# Equivalence between conditions for convergence

Let $(X_k)$ be independant random variables such that $X_k\sim\mathcal{P}(p_k)$ (Poisson distribution with parameter $p_k$). So in particular we have $\sum_{n=1}^NX_k \sim \mathcal{P}(\sum _{n=1}^Np_k)$ and $\mathbb{E}(X_k)=p_k$. By hypothesis $np_n\to \infty$ and can be write : $p_n=b(n)/n$ where $b(n)$ is slowly varying function.

A function $f$ defined on $\mathbb{R}$ is a slowly varying function if for every $\delta >0$, $n^{\delta }f(n)$ is increasing and $n^{\delta }f(n)$ is decreasing if $n$ is large enough. So in particular we have $p_n\to 0$.

I want to show equivalence between the following conditions :

i) a.s. $\forall t \neq 0 \in \mathbb{T}, \, \sum_{k=1}^N\mathbf{1}_{\{ X_k>0\}}e^{2i\pi kt}=o( \sum_{k=1}^N\mathbf{1}_{\{ X_k>0\}})$ when $N\to \infty$.

ii)a.s. $\forall t \neq 0 \in \mathbb{T}, \, \sum_{k=1}^NX_ke^{2i\pi kt}=o( \sum_{k=1}^NX_k)$ when $N\to \infty$.

iii)a.s. $\forall t \neq 0 \in \mathbb{T}, \, \sum_{k=1}^NX_ke^{2i\pi kt}=o( \sum_{k=1}^Np_k)$ when $N\to \infty$.

The notation $\sum_{k=1}^N\mathbf{1}_{\{ X_k>0\}}e^{2i\pi kt}=o( \sum_{k=1}^N\mathbf{1}_{\{ X_k>0\}})$ means that $$\frac{\sum_{k=1}^N\mathbf{1}_{\{ X_k>0\}}e^{2i\pi kt}}{\sum_{k=1}^N\mathbf{1}_{\{ X_k>0\}}}\to 0$$ when $N\to \infty$.

The torus will be denoted by $\mathbb{T}=[0,1)=\mathbb{R}\setminus \mathbb{Z}$.

First observe that $\sum _k p_k=\infty$. But we can obtain a more precise estimation : $\sum _k^N p_k=a_N\log (N)$ where $a_N\to \infty$ when $N\to \infty$.
So by using Chernoff bound we can show (cf Mickael's answer) that $ii)\implies iii)$ since $\sum _k X_k\leq M \sum _k p_k$ with probability one. The same argument can be used to show that with probability one $\sum _k p_k\leq m \sum _k X_k$. So $ii)\Leftrightarrow iii)$.

It remains to show the equivalence between $i)$ and $ii)$ and $iii)\implies ii)$. I suggest the following argument.

Let $Y_k=X_k-\mathbf{1}_{\{ X_k>0\}}$, $\mathbb{P}(Y_k>0)=e^{-p_k}\sum _{n\geq 2}\frac{p_k^n}{n!}$. So $\sum _k\mathbb{P}(Y_k>0)<\sum _n \sum _k \frac{p_k^2}{n!}<+\infty$ (we can assume that $0<p_k<1$ and by hypothesis $\sum p_k^2=\sum b(k)^2/k^2<+\infty$). By Borel-Cantelli lemma $Y_k=0$ a.s. for sufficiently large $k$. Since $\sum _k X_k \to \infty$ the result follows.

• Indeed, $t\in (0,1)$. I didn't see a mistake on indices of summation. – Patissot May 4 '15 at 16:36
• As @Teepeemm mentions, you should change your summation index from $n$ to $k$. Also define: (i) $\mathbb{T}$, (ii) Notation "$\forall t \neq 0 \in \mathbb{T}$," (iii) What is meant by a complex-valued function being $o(\cdot)$, (iv) define a "slowly varying function." – Michael May 6 '15 at 21:52
• Also, how can you "very easily show the sum $\sum_{k=1}^N X_k$ tends to infinity" if you do not specify the $b(n)$ function? For example, if $b(n)=1/n$ then $p(n) = 1/n^2$ and $\sum_{k=1}^{\infty} X_k$ is finite with probability 1. – Michael May 6 '15 at 21:53
• Intuitively, if $b(n)=1$ for all $n$, I would not expect the statements to be true, even if they are equivalent. – Michael May 6 '15 at 22:05
• It is also not clear why we can assume $0 < p_k < 1$ and $\sum_{k=1}^{\infty} b(k)^2/k^2 < \infty$. The latter is not true, for example, if $b(k)=k$. Overall, this problem is not stated clearly. I do agree with your conclusions that if $0 < p_k < 1$ and $\sum_{k=1}^{\infty} b(k)^2/k^2 < \infty$, then with prob 1 we have $X_k = 1_{\{X_k>0\}}$ for all sufficiently large $k$. – Michael May 6 '15 at 22:07

Some related results: Define $\lambda_n = p_n = E[X_n]$.

Claim 1: If $\sum_{n=1}^{\infty} \lambda_n < \infty$, then $\sum_{n=1}^{\infty} X_n < \infty$ with prob. 1.

Proof: Suppose $\sum_{n=1}^{\infty} \lambda_n < \infty$. Then: \begin{align} \sum_{n=1}^{\infty} Pr[X_n>0] &= \sum_{n=1}^{\infty} [1-e^{-\lambda_n}] \leq \sum_{n=1}^{\infty} \lambda_n < \infty \end{align} and hence (with prob 1) $X_n=0$ for all sufficiently large $n$. So $\sum_{n=1}^{\infty} X_n < \infty$ with prob 1. $\Box$

Claim 2: For all $z>0$ and all positive integers $K$ we have: $$Pr\left[\frac{\sum_{n=1}^K X_n}{\sum_{n=1}^{K} \lambda_n} \geq z\right] \leq e^{-(z+1-e)\sum_{n=1}^K\lambda_n}$$

Proof: By the Chernov bound we have: \begin{align} Pr\left[\sum_{n=1}^K X_n \geq z\sum_{n=1}^K\lambda_n\right] &\leq E[e^{\sum_{n=1}^KX_n}]e^{-z\sum_{n=1}^K\lambda_n}\\ &=\left(\prod_{n=1}^KE[e^{X_n}]\right)e^{-z\sum_{n=1}^K\lambda_n}\\ &= \left(\prod_{n=1}^K e^{\lambda_n(e-1)} \right) e^{-z\sum_{n=1}^K\lambda_n}\\ &= e^{-(z+1-e)\sum_{n=1}^K\lambda_n} \end{align} $\Box$

For example, using $z=2$ in Claim 2 gives for all positive integers $K$:

$$Pr\left[ \frac{\sum_{n=1}^KX_n}{\sum_{n=1}^K\lambda_n} \geq 2 \right] \leq e^{-(3-e)\sum_{n=1}^K\lambda_n}$$

and if $\sum_{n=1}^{\infty}\lambda_n=\infty$ then the probability on the left-hand-side goes to $0$ as $K\rightarrow\infty$.

Claim 3: If there is a constant $c>0$ such that $\sum_{n=1}^K \lambda_n \geq c\log(K)$ for all positive integers $K$, then with prob 1 we know $\sum_{n=1}^K X_n \leq \left(e-1+\frac{2}{c}\right)\sum_{n=1}^K\lambda_n$ for all sufficiently large $K$. Thus, with prob 1 we have $\sum_{n=1}^KX_n \leq \Theta(\sum_{n=1}^K \lambda_n)$.

Proof: Define $z = e-1 + 2/c$. By Claim 2 we have: \begin{align} Pr\left[\sum_{n=1}^KX_n \geq z\sum_{n=1}^K \lambda_n\right] &\leq e^{-(z+1-e)\sum_{n=1}^{K}\lambda_n}\\ &\leq e^{-(z+1-e)c\log(K)}\\ &= e^{-2\log(K)} \\ &= 1/K^2 \end{align} Since $1/K^2$ is summable, it follows (with prob 1) that $\sum_{n=1}^KX_n \leq z\sum_{n=1}^K\lambda_n$ for all sufficiently large $K$. $\Box$

In the same way, if there is a constant $d>0$ such that $\sum_{K=1}^{\infty} e^{-d\sum_{n=1}^K\lambda_n} < \infty$, then with prob 1 we have $\sum_{n=1}^KX_n \leq \Theta(\sum_{n=1}^K\lambda_n)$. This follows immediately from Claim 2 by defining $z$ such that $z+1-e=d$.