Birds on a Wire

By Bogdan Lataianu

Let $U_{(i)}$ be the order statistics of uniform distribution, $i=1\ldots n-1$. and $v_r=u_i-u_{i-1}$ with $u_0=0$. Then the pdf of $v_1,\ldots,v_r$ for $\sum_{i=1}^r v_i\leq 1$ is (see [1})

$f(v_1,\ldots,v_r)=\frac{(n-1)!}{(n-r-1)!}(1-v_1-\ldots v_r)^{n-r-1}, r=1\ldots n-1$

So when r=3, $f(v_1,v_2,v_3)=(n-1)(n-2)(n-3)(1-v_1-v_2-v_3)^{n-4}.$ Let $Z_i$ be the painted length of the $i$-th interval after a realization of the order statistics and denote $X=V_{i-1},$ $Y=V_i,$ $Z=V_{i+1},$ $A$ the event ${V_i \lt V_{i-1} \hbox{ or } V_i \lt V_{i+1}}$. Then $Z_i = 1_AY$ and

$\displaystyle E(Z_i)=\int 1_A y f(x,y,z) dx dy dz=\int_A yf(x,y,z) dx dy dz.$ We have

$P(A)=P(Y \lt X)+P(Y \lt Z)-P(Y \lt X \hbox{ and } Y \lt Z)$.

Therefore

$\displaystyle \begin{align} E(Z_i)&=(n-1)(n-2)(n-3)\Bigg(\int_{0}^{1/2}\int_y^{1-y}\int_0^{1-x-y}y(1-x-y-z)^{n-4}dzdxdy\\ &\qquad\qquad\qquad+\int_0^{1/2}\int_0^{1-2y}\int_y^{1-x-y}y(1-x-y-z)^{n-4}dzdxdy\\ &\qquad\qquad\qquad-\int_0^{1/3}\int_y^{1-2y}\int_y^{1-x-y} y(1-x-y-z)^{n-4}dzdxdy\Bigg)\\ &=\frac{1}{4n}+\frac{1}{4n}-\frac{1}{9n}=\frac{7}{18n} \end{align}$

So in our case suppose we have $n$ birds. Then $E$(painted length)=$E$(two intervals never painted)+$E$(two intervals always painted)+$\displaystyle (n-3)E(Z_i)=\frac{2}{n+1}+(n-3)\frac{7}{18n}$ with the limit $\displaystyle \frac{7}{18}.$

Reference

Herbert A. David and Haikady N. Nagaraja, Order statistics, Wiley-Interscience; 3 edition (August 4, 2003)

Geometric Probability

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