ISI MStat 2016 Problem 5 | Order Statistics | PSB Sample

This is a beautiful problem from ISI MStat 2016 Problem 5 (sample) PSB based on order statistics. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem- ISI MStat 2016 Problem 5

Let \( n \geq 2,\) and \( X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed Poisson \( (\lambda) \) random variables for some \( \lambda>0 .\) Let \( X_{(1)} \leq\) \( X_{(2)} \leq \cdots \leq X_{(n)}\) denote the corresponding order statistics.
(a) Show that \( \mathrm{P}\left(X_{(2)}=0\right) \geq 1-n\left(1-e^{-\lambda}\right)^{n-1}\)
(b) Evaluate the limit of \( \mathrm{P}\left(X_{(2)}>0\right)\) as the sample size \( n \rightarrow \infty \) .

Prerequisites

• Pmf of k-th Order Statistic i.e \( x_{(k)} \)
• Binomial Expansion
• Basic Inequality
• squeeze (or sandwich) theorem
• Limit

Solution

(a) Given , \( n \geq 2,\) and \( X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed Poisson \( (\lambda) \) random variables for some \( \lambda>0 .\) Let \( X_{(1)} \leq\) \( X_{(2)} \leq \cdots \leq X_{(n)}\) denote the corresponding order statistics.

Let , F(j) be the CDF of \( X_{1}, X_{2}, \ldots, X_{n}\) i.e CDF of Poisson \( (\lambda) \)

Then , Pmf of k-th Order Statistic i.e \( x_{(k)} \)

\( P(x_{(k)} = j)= F_{k} (j)-F_{k} (j-0) \) , where \( F_{k} (j) =P(x_{(k)} \le j) \) i.e the CDF of k-th Order Statistic

\( F_{k} (j) = \sum_{i=k}^{n} \) \({n \choose i} \) \( {(F(j))}^{i} {(1-F(j))}^{n-i} \)

So, \( P(x_{(k)} = j) = \sum_{i=k}^{n} {n \choose i} [ {(F(j))}^{i} {(1-F(j))}^{n-i}-{(F(j-0))}^{i} {(1-F(j-0))}^{n-i}] \)

Here we have to find , \( P(x_{(2)} = 0)= \sum_{i=2}^{n} {n \choose i} [{(F(0))}^{i} {(1-F(0))}^{n-i} - 0] \)

since , Poisson random variable takes values 0 ,1,2,.... i.e it takes all values < 0 with probabiliy 0 , that's why \( {(F(j-0))}^{i} {(1-F(j-0))}^{n-i} =0\) here for j=0 .

And , \( F(0)=P(x \le 0) = P(X=0)={e}^{- \lambda} \frac{{\lambda}^{0}}{0!} ={e}^{- \lambda} \) , as X follows Poisson \( (\lambda) \) .

So, \( {(F(0))}^{i} {(1-F(0))}^{n-i}= {({e}^{- \lambda})}^{i} {(1-{e}^{- \lambda})}^{n-i} \)

Therefore , \( P(x_{(2)} = 0)= \sum_{i=2}^{n} {n \choose i} [{({e}^{- \lambda})}^{i} {(1-{e}^{- \lambda})}^{n-i} ] \)

\( = {({e}^{- \lambda}+1-{e}^{- \lambda})}^{n} - {n \choose 0}[{({e}^{- \lambda})}^{0} {(1-{e}^{- \lambda})}^{n-0} ]- {n \choose 1}[{({e}^{- \lambda})}^{1} {(1-{e}^{- \lambda})}^{n-1}] =1-{(1-{e}^{- \lambda})}^{n}- n {e}^{- \lambda} {(1-{e}^{- \lambda})}^{n-1} \)

\( =1-{(1-{e}^{- \lambda})}^{n-1}[1-{e}^{- \lambda} +n{e}^{- \lambda}] \)

\( =1-{(1-{e}^{- \lambda})}^{n-1}[1+(n-1){e}^{- \lambda}] \ge 1- n{(1-{e}^{- \lambda})}^{n-1} \) .

Since , \( 1+(n-1){e}^{- \lambda} \le n \Longleftrightarrow {e}^{ \lambda} \ge 1 \) for \( n \ge 2\) and \( \lambda >0 \) which is true hence our inequality hold's true (proved)

Hence , \( \mathrm{P}\left(X_{(2)}=0\right) \geq 1-n\left(1-e^{-\lambda}\right)^{n-1}\) (proved )

(b) \( 0 \le P(x_{(2)} >0) =1-P(x_{(2)}= 0) \) \( \le 1-1+n\left(1-e^{-\lambda}\right)^{n-1}\) ( Using inequality in (a) )

So, \( 0 \le P(x_{(2)} >0) =1-P(x_{(2)}= 0) \) \( \le n\left(1-e^{-\lambda}\right)^{n-1}\) -----(1)

As \( 0< 1-{e}^{- \lambda} <1\) for \( \lambda >0 \) i.e it's a fraction so it can be written as \( \frac{1}{a} \) for some \( a>1\) , Hence \( \lim_{n\to\infty} n\left(1-e^{-\lambda}\right)^{n-1} = \lim_{n\to\infty} \frac{n}{a^n} =0 \) (Proof -Use l'hospital rule or think intutively that as n tends to infinity the exponential functions grows more rapidly than any polynomial function ).

Now taking limit \( n \to \infty \) in (1) , we get by squeeze (or sandwich) theorem

\( \lim_{n\to\infty} P(x_{(2)} >0) =0 \)