the answer. Thanks again. $\endgroup$ Sep 17, 2020 at 7:29

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