Get inspired by the success stories of our students in IIT JAM 2021. Learn More 

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

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 sandwichtheorem

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

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com