Find $\lim_{n\to\infty} P(M_{n} \geq n^{p})$, where $M_{n}=\max_{i \leq n}X_{i}$ and $X_{i} = \sum_{j=1}^{i}Y_{j}$

Question

$Y_{j}$ is such that $P(Y_{j} = 1) = P(Y_{j} = -1) = \frac{1}{2}$

$0\leq p\leq 1$

What I have:

$\displaystyle P(M_{n} \geq n^{p}) = \sum_{k=1}^{n}P(X_{k} \geq n^{p})$

$\displaystyle P(X_{k} \geq n^{p})= P(\sum_{j = 1}^{k}Y_{j} \geq n^{p}) = \frac{1}{2^{k}}\sum_{j=\lceil n^{p} \rceil}^{k}C_{k}^{j} $

And I can't seem to be able to calculate that. Any thoughts?

Answer 1

I found an easier solution. There is Dub's theorem, so:

$P(max_{i \leq n}X_{i}) \leq \frac{EX_{N}^{+}}{n^{p}} = \frac{1}{2n^{p}}$