0
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ 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 $\endgroup$ – TEX Apr 26 '18 at 11:26
  • $\begingroup$ But I have doubts since I get always confused with conditional probabilities. Could you confirm this? $\endgroup$ – TEX Apr 26 '18 at 11:27
  • 1
    $\begingroup$ Sorry, but I don't know how $P(X_n\le b|Y_n)$ is defined. $\endgroup$ – Logic_Problem_42 Apr 26 '18 at 15:59
  • 1
    $\begingroup$ Yes, what you wrote is correct. $\endgroup$ – Landon Carter Apr 26 '18 at 17:00
  • 1
    $\begingroup$ Yes, you are correct. $\endgroup$ – Landon Carter Apr 26 '18 at 18:22
-1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.