4
$\begingroup$

suppose $X_1,X_2,\ldots$ is sequence of independent random variables of $U(0,1)$ if $N=\min\{n>0 :X_{(n:n)}-X_{(1:n)}>\alpha , 0<\alpha<1\}$ that $X_{(1:n)}$ is smallest order statistic and $X_{(n:n)}$ is largest order statistic. how can find $E(N)$

  • 1
    find distribution of $(M_n-m_n)$, I think its $\frac 1 {n(n-1)}x^{n-2}(1-x)$. Then $\mathbb P (T > n) = \mathbb P (M_n - m_n < \alpha)$ because $M_n - m_n$ increasing.2012-07-31
  • 0
    @mike : Where you refer to "$T$", do you mean $N$?2012-07-31
  • 0
    yes, sorry ${}{}$2012-07-31

1 Answers 1

3

Let $m_n=\min\{X_k\,;\,1\leqslant k\leqslant n\}=X_{(1:n)}$ and $M_n=\max\{X_k\,;\,1\leqslant k\leqslant n\}=X_{(n:n)}$. As explained in comments, $(m_n,M_n)$ has density $n(n-1)(y-x)^{n-2}\cdot[0\lt x\lt y\lt1]$ hence $M_n-m_n$ has density $n(n-1)z^{n-2}(1-z)\cdot[0\lt z\lt1]$.

For every $n\geqslant2$, $[N\gt n]=[M_n-m_n\lt\alpha]$ hence $$ \mathrm P(N\gt n)=\int_0^\alpha n(n-1)z^{n-2}(1-z)\mathrm dz=\alpha^{n}+n(1-\alpha)\alpha^{n-1}. $$ The same formula holds for $n=0$ and $n=1$ hence $$ \mathrm E(N)=\sum_{n=0}^{+\infty}\mathrm P(N\gt n)=\sum_{n=0}^{+\infty}\alpha^n+(1-\alpha)\sum_{n=0}^{+\infty}n\alpha^{n-1}=\frac2{1-\alpha}. $$ Edit: To compute the density of $(m_n,M_n)$, start from the fact that $$ \mathrm P(x\lt m_n,M_n\lt y)=\mathrm P(x\lt X_1\lt y)^n=(y-x)^n, $$ for every $0\lt x\lt y\lt 1$. Differentiating this identity twice, once with respect to $x$ and once with respect to $y$, yields the opposite of the density of $(m_n,M_n)$.

  • 1
    The answer is the same as the one we would get if the question were "What is the average number of trials required to observe two occurrences of an event of probability $1-\alpha$?" So I wonder if the result can be obtained via an argument based on the linearity of expectation.2012-07-31
  • 0
    @DilipSarwate : The "trials" of which you speak would not be independent! So there's a complication.2012-07-31
  • 1
    @did : I wonder if this answer should be considered incomplete because it fails to explain why that's the density of the pair.2012-07-31
  • 0
    @MichaelHardy: See Edit.2012-08-01
  • 0
    very very thanks for your reply. why [N>n]=[Mn−mn<α]? please explain me.2012-08-01
  • 0
    Since $N=\min(k:A_k)$ for some $A_k$, $N\gt n$ means $A_1$, $A_2$, ..., $A_n$ do not occur. Here $A_k=[M_k-m_k\gt\alpha]$, hence $A_k\subseteq A_{k+1}$ for every $k$, and "$A_1$, $A_2$, ..., $A_n$ do not occur" is equivalent to "$A_n$ does not occur", that is, equivalent to $M_n-m_n\leqslant\alpha$.2012-08-01