Asymptotic Probability

Question

Suppose $x_n, y_n$ are sequences of discrete (not necessarily independent) random variables and $\mathbb{P}(x_n=a_n)\to 1$ as $n\to \infty$ and $\mathbb{P}(y_n=b_n)\to 1$ as $n\to \infty$. I am trying to convince myself that we have $\mathbb{P}(x_ny_n=a_nb_n)\to 1$ as $n\to \infty$. We can start with the bound

$$\mathbb{P}(x_ny_n=a_nb_n)\geq \mathbb{P}(x_n=a_n\mbox{ and } y_n=b_n)=\mathbb{P}(x_n=a_n\big\vert y_n=b_n)\mathbb{P}(y_n=b).$$ Taking a limit on both sides yields $$\lim_{n\to \infty}\mathbb{P}(x_ny_n=a_nb_n)\geq \lim_{n\to \infty}\mathbb{P}(x_n=a_n\big\vert y_n=b_n).$$ We need to show that $$\lim_{n\to \infty}\mathbb{P}(x_n=a_n\big\vert y_n=b_n)=1.$$ Is this true?

Answer 1

Fix an $\epsilon > 0$. Then since $\lim_{n\rightarrow \infty} \mathbb{P}(x_n = a_n) = 1$ there must be an $N_x$ such that for all $n \geq N_x$ we have $\mathbb{P}(x_n = a_n) > 1-\epsilon/2$. Similarly, there exists an $N_y$ such that for all $n \geq N_y$ we have $\mathbb{P}(y_n = b_n) > 1-\epsilon/2$. Let $N = \max(N_x,N_y)$.

Using this we have for $n \geq N$ \begin{align} 1 - \epsilon/2 &< \mathbb{P}(x_n = a_n) \\ &= \mathbb{P}(x_n = a_n \text{ and } y_n = b_n) + \mathbb{P}(x_n = a_n \text{ and } y_n \neq b_n) \\ &= \mathbb{P}(x_n = a_n \text{ and } y_n = b_n) + \mathbb{P}(x_n=a_n | y_n \neq b_n)\mathbb{P}(y_n \neq b_n) \\ &\leq \mathbb{P}(x_n = a_n \text{ and } y_n = b_n) + \mathbb{P}(y_n \neq b_n) \\ &< \mathbb{P}(x_n = a_n \text{ and } y_n = b_n) + \epsilon/2 \\ \implies1-\epsilon/2-\epsilon/2 &< \mathbb{P}(x_n = a_n \text{ and } y_n = b_n) \\ \implies 1-\epsilon&< \mathbb{P}(x_n = a_n \text{ and } y_n = b_n) \end{align} In particular this implies that for $n \geq N$ $$ 1-\epsilon < \mathbb{P}(x_n = a_n \text{ and } y_n = b_n) \leq \mathbb{P}(x_ny_n = a_nb_n) $$

This shows that indeed $$\lim_{n \rightarrow \infty} \mathbb{P}(x_ny_n = a_nb_n) = 1.$$

Answer 2

Set $A_n = \{x_n = a_n\}$ and $B_n = \{y_n = b_n\}$. Then your assumption is that both $A_n$ and $B_n$ hold with high probability, i.e. $\lim_n \mathbb{P}(A_n) = 1 = \lim_n \mathbb{P}(B_n)$.

Notice that $$\mathbb{P}(x_ny_n = a_n b_n) \geq \mathbb{P}(A_n \cap B_n),$$ so to show the result one can show that $A_n \cap B_n$ holds with high probability. This is true since $$\mathbb{P}((A_n\cap B_n)^c)=\mathbb{P}(A_n^c \cup B_n^c) \leq \mathbb{P}(A_n^c)+\mathbb{P}(B_n^c) \to 0.$$