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# ISI MStat 2016 Problem 5 | Order Statistics | PSB Sample

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

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 $n \rightarrow \infty$ .

## 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

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

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