Series in Probability.

Question

I'm working on a problem which has a bit like

$^{(1)}P(\Sigma_{i\in N}A_i=1)$

where the sequence $A_i\in\{0,1\}, i\in N$, also independent.

$^{(2)}P(A_i=1)=\frac{2}{3i+2}$

How can I simplify $(1)$?

Answer 1

If I understand your question statement correctly, we have $A_i$ as mutually independent Bernoulli trails with the chance of success $p_i = \frac2{3 i + 2}$ for $A_i = 1$. In other words, the chance of $A_i = 0$ is $q_i = 1 - p_i = \frac{3i}{ 3i + 2}$.

Your notation is a bit unclear in that whether $N$ means a general finite integer or $\mathbb{N}$ the entire set of countably infinite integers.

Anyway, let's consider $i = 1,2,\ldots, n$ for a finite $n$.

\begin{align} \mathcal{P}_n \equiv P\left( \sum_{i = 1}^n A_i = 1 \right) &= \sum_{i = 1}^n \left[ P(A_i = 1) \prod_{j \neq i} P(A_j = 0) \right] \\ &= \sum_{i = 1}^n \left[ \frac2{3i + 2} \prod_{j \neq i} \frac{ 3i }{3i + 2} \right] \\ &= \frac23 \frac{\Gamma \left(\frac{5}{3}\right) \Gamma (n+1)}{\Gamma(n+\frac53)} H_n \end{align} where $\Gamma(k)$ is the Gamma function, which equals $(k-1)!$ for integer $k$, and $H_n$ is the harmonic number.

At the limit of $n \to \infty$ to cover all the natural numbers, it can be shown that $\mathcal{P}_n \to 0$.

Actually, if all you care is the $i \in \mathbb{N}$ case with all natural numbers, right off the bat we know the desired probability is zero $P\left( \sum_{i \in \mathbb{N} } A_i = 1 \right) = 0$.

This is because they sum to $1$ but only one of the $A_i$ get to be one and all the infinitely many others has to be zero. The product of infinitely many such $q_i = \frac{3i}{3i+2} < 1$ is trivially zero.

Answer 2

There must be a typo in the question. Since $\sum P\{A_i=1\} =\sum_i \frac 2 {3i+2} =\infty$ it follows by Borel Cantelli Lemma that with probability $1$, infinitley many $A_i$'s are equal to $1$. Hence $P\{\sum_i A_i=\infty\}=1$.