# Tossing a fair coin and denote his probability

I am trying to calculate the probability of an infinitely tossed coin. The exercise is:

A coin is tossed infinitely often, where the probability of success is $$p$$. The variable $$Y_k$$ is $$1$$, if the $$k$$-th litter is successful, and $$0$$, in the event of failure. The variables $$\{Y_k\}$$ are independent. We define the variables $$N_n := \min \{ k \geq n : Y_k =0 \} -n$$ as the length of a winning streak with start in n. We define the function $$f(n) = \lfloor a \frac{\log n }{ - \log p} \rfloor$$, with $$\lfloor x \rfloor = \max \{ n \in \mathbb{Z} : n \leq x \}$$ and $$a > 0$$. Let $$A_n$$ be the event $$A_n = \{ N_n \geq f(n) \}$$

1. Calculate $$P[A_n]$$ and find an upper limit for $$P[A_n]$$.
2. Let $$a>1$$. Does the $$A_n$$ enter infinitely or finally often?
3. Can you also answer the question in case $$a<1$$ with the same argument? Enter a reason.

my argumentation for 1)

the probability of a success is $$p$$. We want to find the probability that a winning streak is larger or equal to $$f(n)$$. That means the tosses from $$1$$ to $$f(n)$$ had to be successfull so we have that

$$P[A_n] = p^{f(n)}$$

and the upper limit should be:

$$p^{f(n)} = p^{\lfloor a \frac{\log n }{ - \log p} \rfloor} \leq p^{ a \frac{\log n }{ - \log p} } = p^{a \log n \, (- \log p)^{-1} } = p^{ - a \log n} p^{- \log p} \leq e^{-a \log n} e^{- \log p} = p^{-1} n^{-a}$$

can i just say that $$p^{- a \log n} \leq e^{- a \log n}$$ ? Does it always hold?

2.) It follows by Borell - Cantelli:

$$\sum_{n=1}^{\infty} P[A_n] \leq p^{-1} \sum_{n=^1}^{\infty} n^{-a} < \infty$$ and this sum converges since $$a>1$$. So by Borell - Cantelli the event occurs finally often.

3.) for $$a \leq 1$$ the sum dont converges, so we can not say anything about it.

Does someone has a better argue for 3.) ?

• For the upper limit, notice that $\log n / \log p = \log_p(n)$ so we can settle $n^{-a}$. For $(3)$, even if the sum converged via a lower bound, you'd need pairwise independence to apply second Borell-Cantelli lemma, so you definitely can't use the same argument. – Fimpellizieri Dec 5 '19 at 17:25