Show that $\Pr\big(X_t=x \, | \, X_{1},\dots, X_{t-1}\big) \geq \nu_x$ implies $\Pr\big(X_t=x \, | \, Y_{1},\dots, Y_{t-1}\big) \geq \nu_x$

Question

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

Consider a sequence of binary random variables $\{Y_t\}_t$ such that If $Y_t=1$, then $X_t=x$, i.e., $Y_t\leq 1\{X_t=x\}$, for all $t=1,2,\dots$

Could you help me to show that \begin{equation} \\Pr\big(X_t=x \, | \, Y_{1},\dots, Y_{t-1}\big) \geq \nu_x \quad \text{ for all } t=1,2,\dots \quad ? \end{equation}

Answer 1

The claim is false, as can be disproved by the following counterexample.

Set the probability space $\Omega = \{1, 2, 3, 4\}$ and $P(\{i\}) = 1/4$ for $i = 1, 2, 3, 4$. Let $A_1 = \{1\}$, $A_2 = \{2\}$, $B_1 = \{1, 2\}$, $B_2 = \{2, 3\}$. Define $X_t = I_{B_t}, Y_t = I_{A_t}, t = 1, 2$ and take $x = 1$, $\nu_x = 1/2$.

It can be easily verified that $Y_t \leq X_t = I_{\{X_t = 1\}}$, hence the condition in the claim is satisfied. However, while \begin{align} P(X_2 = 1|X_1) = P(B_2|B_1)I_{B_1} + P(B_2|B_1^c)I_{B_1^c} = \frac{1}{2} \geq \nu_x, \end{align} we have \begin{align} P(X_2 = 1|Y_1) = P(B_2|A_1)I_{A_1} + P(B_2|A_1^c)I_{A_1^c} = \frac{2}{3}I_{A_1^c}. \end{align} This means that for $\omega = 1$, $P(X_2 = 1|Y_1)(\omega) = 0$, fails to be greater than $1/2$.