How to compute the variance for this biased random walk.

Question

Suppose over each time increment $\Delta t$ our process X increases by $h$ with probability $p$ and decreases by $h$ with probability $1-p$. Lets call $\Delta X=X(t+\Delta t)-X(t)$ the increment of X. Then it is clear to me that the variance will be $E(\Delta X)=(2p-1)h$. I don't see how the variance is $$var(\Delta X)= (1-(2p-1)^2)h^2.$$

I tried using the variance formula to recover the above, however I get stuck when I have to compute $E((\Delta X)^2)$.

Answer 1

It holds $\Delta X \in \{-h,h\}$ and so $(\Delta X)^2 \equiv h^2$ and it follows $$var(\Delta X) = E((\Delta X)^2) - E^2(\Delta X) = h^2 - (2p-1)^2h^2 = (1-(2p-1)^2)h^2$$

Answer 2

Since squaring $h$ and $-h$ gives the same thing, $\Delta X^2 = h^2$ with probability 1. So $E(\Delta X^2) = h^2.$

You can also calculate it like $$ E(\Delta X^2) = ph^2 + (1-p)h^2 = h^2.$$

Then you can use the formula $\mathrm{Var}(\Delta X) = E(\Delta X^2)-E(\Delta X)^2$ to get the answer.