# Pointwise convergence in limit of conditional probabilities implies almost sure convergence?

Consider two sequences of real valued random variables $\{X_n\}_{n\in \mathbb{N}}$ and $\{Y_n\}_{n\in \mathbb{N}}$, with, $\forall n \in \mathbb{N}$, $X_n$ and $Y_n$ defined on the probability space $(\Omega, \mathcal{F}, P),$ respectively with support $\mathcal{X}, \mathcal{Y}$. Consider a sequence of real numbers $\{a_n\}_{n\in \mathbb{N}}$.

Assume that for some $b\in \mathbb{R}$ $$\lim_{n\rightarrow \infty} | \overbrace{P(X_n\leq b| Y_n=y)}^{\text{scalar}}-a_n|=0 \text{ }\text{ \forall y \in \mathcal{Y}}$$

Does this imply $$\overbrace{P(X_n\leq b| Y_n)}^{\text{Random variable because of Y_n (unless X_n\perp Y_n)}}-a_n \rightarrow_{a.s.}0 \text{ as n\rightarrow \infty}$$ ?

• I think the answer is yes because almost sure convergence by definition means that $P(\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} |P(X_n\leq x| Y_n=Y_n(\omega))-a_n| =0)=1$ which I think it is implied by pointwise convergence in limit – TEX Apr 26 '18 at 11:26
• But I have doubts since I get always confused with conditional probabilities. Could you confirm this? – TEX Apr 26 '18 at 11:27
• Sorry, but I don't know how $P(X_n\le b|Y_n)$ is defined. – Logic_Problem_42 Apr 26 '18 at 15:59
• Yes, what you wrote is correct. – Landon Carter Apr 26 '18 at 17:00
• Yes, you are correct. – Landon Carter Apr 26 '18 at 18:22

Given:

$$\forall y \in \mathcal Y, \lim_n \frac{E[1_{B_n} 1_{Y_n=y}]}{P(Y_n=y)}= \lim_n a_n$$

Conjecture:

$$\lim_n E[1_{B_n}| Y_n]= \lim_n a_n$$

Ugh, the fact that we can say '$Y_n=y$' means $Y_n$ is discrete? In that case:

$$E[1_{B_n}| Y_n] = \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$\to \lim E[1_{B_n}| Y_n] = \lim \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$= \sum_{y \in \mathcal Y} \lim E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$= \sum_{y \in \mathcal Y} [\lim E[1_{B_n}| Y_n = y]][\lim[1_{\{Y_n=y\}}]]$$

$$= \sum_{y \in \mathcal Y} [\lim a_n] \lim[1_{\{Y_n=y\}}]$$

$$= [\lim a_n] \sum_{y \in \mathcal Y} \lim[1_{\{Y_n=y\}}]$$

Well if I proved for the discrete case, then it would not have been sufficient to disprove as a whole. However, I disproved for the discrete case which is sufficient to disprove as a whole.

Observation: In the discrete case, this sounds like:

$$\lim_n b_{n,k} = \lim_n a_n \to \lim_n \sum_{k} b_{n,k}c_k = \lim_n a_n$$