Hi here is my problem:
We perform infinitely many independent experiments. The $n^{th}$ one is successful with probability $n^{−α}$ and fails with probability $1 − n^{−α}$, $0 < α$. Let $k ≥ 1$. We are happy if we see $k$ consecutive successes infinitely often. What is the probability of this?
I tried this:
Let $A_i$ be the event "$k$ consecutive successes happen from the $i^{th}$ trial". Then $$\Pr\{A_i\}=\Pr\{\sum_{j=i}^{i+k}X_j=k\}$$ with $X_1,X_2,...$ iid random variables having Bernoulli distribution with parameter $n^{-α}$.
So $$\Pr\{A_i\}=i^{-αk}$$
Then, using Borel-Cantelli Lemmas, we have that $$\sum_{i}^{\infty}i^{-αk}=\infty$$ if $αk<1$ thus $$\Pr\{\bigcap^{\infty}_{j=i}\bigcup^{\infty}_{i=1}A_i\}=1$$
But I'm sure it's wrong (not coherent) and I don't know where to look for a good way to find the answer...
Thanks, Herosix
You have done almost all the work, I'll repeat. For the event $A_i$
$$ \Pr\{A_i\}=\Pr\{X_i=1,X_{i+1}=1,\ldots,X_{i+k-1}=1\}=i^{-\alpha}(i+1)^{-\alpha}\cdot\ldots\cdot(i+k-1)^{-\alpha} = \frac{1}{\left[\,i\,(i+1)\ldots(i+k-1)\,\right]^\alpha} $$ For $\alpha k>1$ the series $\sum_{i=1}^\infty \Pr\{A_i\}<\sum_{i=1}^\infty i^{-\alpha k}$ converges. By Borel-Cantelli lemma, events $A_i$ occure infinitely often with zero probability.
For $\alpha k\leq 1$ the series diverges. But please note that $A_i$ are not independent. To use second part of Borel-Cantelli lemma, we need in independent events. We can define subsequence of events $B_n=A_{nk+1}$, $n\geq 0$. This events are independent and the event $\{B_n \text{ i.o.}\}$ implies the event $\{A_n \text{ i.o.}\}$ (i.o.=infinitely often). $$ \Pr\{B_n\}=\frac{1}{\left[\,(nk+1)\,(nk+2)\ldots((n+1)k)\,\right]^{\alpha}}>\left((n+1)k\right)^{\alpha k} $$ and $\sum_{n=1}^\infty \Pr\{B_n\}=\infty$. Then the events $B_n$ occure i.o. with probability $1$. Then the probability that $A_i$ i.o. is also equals to $1$.