Let $X_i$ be the $i$-th element of the vector $X=(X_1, ..., X_m)$ of i.i.d. random variables. I am looking for a lower bound for the expression
$\mathbb{P}((\sum^n_{i=1}\prod^{m_i}_{j=1}(X_j))^2 \geq 1)$
where $m_i$ can be different for each $i$. The elements of $X$ have finite first and second moments, other than that I would like to place minimal assumptions on them (in particular, they could or could not be negative, bounded, centered). However, if there was some mild assumption that would greatly simplify the problem, that could be interesting.
I will have to compute the bound for many different $m$ and $n$. Therefore, I would prefer a solution that builds on e.g. the expected value or variance of the elements of $X$ directly, rather than those of their products or sums of products as their calculation might not be feasible.
Without the sum, I can apply the logarithm and work with a sum of i.i.d. random variables and then use e.g. Cantelli's inequality to get what I am looking for. But with the sum being there, it proves tricky.
Let $Z := (\sum^n_{i=1}\prod^m_{j=1}(X_j))^2$.
$\mathbb{P}(Z \geq 1) = 1 -\mathbb{P}(Z < 1)$
Now I am looking for upper bounds for the probability on the right side to get a lower bound of the overall expression. I would like to apply Markov's inequality since it only relies on the expected value of the $X_j$ but can not do so since the inequality is the wrong way. So I could try Chebyshev:
$= 1 -\mathbb{P}(Z - \mathbb{E}(Z) < 1 - \mathbb{E}(Z))$
$= 1 -\mathbb{P}(\mathbb{E}(Z) - Z \geq \mathbb{E}(Z) - 1)$
$\geq 1-\frac{\text{Var}(Z)}{(\mathbb{E}(Z) - 1)^2},\quad$ assuming $\mathbb{E}(Z) - 1 > 0$.
However, I do not see a feasible way of obtaining the variance or expectation in the last expression for a wide range of $n$ and $m$.
Is there perhaps an inequality, or a trick or transformation that I am overlooking?
