Say I have real-symmetric random walk $S_n = X_1+X_2+...+X_n$, where $P(X_k=1)=P(X_k=-1)=\dfrac{1}{2}.$ I showed that $E(S_n^+)$ is a non-decreasing sequence so it converges (possibly to infinity). But how does one actually describe the limit? Here $S_n^+ = \max{\{S_n, 0\}}.$
Furthermore, intuition suggests that $S_n^+(\omega)$ itself cannot converge to a finite limit, but I cannot get an estimate.
Can anybody give me a hint on above two questions?
On
One idea to tackle this question could be to compute the Doob decomposition of the submartingale $S_n^+$. (That this process is a submartingale follows from the Jensen inequality for conditional expectation.)
It is relatively straighforward to prove the the compensator (the local trend) of $S_n^+$ is given by $\frac 12 1_{\{S_{n-1} = 0\}}$, which means that $S_n^+ - S_{n-1}^+$ is conditionally on $S_{n-1}$ always zero, unless $S_{n-1} = 0$. In this case $S_n^+ = S_n^+ - S_{n-1}^+$ is either 0 or 1.
In particular, the expectation of $S_n^+$ should correspond to $1/2$ times the expected number of visits in 0 until time $n-1$. (So the expectation should converge to infinity, using the recurrence of the random walk.)
$$\begin{eqnarray*}\mathbb{P}[S_n=m]&=&\frac{1}{2^n}\left|\left\{v\in\{0,2\}^n:\sum v=m+n\right\}\right|\\&=&\frac{1}{2^n}[z^{m+n}](1+z^2)^n\end{eqnarray*}$$
$$\mathbb{E}[S_n^+]=\sum_{k\geq 1}\mathbb{P}[S_n^+\geq k]=\sum_{k\geq 1}k\cdot\mathbb{P}[S_n=k]$$ $$\mathbb{E}[S_{2n}^+]=\frac{1}{4^n}\sum_{k\geq 1}(2k)[z^{2n+2k}](1+z^2)^{2n}=\frac{1}{4^n}\sum_{k\geq 1}2k\binom{2n}{n+k}=\frac{n\binom{2n}{n}}{4^n} $$ diverges to $+\infty$ like $\sqrt{\frac{n}{\pi}}$. In general, $$ \mathbb{E}[S_n^+]\approx \color{red}{\sqrt{\frac{n}{2\pi}}} \to +\infty $$