Problem 8.17 Easy Free 1 variation

Exponential Race After Scaling

An ellipse centred at the origin has horizontal radius AA and vertical radius BB, where A,BA,B are independent Exp(1)\mathrm{Exp}(1) random variables. Recall that an ellipse with radii aa and bb has area πab\pi ab. A circle is drawn whose diameter equals AA. Find the probability that the circle’s area is larger than the ellipse’s area.

Work it out, then compare with the answer.

SolutionShow the worked solution

The circle has radius A/2A/2, so its area is C=π4A2C=\frac{\pi}{4}A^2, while the ellipse has area E=πABE=\pi AB. Since A>0A>0 almost surely, we may divide by πA\pi A:

C>E  ⟺  14A2>AB  ⟺  A>4B.C>E \iff \tfrac14 A^2 > AB \iff A>4B .

A scaled exponential is again exponential. Indeed P(4B>t)=e−t/4\mathbb{P}(4B>t)=e^{-t/4}, so 4B∼Exp(14)4B\sim\mathrm{Exp}(\tfrac14), and it is independent of A∼Exp(1)A\sim\mathrm{Exp}(1).

The event A>4BA>4B says that 4B4B wins the race against AA. For independent Exp(λ)\mathrm{Exp}(\lambda) and Exp(μ)\mathrm{Exp}(\mu) variables, the first is the smaller with probability λ/(λ+μ)\lambda/(\lambda+\mu). Therefore

P(A>4B)=1/41/4+1=15.\mathbb{P}(A>4B)=\frac{1/4}{1/4+1}=\frac15 .
15\boxed{\tfrac15}

Key idea. Rescaling turns the comparison into an exponential race with rates 11 and 14\tfrac14.

Variations

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

Variation 8.17a

Let X1,X2X_1,X_2 be i.i.d. Exp(1)\mathrm{Exp}(1) random variables, with order statistics O1=min⁡(X1,X2)O_1=\min(X_1,X_2) and O2=max⁡(X1,X2)O_2=\max(X_1,X_2). Find the probability that the range O2−O1O_2-O_1 exceeds the midpoint (X1+X2)/2(X_1+X_2)/2.

Work it out, then compare with the answer.

Solution

Since {X1,X2}={O1,O2}\{X_1,X_2\}=\{O_1,O_2\}, the midpoint equals (O1+O2)/2(O_1+O_2)/2. Write G=O2−O1G=O_2-O_1 for the gap. Then

O2−O1>12(O1+O2)  ⟺  O2>3O1  ⟺  G>2O1.O_2-O_1>\tfrac12(O_1+O_2)\iff O_2>3O_1\iff G>2O_1 .

The minimum O1O_1 is Exp(2)\mathrm{Exp}(2). By memorylessness, the gap GG is Exp(1)\mathrm{Exp}(1) and independent of O1O_1.

Rescaling gives 2O1∼Exp(1)2O_1\sim\mathrm{Exp}(1). So GG and 2O12O_1 are i.i.d. Exp(1)\mathrm{Exp}(1), and by symmetry each is the larger with probability one half:

12\boxed{\tfrac12}