Let $X$ and $Y$ be independent random variables, uniformly distributed in the interval $[0, 1]$. Find the CDF and the PDF of $|X − Y|$.

Question

So this what I have: \begin{align} \ F_{z}(Z) &= P (Z<z) \\\ \ &= P( |X-Y| < z)\\ \ &= P(-z < X-Y < z) \\ \ &= P(-z+y<X<z+y)\\ \ &=F_x(z+y)-F_{x}(-z+y) = ? \end{align}

As for the pdf it is just:

\begin{align} f_{z}(Z)=f_{x}(z+y)+f_{x}(-z+y) \end{align}

Duplicate of CDF of absolute value of difference in random variables. — StubbornAtom, Jun 28 '20 at 06:27

GNUSupporter 8964民主女神地下教會 · Accepted Answer · 2019-05-08T12:49:50.923

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Consider the unit square $\Omega = [0,1]^2$. $|X-Y|$ is directly proportional to the distance from the diagonal from the diagonal $X = Y$. The event $\{|X - Y| \le z\}$ is represented by the red region. For all $d \in [0,1]$,

The CDF is the area of the $\color{red}{\text{red region}}$. $$F_Z(d) = P(\color{red}{|X - Y| \le d}) = 1 - (1-d)^2 = 2d - d^2$$

Differentiating the CDF with respect to $d$ on $(0,1)$ gives the PDF. $$f_Z(d) = F_Z'(d) = 2 - 2d.$$

edited May 08 '19 at 12:49

answered May 05 '19 at 10:02

GNUSupporter 8964民主女神地下教會

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Thank you very much. – Aditya Lahiri May 08 '19 at 00:42

Let $X$ and $Y$ be independent random variables, uniformly distributed in the interval $[0, 1]$. Find the CDF and the PDF of $|X − Y|$.

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