The expected range covered by a random walk The question that I have been struggling with lately is:
If we have a one-dimensional random walk of length $n$ (consisting of $n$ steps) with discrete steps $1$ and $-1$, with probabilities of occurrence of $p_{1}$ and $p_{-1}$ respectively, what would the expected range covered by the walk be? In other words, what would the expected difference between the highest and the lowest point the walk reaches be?
Furthermore, if it is even possible to find the result for the first case, what would the expected difference be for a walk with some other steps (possibly more than two, f.ex 1, -1 and -2)? Or if the step was a continuous variable?

Let $r=\frac{p_1}{p_{-1}}$, $M_n=\max_{k\le n} S_k$, $m_n=\min_{k\le n}S_k$. Then by reflection principle $$P(M_n\ge t)=P(S_n\ge t)+r^{t}P(S_n\le -(t+1))$$
$$P(-m_n\ge t)=P(S_n\le -t)+r^{-t}P(S_n\ge t+1)$$ and hence you get compute $E(M_n)$ and $E(-m_n)$. Case $r=1$ is especially easy $(E(M_n-m_n)=\Theta(\sqrt n))$.
Similarly, for Brownian motion with drift $\mu$ and $\sigma^2=1$:
$$P(M_n>t)=P(B_n>t)+e^{2t\mu}P(B_n\le -t)$$
$$P(-m_n>t)=P(B_n<-t)+e^{-2t\mu}P(B_n\ge t)$$
with an easy computation for $\mu=0$. If $\mu>0$, then $E(-m_n)$ converges to a constant, while $E(M_n)\sim C(\mu)n+O(\sqrt n)$. All other random walks (with finite variance) converge to Brownian motion asymptotically. 
