In a random integer walk along a number line (each step 0.5 probability of moving right and 0.5 probability of moving left), what is the average distance from the origin during the walk?
Other questions on here ask something that sounds like this, but I'm not sure if they are talking about the average during the walk, or just your expected distance at the end of the walk.
Follow-up question: if you have the answer for the end of a walk of length $n$ (call it $w(n)$), could you get the during walk average by doing the average $\sum_1^n{w(i)}/n$, or would that not work because the distance at each step is not independent of the distance of the previous steps?
Assume without loss of generality that the random walk starts at 0. Let $X_i$ be a random variable with $P(X_i = 1) = P(X_i = -1) = 1/2$, so that the random walk is represented by the random variable $S_n = X_1 + \cdots + X_n$. At time $m$, the distance from the origin is $|S_m|$. Let us compute $E[|S_{2m}|] = E[|X_1 + \cdots + X_{2m}|]$.
The random walk can only be at an even position, so $P(|S_{2m}| = j) = 0$ for $j$ odd. Now let $2k$ be an even integer in $[0,2m]$. By symmetry, $$P(X_1 + \cdots + X_{2m} = 2k) = P(X_1 + \cdots X_{2m} = -2k)$$ Let us compute the probability that the walker is at $2k$. If the walker moves $j$ steps forward and $2m - j$ steps backwards, he is at $2j - 2m = 2k$. Thus we require $k+m$ many $+1$'s. The sum of the $X_i$ is a binomial distribution with parameters $p = 1/2$ and $n = 2m$, so that $$P(S_{2m} = 2k) = \binom{2m}{m+k} \left(\frac12\right)^{2m}$$ and $$P(|S_{2m}| = 2k) = \binom{2m}{m+k} \left(\frac12\right)^{2m-1}.$$ Computing the expectation, \begin{align*} E[|S_{2m}|] &= \sum_{k=0}^{2m} k P(|S_{2m}| = k) \\ &= \sum_{k=0}^{m} 2k P(|S_{2m}| = 2k) \\ &= \sum_{k=0}^{m} k \left(\frac12\right)^{2m-2}\binom{2m}{k+m} \\ &= \frac{m+1}{2^{2m-1}}\binom{2 m}{m+1} \\ &= \frac{m+1}{2^{2m-1}} \cdot \frac{(2m)!}{(m+1)!(m-1)!} \\ &= \frac{m+1}{2^{2m-1}} \cdot \frac{(2m)!}{m!m!} \cdot \frac{m}{m+1} \\ &= \frac{2m}{2^{2m}} \binom{2m}{m}. \end{align*} Similarly, one can show that $$E[|S_{2m+1}|] = \frac{2m+1}{2^{2m}} \binom{2m}{m}.$$ Now we can compute the average distance. \begin{align*} E\left[\frac{|S_1| + \cdots + |S_{2n}|}{2n} \right] &= \frac{1}{2n}(E[|S_1|] + \cdots + E[|S_{2n}|]) \\ &= \frac{1}{2n}\left(\sum_{k=0}^{n-1} E[|S_{2k+1}|] + \sum_{k=1}^n E[|S_{2k}|]\right) \end{align*}