Let $p,\mu_c \in [0,1]$. Let $(X_i)_{i \in \mathbb{N}}$ be a sequence of i.i.d random variables, each following a Bernoulli distribution $\mathcal{B}(p)$. For $n \in \mathbb{N}$, let $ \mu_n = \frac{1}{n+1} \sum_{i=0}^n X_i $ and $A_n$ the event: "$\mu_n \leq \mu_c$". I am looking for a way to compute $$ \mathbb{P} \left(\bigcap_{n \in \mathbb{N}} A_n \right), $$ that is, the probability that the mean of the sequence never exceeds the critical value $\mu_c$.
You can assume $\mu_c > p$, since this probability is $0$ when $\mu_c \leq p$.
I first wanted to compute it using a martingale, but the optional stopping theorem does not seem to help. I don't know if computing it "by hand" (combinatorially) is possible, but it would be quite unsatisfying anyway. Eventually, it may be somehow linked to this post, but I don't understand Brownian motion well enough to use it correctly.
Thank you for any advice!
Edit: When $p = \frac{1}{2}$ and $\mu_c = \frac{3}{4}$, simulations give $$\mathbb{P} \left(\bigcap_{n \leq 1000} A_n \right) \approx 0.456.$$ I think that this is a very good approximation of $\mathbb{P} \left(\bigcap_{n \in \mathbb{N}} A_n \right)$ (going up to 2000 does not affect the result). This fact may be used to check any suggested answer. I find it surprisingly high, since $\mathbb{P}(A_0)$ alone is equal to $\frac{1}{2}$.
from Hot Weekly Questions - Mathematics Stack Exchange
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