# Random walk with probability $p \neq 1$ of stepping at each $\Delta t$

I believe the classical 1D random walk requires the walker to move distance $d= \pm 1$ at each time step $\Delta t$.

What is the variance $\sigma^2(t)$ of the random walker if it moves a distance $d=\pm 1$ with probability $p< 1$ (staying put with probability $1-p$) at each time step $\Delta t$?

Attempt: I would guess $\sigma^2(t) = \frac{pt}{\Delta t}$. I'm familiar with the proof of the $p=1$ case, in which we have $$\sigma_{p=1}^2(\Delta t) = \langle d^2 \rangle = 1$$ and then, assuming $\sigma^2(t) = \frac{t}{\Delta t}$ holds for a base case, we can show that $$\sigma^2(t+\Delta t) = \langle (x_t + d)^2\rangle = \langle x_t^2\rangle + 2\langle x_t d\rangle + \langle d^2 \rangle = \sigma^2(t)+1 = \frac{t+\Delta t}{\Delta t}$$ which proves it for all $t$. Generalizing to some $p<1$, the base case $\sigma^2(\Delta t) = \frac{p \Delta t}{\Delta t}$ is clear but $$\sigma^2(t+\Delta t) = p\langle(x_t + d)^2\rangle + (1-p)\langle x_t^2\rangle = \frac{p(t+\Delta t)}{\Delta t} + \frac{(1-p)t}{\Delta t}$$ $$= \frac{t+p\Delta t}{\Delta t} \neq \frac{p(t+\Delta t)}{\Delta t}.$$

• Hint: what will $p=0$ imply? Jan 18, 2018 at 1:59
• @karakfa $p=0$ implies $\sigma^2(t) = 0, \forall t$, consistent with my guess. I just cannot prove it. Jan 18, 2018 at 13:30

Your mistake comes in assuming that $\langle x^2_t\rangle = \frac{t}{\Delta t}$, whereas it should be $p \frac{t}{\Delta t}$. Substituting this in, your guess is correct.
$$\sigma^2(t + \Delta t) = p \langle (x_t + d)^2 \rangle + (1-p) \langle x_t^2 \rangle = p \frac{p t + \Delta t}{\Delta t} + (1-p) \frac{pt}{\Delta t} = p \frac{t + \Delta t}{\Delta t}$$