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I have been given the following question,

Let $n ≥ 2$, and $X_1, X_2, . . . ,X_n$ be independent and identically distributed $Poisson (λ)$ random variables for some $λ > 0$. Let $X_{(1)} ≤ X_{(2)} ≤ · · · ≤ X_{(n)}$ denote the corresponding order statistics.

(a) Show that $P(X_{(2)} = 0) ≥ 1 − n(1 − e^{−λ})^{n−1}$.

(b) Evaluate the limit of $P(X_{(2)} > 0)$ as the sample size $n → ∞$.

I tried solving the question on my own and I have also been able to obtain the following expression; $P(X_{(2)}=0) = 1 - (1-e^{-\lambda})^n-ne^{-\lambda}(1-e^{-\lambda})^{n-1}$

$= 1-(1-e^{-\lambda})^{n-1}(1+e^{-\lambda}(n-1))$ ;

and it can be then shown that

$(1+e^{-\lambda}(n-1)) \le n \quad \text{for all } \lambda > 0 \text{ and } n \ge 2$

and thus

$P(X_{(2)} = 0) ≥ 1 − n(1 − e^{−λ})^{n−1}$

but I want to ask that is there any meaning of this statement. I mean is there any significance of the quantity on the left hand side of the inequality so that the inequality can be derived intuitively or by any other method?

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  • $\begingroup$ Left hand side of which equation? $\endgroup$ – whuber Apr 15 at 20:28
  • $\begingroup$ @whuber Of the inequality that needs to be proven, i.e. part a) of the question. $\endgroup$ – Sanket Agrawal Apr 16 at 6:17
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Given: $(X_1, ...,X_n)$ denotes a random sample of size $n$ drawn on $X$, where $X \sim \text{Poisson}(\lambda)$ with pmf $f(x)$:

enter image description here

Then, the pmf of the $2^{\text{nd}}$ order statistic, in a sample of size $n$, is $g(x)$:

enter image description here

... where:

  • I am using the OrderStat function from the mathStatica package for Mathematica to automate the nitty-gritties, and

  • Beta[z,a,b] denotes the incomplete Beta function $\int _0^z t^{a-1} (1-t)^{b-1} dt$

  • Gamma[a,z] is the incomplete gamma function $\int _z^{\infty } t^{a-1} e^{-t} dt$

The exact desired probability $P(X_{(2)}=0)$ is simply:

enter image description here

The following diagram plots and compares:

  • the exact solution to $P(X_{(2)}=0)$ just derived (red curve)
  • to the bound $P(X_{(2)} = 0) ≥ 1 − n(1 − e^{−λ})^{n−1}$ proposed in the question

enter image description here

... plotted here when $\lambda =3$.

The bound appears useless for any proper purpose - even a drunk monkey could do better by simply choosing 0 than using the bound proposed that is negative over a huge chunk of the domain (and will get worse as $\lambda$ increases).

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  • $\begingroup$ Thank you for the answer. It clears my doubts. $\endgroup$ – Sanket Agrawal Apr 18 at 17:33
  • $\begingroup$ If I may ask regarding the b) part of the same question, am I correct in deducing that the required limit is equal to 0? $\endgroup$ – Sanket Agrawal Apr 18 at 17:39
  • $\begingroup$ @SanketAgrawal Yes ... even by intuition, as the sample size $n$ tends to infinity, the probability that the second smallest value will be a zero becomes certain, so $P(X_{(2)}=0)$ tends to 1 (as in the picture above), and therefore $P(X_{(2)}>0)$ must tend to 0. $\endgroup$ – wolfies Apr 21 at 18:13
  • $\begingroup$ okay. Thank you. $\endgroup$ – Sanket Agrawal Apr 21 at 18:19
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$\mathbb{P}(X_{(2)} = 0)$ asked for the probability where the second least r.v. is zero. In other words, it asked for the probability where at least two of $X_1, \cdots X_n$ are zero.

The statement "there are at least two zeros among $X_1, \cdots, X_n$" is false when we have an event where "at least $n-1$ of $X_1, \cdots, X_n$ being greater than zero". The event happens with a probability of at most $$n(1 - e^{-\lambda})^{(n-1)}$$ where you have $n$ choices (which $n-1$ r.v.s to choose, or alternatively, which one to leave out), and for each choice you need all $n-1$ of them to be greater than zero, and assume nothing of the one you left out (i.e. adding another multiplier of $1$).

The quoted probability is the upper bound as the event is the least restrictive event among all events that falsify the statement, and any other events will be at least as, or more, restrictive. For example, one can condition on the one variate left out in the selection process, but that involve making more assumptions on the variate's value. Reducing the cardinality of an event (to its subset) will not increase the probability of that event happening, as shown in this Maths.SE question.

To obtain the required probability it is sufficient to exclude the event from the entire event space from consideration. The RHS of the question ($1 - n(1 - e^{-\lambda})^{(n-1)}$) thus forms the lower bound to the required probability.

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  • $\begingroup$ I understand the fact that "having at least n-1 variates greater than 0" renders the statement that "there are at least two variates equal to zero" false and that is what I have made use of to derive the inequality. But how is the probability that at least n-1 variates are greater than zero equal to the quantity you have mentioned? $\endgroup$ – Sanket Agrawal Apr 17 at 18:49
  • $\begingroup$ I mean the probability that there are exactly n-1 variates greater than zero will be $ne^{-\lambda}(1-e^{-\lambda})^{(n-1)}$ and the probability that there are exactly n variates greater zero will be $(1-e^{-\lambda})^n$. Shouldn't the probability that there are at least n-1 variates greater than zero be equal to sum of these two? $\endgroup$ – Sanket Agrawal Apr 17 at 18:54
  • $\begingroup$ @SanketAgrawal Yes, you are absolutely right on the probability should be the sum of the two. I guess I meant to say the probability that at least $n-1$ variates are greater than zero is intuitively at most $n(1-e^{-\lambda})^{(n-1)}$, when you assume nothing about the value of the one variate you left out. Clearly I got myself confused as well - will try to fix the answer tomorrow. $\endgroup$ – B.Liu Apr 18 at 1:53
  • $\begingroup$ When you fix the answer, will you please also explain how the probability that at least n-1 variates are greater than zero will be at most the given quantity. $\endgroup$ – Sanket Agrawal Apr 18 at 10:32
  • $\begingroup$ @SanketAgrawal I have attempted. $\endgroup$ – B.Liu Apr 18 at 12:51

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