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Let $x$ and $y$ be uniformly distributed, independent random variables on $[0,1]$. What is the probability that the distance between $x$ and $y$ is less than $1/2$?

My continuous probability knowledge is very slim. I know that $P(x\leq \frac{1}{2})=\frac{1}{2}$, but I don't know how to work in the $y$. I would like to see an answer involving integrals, even if its not necessary, so I can try I pick up some of this continuous probability stuff.

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  • $\begingroup$ Draw a picture of the locus where $|x-y|<1/2$. You don't need integrals if you remember basic geometry. $\endgroup$
    – symplectomorphic
    Commented Jul 15, 2017 at 16:07
  • $\begingroup$ (You've edited to ask for answers involving integrals. To calculate $\mathbb{P}[(x,y)\in A]$ you compute $\int_A f(x,y)$, where $f$ is the joint pdf of $X$ and $Y$. So you still need to consider the region $A$, the set of points satisfying $|x-y|<1/2$, to set up the limits of integration.) $\endgroup$
    – symplectomorphic
    Commented Jul 15, 2017 at 16:18
  • $\begingroup$ Is the joint pdf of $x$ and $y$ just the pdf of the uniform distribution? $\endgroup$
    – SihOASHoihd
    Commented Jul 15, 2017 at 16:20
  • $\begingroup$ By independence, the joint pdf $f_{X, Y}(x,y)$ factors into the product of the univariate pdfs, $f_X(x)$ and $f_Y(y)$. What are the univariate pdfs here? $\endgroup$
    – symplectomorphic
    Commented Jul 15, 2017 at 16:23
  • $\begingroup$ It is just 1 right? $\endgroup$
    – SihOASHoihd
    Commented Jul 15, 2017 at 16:37

1 Answer 1

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It helps if you plot the support, X on the x-axis, Y on the y-axis. The support is $[0,1]\times[0,1]$.

$P(|X-Y|<1/2)$ is the area between the line $y=x-1/2$ and $y=x+1/2$ in the unit square. For independent uniform, you could use geometry. For other distributions, you'll have to integrate.

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