Problem 13.17 Easy Free 1 variation

Dart Distance From the Centre

A dart lands uniformly at random on a circular dartboard of radius $1$ centred at the origin. Find the expected distance from the landing point to the centre.

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $R$ be the distance from the landing point to the centre. For $0\le r\le 1$ , the event $\{R\ge r\}$ means the dart lands in the annulus between radius $r$ and radius $1$ . The landing point is uniform on the disc of area $\pi$ , so

\mathbb{P}(R\ge r)=\frac{\pi-\pi r^2}{\pi}=1-r^2 .

Since $R$ is non-negative and $R\le 1$ , the tail-sum formula gives

\mathbb{E}[R]=\int_0^\infty \mathbb{P}(R\ge r)\,dr=\int_0^1 (1-r^2)\,dr=1-\frac13 .

\boxed{\mathbb{E}[R]=\tfrac23}

Alternative. The distribution function is $\mathbb{P}(R\le r)=r^2$ , so $R$ has density $2r$ on $[0,1]$ . Then $\mathbb{E}[R]=\int_0^1 2r^2\,dr=\tfrac23$ . The answer exceeds $\tfrac12$ because there is more area far from the centre than near it.

Variations

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

Variation 13.17a

A point $(X_1,X_2,\dots,X_{10})$ is chosen uniformly at random from the solid $10$ -dimensional ball of radius $12$ centred at the origin. Find $\operatorname{Var}(X_1)$ .

Work it out, then compare with the answer.

Solution

Let $R^2=X_1^2+\dots+X_{10}^2$ . The ball is symmetric under reflections and under permutations of the coordinates. Hence $\mathbb{E}[X_1]=0$ and all the $\mathbb{E}[X_i^2]$ are equal, so

\operatorname{Var}(X_1)=\mathbb{E}[X_1^2]=\frac{\mathbb{E}[R^2]}{10}.

Volume in $10$ dimensions scales as $r^{10}$ , so $\mathbb{P}(R\le r)=(r/12)^{10}$ for $0\le r\le 12$ . The density of $R$ is therefore $10r^9/12^{10}$ , and

\mathbb{E}[R^2]=\int_0^{12} r^2\cdot\frac{10r^9}{12^{10}}\,dr=\frac{10}{12}\cdot\frac{12^{12}}{12^{10}}=\frac{10}{12}\cdot 144=120 .

Dividing by $10$ :

\boxed{\operatorname{Var}(X_1)=12}

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