Implications of $\Pr\big(X_n=x \, | \, X_{1},\dots, X_{n-1}\big) \geq a$

Question

Let $(X_n)_n$ be a sequence of discrete random variables and assume that \begin{equation} \Pr\big(X_n=x \, | \, X_{1},\dots, X_{n-1}\big) \geq a \quad \text{ for all } n=1,2,\dots \end{equation} where $a\in [0,1]$.

Consider a sequence of random variables $(Y_n)_n$ such that $$ Y_n=f(X_n, Z_n) $$ where $(Z_n)_n$ is a sequence of random variables, uniform in $[0,1]$ and independent of $(X_n)_n$ and $$ f(X_n, Z_n)=\begin{cases} 1 & \text{if $X_n=x$ and $Z_n\leq 0.3$}\\ 0 & \text{otherwise } \end{cases} $$

Is it true that \begin{equation} \Pr\big(X_n=x \, | \, Y_{1},\dots, Y_{n-1}\big) \geq a \quad \text{ for all } n=1,2,\dots \quad ? \end{equation}

Answer 1

Given this strengthened condition (compared with this question), the proposition is true. To prove, note that

1. $\sigma(Y_1, \ldots, Y_{n - 1}) = \sigma(f(X_1, Z_1), \ldots, f(X_{n - 1}, Z_{n - 1})) \subset \sigma(X_1, Z_1, \ldots, X_{n - 1}, Z_{n - 1})$.
2. $\sigma(X_1, X_2, \ldots)$ is independent of $\sigma(Z_1, Z_2, \ldots)$.

These two implications are sufficient to derive the desired inequality (the discreteness of $X$, the uniformity of $Z$ and the specific form of $f(X, Z)$ are not really needed). First, property 1 and the tower property of conditional expectation imply that \begin{align} P(X_n = x|Y_1, \ldots, Y_{n - 1}) = E[P(X_n = x |X_1, \ldots, X_{n - 1}, Z_1, \ldots, Z_{n - 1})|Y_1, \ldots, Y_{n - 1}]. \tag{1}\label{1} \end{align}

Next, property 2 implies (use the corollary in this answer) that the inner conditional probability in the right hand side of $\eqref{1}$ is $P(X_n = x|X_1, \ldots, X_{n - 1})$, which by condition is bounded below by $a$. Hence the left hand side of $\eqref{1}$ is also bounded below by $a$ (here we used the monotonicity of conditional expectation: if $\xi \geq \eta$, then $E[\xi|\mathscr{G}] \geq E[\eta|\mathscr{G}]$). This completes the proof.