Let $X$ be a random variable such that $X = X_{1}+X_{2}+\dots+X_{n}$, where $$ X_{i} = \begin{cases} 1\quad \text{with probability $p$}\\ 0\quad \text{otherwise}, \end{cases},\quad 1\leq i\leq n. $$
Also, $X_{i}$ and $X_{j}$ $(1\leq i<j\leq n)$ are dependent.
My question is: how to calculate
$\Pr(X=0)$ $\quad$ or $\quad$ $\Pr(X<1)$
I tried to solve this problem by using Chernoff Bound, Janson's Inequality, application of negatively dependent random variables. But
- If $X_{i}$ and $X_{j}$ $(1\leq i<j\leq n)$ are independent, then I use Chernoff Bound, but they are dependent.
- Unfortunately, $\sum_{1\leq i<j\leq n} \Pr(X_{i}=1=X_{j})$ is not $o(E(X))$, so I can't use Janson's Inequality.
- $X_{i}$ and $X_{j}$ $(1\leq i<j\leq n)$ are not negatively dependent.
Can you suggest a way to solve this problem?
your answer depends on the exact depends structure (i.e. the underlying Copula) of the $(X_1, \dots, X_n)$. With no further information, $\Bbb P (X=0)$ cannot be further determined. To illustrate this, considere these two examples for $n =2$.
1.) if $X_1 = X_2$ then $\Bbb P (X=0) = \Bbb P (X_1=0) = 1-p$.
2.) if $p = \frac 12$ and $X_1 = 1 - X_2$ then $\Bbb P (X=0) = 0$.
In general if $F$ is the joint distribution of $(X_1, \dots, X_n)$, i.e. $$ F(x_1, \dots, x_n) = \Bbb P (X_1 = x_1, \dots, X_n = x_n), $$ then $$ \Bbb P (X=0) = F(0, \dots, 0). $$