If a ransom walk is binomial (1/2 probability of going forward, 1/2 backward) why isn;t the variance
a) $\sigma=(\frac{n}{4})^.5$
b) instead of $\sigma=(n)^.5$
these sources seem to give conflicting answers for the variance.
This source defines the variance as b) http://www.pma.caltech.edu/~mcc/Ph127/a/Lecture_8.pdf
This source says it's a) http://mathworld.wolfram.com/RandomWalk1-Dimensional.html
Let's calculate. Let $X_i=1$ with probability $1/2$, and $-1$ with probability $1/2$. Let $Y=X_1+X_2+\cdots +X_n$. We want the variance of $Y$.
Note that $E(X_i^2)=1$, for $X_i^2$ is certain to be $1$. Since $E(X_i)=0$, the variance of $X_i$ is $1$.
Since $Y$ is the sum of $n$ independent random variables with variance $1$, the variance of $Y$ is $n$, so the standard deviation is $\sqrt{n}$.
Formula a) gives the standard deviation of the number of steps to the right, say, at time $n$. Formula b) gives the standard deviation of the position at time $n$. Since the position at time $n$ is twice the number of steps to the right at time $n$ minus $n$, both options are compatible (and true).
