Let ${X_i}$ be iid with $E[X]=0$ and $Var(X)=\sigma^2$ Let $S_0=0$ and $S_n=X_1+...+X_n$ for all $n \ge 1$. Show that $\lim_{n\rightarrow \infty}P(S_k>0 \space for \space k=n,n+1,\dots,2n)=1/4 $.
I try to apply Donsker's theorem with the continuous function $\omega\mapsto\min_{1/2\leq t\leq1}\omega\left( t\right) $. After this, I got the following expression.
$$ \lim_{n\rightarrow\infty}P\left( \min_{n\leq k\leq2n}\frac{S_{k}}{\sqrt{2n}% }>x\right) =P\left( \min_{1/2\leq t\leq1}B_{t}>x\right) $$
Anyone knows what should I do next?
An approach is to condition on $B_{1/2}$. Slightly more generally, fix some $s\gt0$ and let $$ A=\left[\min\limits_{s\leqslant t\leqslant2s}B_t\gt0\right], $$ then $P[A\mid B_s=x]=0$ for every $x\leqslant0$ and $P[A\mid B_s=x]=P[T_x\gt s]$ for every $x\gt0$, where, for every $x$, $T_x=\min\{t\geqslant0\mid B_t=x\}$.
By the reflexion principle, $P[T_x\lt s]=2P[B_s\gt x]$ hence $P[T_x\gt s]=P[|B_s|\lt x]$. Considering a Brownian motion $W$ independent of $B$, one gets $$ P[A]=P[|B_s|\lt W_s]=\tfrac12P[|B_s|\lt |W_s|]=\tfrac12\cdot\tfrac12, $$ first because the distribution of $W_s$ is symmetric and then because $(B,W)$ and $(W,B)$ are equidistributed.
More generally, for every positive $s$ and $u$, $$ P[\min\limits_{s\leqslant t\leqslant s+u}B_t\gt0]=\tfrac12P[|B_u|\lt |W_s|]=\tfrac12P[|B_1|\lt v\cdot|W_1|],\qquad v=\sqrt{s/u}. $$ The ratio $W_1/B_1$ is standard Cauchy hence $$ P[\min\limits_{s\leqslant t\leqslant s+u}B_t\gt0]=\int_v^{+\infty}\frac{\mathrm dx}{\pi (1+x^2)}=\frac1\pi \arctan v. $$