Showing Lyapunov (Lindeberg) condition holds for sum of independent bernoulli distribution with poisson tail probability

Question

I'm trying to find an asymptotic distribution of the follwing random variable

$$Z_n=\sum_{i=0}^n Y_i$$ where $Y_i = I[T_i<t]$ with $T_i \text{~} Gamma(i, \lambda)$.

Here $t$ is a fixed number.

My initial trial was using Linderberg CLT by using the Lyapunov CLT ( or Linderberg CLT). First I noticed that by using integral by parts,

$$E[Y_i]=P[Y_i=1]=P[T_i<t]=\sum_{k=i}^\infty \frac{(t/\lambda)^ie^{-t/\lambda}}{i!}$$

which is the tail probabilty of the poisson random variable with mean $t/\lambda$.

Let $p_i:=E[Y_i], R_i:=Y_i-p_i, S_n:=\sum_{i=1}^nR_i, \sigma_n^2:=Var(S_n)=\sum_{i=1}^np_i(1-p_i)$.

I'm trying to show that $R_i$ satisfies the Lyapunov condition where

$$ \sigma_n^{-(2+\delta)}\sum_{i=1}^nE[|R_i|^{(2+\delta)}]\to0\ \text{as $n\to \infty$ }$$.

With $\delta=1$,

I get

$$E[|R_i|^3]=(1-p_i)^3p_i + p_i^3(1-p_i)^3\le p_i(1-p_i)$$ therefore

$$\sigma_n^{-3}\sum_{i=1}^nE[|R_i|^{3}] \le \sigma_n^{-1}.$$

As long as I can show that $$\sigma_n \to \infty \ \text{as $n\to \infty$ }$$

I'm done with the proof. But I'm stuck with showing the divergence of $\sigma_n$.

In other words, I need to show that

$$ \sum_{i=1}^n p_i(1-p_i) \to \infty \ \text{as $n\to \infty$ }.$$

I sincerely ask to help me with the proof of the last statement. Or Am i doing something wrong?

Thank you.

Answer 1

Your treatment appears sound.

The following is improvisation, so most likely it is inefficient. It transforms your question to a simpler one, that of convergence of a sequence.

I will assume $\lambda =1$ to simplify notation. The assumption is inconsequential.

The symbol $p_i$ represents the cumulative distribution function of a Gamma distribution with integer shape parameter equal to $i$, i.e. of an Erlang distribution. We have

$$p_i = e^{-t}\sum_{m=i}^\infty \frac{t^m}{m!}, \;\;p_{i+1} = e^{-t}\sum_{m=i+1}^\infty \frac{t^m}{m!},$$

$$p_{i+1} = p_i - \pi_i,\;\;\; \pi_i \equiv e^{-t}\frac{t^i}{i!},\;\;\; \pi_i\to 0 \;\text {as}\; i\to \infty$$

By the monotone converge Theorem, it also follows that $p_i$ will go to zero at the limit, and one can deduce that $p_i(1-p_i)$ will go to zero too.

So

$$p_{i+1}\left(1-p_{i+1}) = ( p_i - \pi_i ) (1- p_i + \pi_i) = p_i(1-p_i) -\pi_i(1-2p_i+\pi_i\right)$$

Using this we can decompose the sum we are interested in in two different ways, and we will use both: \begin{align} p_2(1-p_2) &= p_1(1-p_1) - \pi_1(1-2p_1+\pi_1) \\ p_3(1-p_3) & = p_2(1-p_2) - \pi_2(1-2p_2+\pi_2) \\ &= p_1(1-p_1) - \pi_1(1-2p_1+\pi_1) - \pi_2(1-2p_2+\pi_2)\\ etc \end{align} Using the first equality, i.e. simply going one index value back we get

$$\sum_{i=1}^n p_i(1-p_i) = p_1(1-p_1) + \sum_{i=1}^{n-1} p_i(1-p_i) - \sum_{i=1}^{n-1}\pi_i(1-2p_i+\pi_i)$$

and canceling out and rearranging, we get the expression

$$p_1(1-p_1) = p_n(1-p_n) + \sum_{i=1}^{n-1}\pi_i(1-2p_i+\pi_i) \tag{1}$$

Now if we decompose every $p_i(1-p_i)$ element down to an expression involving $p_1(1-p_1)$, as we indicatively did for $p_3(1-p_3)$ previously we get \begin{align} \sum_{i=1}^n p_i(1-p_i) &= np_1(1-p_1) - (n-1)\pi_1(1-2p_1+\pi_1) \\ &- (n-2)\pi_2(1-2p_2+\pi_2)-...-\pi_{n-1}(1-2p_{n-1}+\pi_{n-1}) \tag{2} \end{align}

Use expression $(1)$ to substitute for $p_1(1-p_1)$ in the right-hand-side of expression $(2)$:

\begin{align} \sum_{i=1}^n p_i(1-p_i) &= n\cdot\Big[p_n(1-p_n) + \sum_{i=1}^{n-1}\pi_i(1-2p_i+\pi_i)\Big] - (n-1)\pi_1(1-2p_1+\pi_1) \\ &- (n-2)\pi_2(1-2p_2+\pi_2)-...-\pi_{n-1}(1-2p_{n-1}+\pi_{n-1}) \\ \\\implies \sum_{i=1}^n p_i(1-p_i) &= np_n(1-p_n) + \sum_{i=1}^{n-1}\Big(i\cdot \pi_i(1-2p_i+\pi_i)\Big) \tag{3} \end{align}

We are interested in

$$\text{lim}_{n\to \infty}\sum_{i=1}^n p_i(1-p_i) = \text{lim}_{n\to \infty}\Big[np_n(1-p_n)\Big] + \sum_{i=1}^{\infty}\Big(i\cdot \pi_i(1-2p_i+\pi_i)\Big) \tag{4}$$

Use selectively the expression for $\pi_i$ and apply the Ratio Test for the convergence of the infinite series:

$$\text{lim}_{i\to \infty}\left(\frac{(i+1)\cdot e^{-t}\frac{t^{i+1}}{(i+1)!}(1-2p_{i+1}+\pi_{i+1})}{i\cdot e^{-t}\frac{t^i}{i!}(1-2p_i+\pi_i)}\right) $$

$$t\cdot\text{lim}_{i\to \infty}\left(\frac{(i+1)\cdot (1-2p_{i+1}+\pi_{i+1})}{i\cdot(i+1) (1-2p_i+\pi_i)}\right) \to 0 <1. $$

...because the terms in parentheses go to unity. So the infinite series in expression $(4)$ is absolutely convergent. Therefore, whether the sum of interest converges or diverges hints on the limit of the sequence $\{np_n(1-p_n)\}$.

Comment: after obtaining that the series $\{p_i(1-p_i)\}$ is monotonically declining, it would be trivial to realize that

$$np_n(1-p_n) < \sum_{i=1}^n p_i(1-p_i)$$ and so that if we could determine that $\{np_i(1-p_i)\}$ diverges the sum of interest would also diverge. But if we were to find out that $\{np_i(1-p_i)\}$ converges, we would not be able to conclude on the divergence or convergence of the sum of interest. All the previous work was done to show that both divergence and convergence of it depend on what happens with $\{np_n(1-p_n)\}$:What is now needed is to examine whether $\{np_n(1-p_n)\}$ is Cauchy or not.