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Question Let $X_i \sim\left(i . i\right.$. $d$.) Bernoulli $\left(\frac{\lambda}{n}\right), n \geq \lambda \geq 0$. $Y_i \sim\left(i\right.$ i. d.) Poisson $\left(\frac{\lambda}{n}\right),\left\{X_i\right\}$ and $\left\{Y_i\right\}$ are independent. Let $\sum_{i=1}^{n^2} X_i=T_n$ and $\sum_{i=1}^{n^2} Y_i=S_n$ (say). Find the limiting distribution of $\frac{T_n}{S_n}$ as $n \rightarrow \infty$.

My attempt:

Let $F$ be the cdf of $\frac{T_n}{S_n}$. Then, $F(\frac{a}{b})$ = $P(T_{n} \leq a) \times P(S_{n} \geq b)$ where $a \in \{0,1,2,...,n^{2}\}$ and $b \in \{1,2,3,4....\}$.

Now, $P(T_{n}\leq a)$ = $\sum_{i=0}^{a}$ $\binom {n^{2}}{i} ({\frac{\lambda}{n}})^{i}(1-\frac{\lambda}{n})^{n^{2}-i}$ and, $P( S_{n} \geq b)$ = $1- \sum_{k=0}^{b}\frac{(n^{2}\lambda)^{k}}{k!}e^{-n^{2}\lambda}$. As the distribution of $S_{n}$ came out to be Poisson with parameter $n^{2}\lambda$.

When $n$ tends to infinity, I can see that $\frac{n^{2}}{e^{n^{2}}}$ goes to 0, hence, $P(S_{n}\geq b)$ goes to 1. But I am not able to find the limit of other expression.

The question has been discussed here:- Find the limiting distribution of $T_n/S_n$ as $n\rightarrow \infty$. There were no answers and I couldn't benefit from the comment over there.

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  • $\begingroup$ Is the constant $n$ in the expression $\sum_{i = 1}^{n^2}$ and in $\text{Bernoulli}(\lambda/n), \text{Poisson}(\lambda/n)$ the same ? $\endgroup$ Commented Apr 8 at 9:13
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    $\begingroup$ Yes. They are same. @ThànhNguyễn $\endgroup$
    – Debu
    Commented Apr 8 at 9:14

1 Answer 1

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First note that $S_{n}$ is the sum of $n^{2}$ many iid $Poi(\lambda/n)$ variates and hence has $Poi(n\lambda)$ distribution.

Now see that by Chebycheff's inequality, $$P(|\frac{S_{n}}{\lambda n}-1|\geq \epsilon)\leq\frac{Var(S_{n})}{\epsilon^{2}\lambda^{2} n^{2}}=\frac{\lambda n}{\epsilon^{2}n^{2}\lambda^{2}}\to 0 $$

Hence $\dfrac{S_{n}}{\lambda n}\xrightarrow{P} 1$

Similarly, you have that $$P(|\frac{T_{n}}{\lambda n}-1|>\epsilon)\leq \frac{Var(T_{n})}{\epsilon^{2}n^{2}\lambda^{2}}=\frac{n^{2}\frac{\lambda}{n}(1-\frac{\lambda}{n})}{n^{2}\lambda^{2}\epsilon^{2}}\to 0$$

(where we have used that the variance of a binomial$(n,p)$ variate is $np(1-p)$)

Hence also $\dfrac{T_{n}}{\lambda n}\xrightarrow{P}1$

And you have $\dfrac{T_{n}}{S_{n}}=\dfrac{T_{n}}{n\lambda}\cdot \dfrac{n\lambda}{S_{n}}\xrightarrow{P}1$

And thus also, $\frac{T_{n}}{S_{n}}\xrightarrow{d}1$. That is, the limiting distribution is $\delta_{1}$

NOTE: There's a small argument as to the well-definedness of the quantity $\frac{1}{S_{n}}$. i.e., we have to show that almost surely, $S_{n}>0$ for large enough $n$.

So note that $P(S_{n}=0)=e^{-\lambda n}$ which is summable. Hence, Borel-Cantelli lemma gives you that almost surely, only finitely many $S_{n}$ will be $0$.

Thus, the expression $\lim_{n\to\infty}\frac{T_{n}}{S_{n}}$ is almost surely well defined.

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  • $\begingroup$ could you also tell me the thinking process of yours, how did you feel, it needs to be done this way and not how I mentioned in the comments, by writing down the distributions explicitly and taking n limiting to infinity. $\endgroup$
    – Debu
    Commented Apr 10 at 11:41
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    $\begingroup$ @Debu Sure. One should first notice that you are summing iid variates, so something akin to the strong law or weak law should apply. The problem is just that the distribution of the summands depend on $n$. But still, it is reasonable to expect that $T_{n}/E(T_{n})\xrightarrow{P} 1$. And once you have this intuition, you just need to check what the variance behaves like. Often a time, when you see ratios, it is better to first think about what the limiting behaviour of the numerator and the denominator is. If you figure that both are of the same order, then you can multiply and divide. $\endgroup$ Commented Apr 10 at 11:45
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    $\begingroup$ @Debu More precisely, you can look at the weak/strong laws for triangular arrays here in Durrett. I believe it's in section 2.2.3. I'll leave you to finding it. Also, if you have a ratio of two complicated random variables, it most likely won't have a nice distribution and so nothing really can be said from characteristic functions as they will be too complicated to evaluate. One usually develops a sense/intuition for these rather than seeing them rigorously. Hope these were useful to you. $\endgroup$ Commented Apr 10 at 11:53
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    $\begingroup$ @Debu Additionally, one should also note that the poisson and binomial variates asymptotically behave in the same way. This is due to the Poisson Central Limit Theorem. Although that is different from what you are asking here, it's still an useful thing to keep in mind. $\endgroup$ Commented Apr 10 at 11:55
  • $\begingroup$ That's very insightful comments of yours. Yes, i found that section in Durret. I will go through it. Thanks a lot for your help. $\endgroup$
    – Debu
    Commented Apr 10 at 12:00

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