Problem 8.1 Easy Free 2 variations

Ratio of Two Uniforms in a Band

Let $X$ and $Y$ be independent $\mathrm{Unif}(0,1)$ random variables. Find $\mathbb{P}\left[1 \le \dfrac{X}{Y} \le 2\right]$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. An event $\{X/Y>m\}$ is a wedge through the origin of the unit square, so its probability is an area.

For $m\ge 1$ the event $\{X > mY\}$ is the region of $[0,1]^2$ below the line $y=x/m$ . Because $1/m\le 1$ , this line leaves the square through its right edge, at $(1,1/m)$ . The region is therefore the triangle with vertices $(0,0)$ , $(1,0)$ and $(1,1/m)$ , and

\mathbb{P}\!\left[\frac{X}{Y}>m\right]=\frac{1}{2m}\qquad(m\ge 1).

The boundary lines have probability zero, so strict and non-strict inequalities give the same value. Hence

\mathbb{P}\!\left[1\le \frac{X}{Y}\le 2\right]=\mathbb{P}\!\left[\frac{X}{Y}\ge 1\right]-\mathbb{P}\!\left[\frac{X}{Y}> 2\right]=\frac12-\frac14 .

Geometrically, this is the triangle with vertices $(0,0)$ , $(1,1/2)$ and $(1,1)$ lying between the lines $y=x/2$ and $y=x$ .

\boxed{\tfrac14}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 8.1a

A right triangle has legs of lengths $A$ and $B$ , where $A,B$ are independent $\mathrm{Unif}(0,1)$ random variables. Let $\theta$ be the angle opposite leg $A$ . Find $\mathbb{P}\left[\theta > \dfrac{\pi}{3}\right]$ .

Work it out, then compare with the answer.

Solution

In the right triangle, $\tan\theta = A/B$ . Since $\tan$ is increasing on $(0,\pi/2)$ , we have $\theta>\pi/3$ if and only if $A/B>\sqrt3$ .

Taking $m=\sqrt3\ge 1$ in the wedge formula $\mathbb{P}[A/B>m]=1/(2m)$ gives

\mathbb{P}\!\left[\theta>\frac{\pi}{3}\right]=\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}\approx 0.289 .

\boxed{\tfrac{1}{2\sqrt3}=\tfrac{\sqrt3}{6}}

Variation 8.1b

Let $X,Y \sim \mathrm{Unif}(0,1)$ be independent. Find $\mathbb{P}[X+Y \ge 2|X-Y|]$ .

Work it out, then compare with the answer.

Solution

First remove the absolute value by cases.

If $Y\ge X$ , the condition reads $X+Y\ge 2(Y-X)$ , that is, $Y\le 3X$ .
If $X\ge Y$ , it reads $X\le 3Y$ .

Together the event is $\{\tfrac13\le Y/X\le 3\}$ .

Its complement is the union of two disjoint wedges, $\{Y>3X\}$ and $\{X>3Y\}$ . By the wedge formula with $m=3$ , each has probability $\tfrac16$ . Therefore

\mathbb{P}[X+Y\ge 2|X-Y|]=1-2\cdot\frac16 .

\boxed{\tfrac23}

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