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Suppose $X_1,X_2,\ldots,$ is a sequence of independent random variables with uniform distribution on unit interval $(0,1)$. Let $N=\min\{n>0\mid X_{(n)}-X_{(1)}>\alpha\}$ where $0<\alpha<1$. How can find $\mathbb{E}(N)$? Note: $X_{(1)},X_{(n)}$ are the smallest and the largest order statistics elements of the sequence $X_1,X_2,\ldots,$ and $\alpha$ is fixed.

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Presumably $X_{(n)}$ and $X_{(1)}$ stand for the smallest and the largest element of the sequence $X_1,X_2,\ldots,X_n$? –  Sasha Mar 9 '12 at 16:15
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Please consider accepting answers to those of your 10 already asked questions, that you feel are helpful. You can accept the answer by clicking on the tick to the left of the answer post. –  Sasha Mar 9 '12 at 16:16
    
Is $\alpha$ fixed, or do you just want $X_n-X_1>0$? Because if $\alpha$ is fixed, then the expected value is infinite (since if $X_1>1-\alpha$, $X_n-X_1>\alpha$ means $X_n>1$.) But if $\alpha$ is not fixed, it is unclear what it is doing in problem. –  Thomas Andrews Mar 9 '12 at 16:27
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The answer is $\mathrm E(N_\alpha)=2/(1-\alpha)$. –  Did Mar 9 '12 at 18:24
    
@DidierPiau I have added my solution with derivation of the result you posted in the comment above. Would appreciate any comments. Thanks. –  Sasha Mar 10 '12 at 4:32

1 Answer 1

up vote 2 down vote accepted

The joint probability density function for $(X_{(1)}, X_{(n)})$ is well known: $$ f_{X_{(1)}, X_{(n)}}(x,y) = (y-x)^{n-2} n(n-1) \cdot [ 0 < x \leqslant y <1] $$ The probability of the event $N_\alpha = n$ can be computed as probability that $X_{(n)}-X_{(1)} < \alpha$, but addition new uniform variable makes it greater than $\alpha$: $$ \mathbb{P}(N_\alpha = n+1) = \mathbb{P}\left( X_{(n)} - X_{(1)}<\alpha, \max(U, X_{(n)}) - \min(U, X_{(1)}) > \alpha \right) $$ Transcribing into the integral: $$ \begin{eqnarray} \mathbb{P}(N_\alpha = n+1) &=& n(n-1) \int_0^1 \mathrm{d} u \int_0^1 \mathrm{d} y \int_0^y \mathrm{d} x (y-x)^{n-2} [ y-x < \alpha \land (y-u > \alpha \lor u - x > \alpha) ] \end{eqnarray} $$

I used Mathematica to evaluate the probability:

In[60]:= Integrate[
 n (n - 1) (y - x)^(n - 2) Boole[
   y - x < al && (u - x > al || y - u > al)], {u, 0, 1}, {y, 0, 
  1}, {x, 0, y}, Assumptions -> 0 < al < 1 && n >= 2]

Out[60]= (-1 + al)^2 al^(-1 + n) n

That is $$ \mathbb{P}(N_\alpha = n) = (n-1) \left(1-\alpha\right)^2 \alpha^{n-2} $$ The expectation is then $$ \mathbb{E}(N_\alpha) = \sum_{n=1}^\infty n \cdot (n-1) \left(1-\alpha\right)^2 \alpha^{n-2} = \left(1-\alpha\right)^2 \frac{\mathrm{d}^2}{\mathrm{d} \alpha^2} \frac{1}{1-\alpha} = \frac{2}{1-\alpha} $$

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