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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.) ?

Thanks in advance.

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  • $\begingroup$ 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. $\endgroup$ – Fimpellizieri Dec 5 '19 at 17:25

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