Suppose we have a sequence of $n$ Bernoulli trials: $$X_1, X_2, \ldots, X_n,$$ where the success probability of $X_i$ depends on the results of $X_1, \ldots, X_{i-1}$. However, suppose we know that $$Pr( X_i = 1 | X_1 = x_1, X_2 = x_2, \ldots, X_{i-1} = x_{i-1} ) \ge \eta,$$ where $\eta > 0$ is a constant and $x_1,\ldots,x_{i-1}$ are arbitrary.
Then it seems intuitively obvious that, if $Y_1,\ldots, Y_n$ are independent Bernoulli trials, each with probability $\eta$ of success, then $$Pr \left( \sum_{i = 1}^n X_i \le b \right) \le Pr \left( \sum_{i = 1}^n Y_i \le b \right), $$ where $b$ is an arbitrary integer.
But I have been unable to prove this statement. Could anyone provide a suggestion for how to prove this?
I was able to find the following proof.
Let $\Omega$ be the set of all binary sequences of length $n$ with at least $b + 1$ ones. For $x \in \Omega$, let $Pr_{1}(x)$ denote the probability of $x$ occurring as a result of the trials $X_1, X_2, \ldots, X_n$. Suppose the last Bernoulli trial $X_n$ is replaced with an independent Bernoulli trial $Y_n$ with success probability $\eta$ -- denote the resulting probability of a sequence $x$ as $Pr_{2}(x)$. Any sequence in $\Omega$ whose probability increases as a result of the change must end in a $0$. Let $x0 \in \Omega$; then it also holds that $x1$ is in $\Omega$. Further, $$ Pr_2( x1 ) + Pr_2( x0 ) = Pr_2(x) = Pr_1(x) = Pr_1(x1) + Pr_1(x1). $$ Therefore, $Pr_2( \Omega ) \le Pr_1( \Omega )$. Continue this argument inductively to replace $X_{n-1}, X_{n-2}, \ldots X_1$ with independent Bernoulli variables $Y_{n-1}, \ldots, Y_1$ to obtain probability distributions $Pr_i$ such that $Pr_{i+1}( \Omega ) \le Pr_i(\Omega)$. Finally,
\begin{align*} Pr \left( \sum_{i = 1}^n Y_i \le b \right) &= 1 - Pr_n ( \Omega ) \\\\ &\ge 1 - Pr_1( \Omega ) \\\\ &= Pr \left( \sum_{i = 1}^n X_i \le b \right). \end{align*}