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# 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 $$n \rightarrow \infty$$ .

## 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

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