What's the probability for k consecutive 1s? I can use Borel-Cantelli Let ${X_{i}}$ be an infinite sequence of independent random variables with $p = P(X_{i} = 1) = 1−P(X_{i} = −1)$. Let $A_{k}$ be the event that $(X_{2^k},X_{2^k+1},X_{2^k+2},...,X_{2^{k+1}−1})$ contains a sequence of k (or more) consecutive 1s. Prove that
$P(\limsup A_{k}) = \mathcal{1}_{[p ≥ 1/2]}.$ (the indicator function of 1)
be happy for some help... I tried to calculate $A_{k}$ in order to use the Borel-cantelli lemmas but it didn't worked out.
thank you everyone!
 A: If there are at least $k$ consecutive ones, then there are exactly $k$ ones starting from some place $j$ between $2^k$ and $2^{k+1}-k-1$. Each such sequence has probability $p^k$, so by union bound the probability of at least one such sequence is at most $(2^{k}-(k-1))p^k<(2p)^k$. So  $A_k<(2p)^k$, and if $p<1/2$ then $A_k$s  are bounded by geometric progression of ratio below 1, and hence has finite sum.
From the other end, the probability that there is a sequence of at least $k$ ones is at least as big as the probability that first $k$ are ones, or that second $k$ are ones or ... or the $\left \lfloor \frac{2^k}{k} \right \rfloor $th $k$ are ones. Each of those have probability  $p^{k}$ and they are independent, so the probability of all of them not happening is $(1-p^k)^{\left \lfloor \frac{2^k}{k} \right \rfloor }$. (We want this to be small.)
This is decreasing in $p$, so if $p\geq 1/2$ then we can take the value at $p=1/2$ as a bound. Further, the expression is monotone in the power, and the floor is at least the exponent minus 1, so the expression at most $(1-\frac{1}{2^k})^{-1}(1-\frac{1}{2^k})^\frac{2^k}{k}$.
The second term $(1-\frac{1}{2^k})^\frac{2^k}{k}$ is of the form $((1-1/x)^{x})^{1/k}$, and since $((1-1/x)^{x})<1/e$ the term is  $<e^{-1/k}$. This last is in turn $< (1-\frac{1}{2k})$ (because $e^{-x}<1-\frac{1}{2}x$ for $0<x\leq 1$ since the function  $1-\frac{1}{2}x-e^{-x}$ is concave, is zero at zero with positive derivative, and is positive at $x=1$; then take $x=1/k$).
Overall we get an upper bound of $(1-\frac{1}{2k})(1-\frac{1}{2^k})^{-1}$ which us  $<1-\frac{1}{4k}$ as soon as $2^k>4k$ i.e. for $k>4$. So the complementary probability is $>\frac{1}{4k}$. Thus we have a bound from below by divergent series, as wanted.
