## Random Walk and Differential Equations (I)

Consider discrete points on the real line, indexed by the integers … -3, -2, -1, 0, 1, 2, … . A drunken man starts at position 0 and time 0. At each time step, he may move to the left or right (each with probability p<1/2), or stay still (with probability 1-2p).

This goes on and on, and we wish to find the probability distribution of his location at time t, i.e. for each m what’s the probability u(mt) that he’s found at position m during time t? Clearly, for each t, $\sum_{m\in\mathbf{Z}} u(m,t) = 1$ since the man must be found somewhere.

To solve it exactly, one can use generating functions. Indeed, the recurrence relation in time gives:

$u(m, t+1) = p\cdot u(m-1, t) + p\cdot u(m+1, t) + (1-2p) u(m, t),$  (*)

for each integer m, non-negative integer t. So if we consider the power series

$P_t(x) = \ldots + u(-1, t)x^{-1} + u(0, t) + u(1, t)x + u(2,t)x^2 + \ldots = \sum_{m=-\infty}^\infty u(m,t)x^m,$

then the above recurrence relation in u(mt) can be expressed succinctly as a recurrence relation in Pt(x):

$P_{t+1}(x) = (p\cdot x + p\cdot x^{-1} + (1-2p)) P_t(x),$

with $P_0(x) = 1$. This expression has certain advantages in computation:

Example 1

Let x=1: this gives $P_{t+1}(1) = P_t(1)$. But we already know this since Pt(1) represents the sum of all u(mt) across m, i.e. it equals 1.

Example 2

Differentiate to obtain:

$P_{t+1}'(x) = (p - px^{-2})P_t(x) + (px + px^{-1} + (1-2p)) P_t'(x).$

Substitute x=1 to obtain $P_{t+1}'(1) = P_t'(1)$. Since $P_0'(1) = 0$, we also have $P_t'(1) = 0$ for all t. Again, we already know this since $P_t'(1) = \sum_m m\cdot u(m,t)$ represents the expected position of our friend, which is always 0 since he’s just as inclined to go left as he is to go right. In probability notation, we have $P_t'(1) = E(X_t)$, where Xt is the random variable for our friend’s location at time t.

Example 3

Differentiate  yet again to obtain:

$P_{t+1}''(x) = 2px^{-3} P_t(x) + 2(p-px^{-2})P_t'(x) + (px + px^{-1} + (1-2p))P_t''(x).$

Substitute x=1 to obtain $P_{t+1}"(1) = 2p P_t(1) + P_t''(1) = P_t''(1) + 2p$. Since $P_0''(1) = 0$ we have $P_t''(1) = 2pt$. But

$P_t''(x) = \sum_{m=-\infty}^\infty u(m,t)m(m-1)x^{m-2} \implies P_t''(1) = E(X_t(X_t-1)),$

so from $E(X_t) = 0$ we can calculate the variance via

$\text{Var}(X_t) = E(X_t^2) - E(X_t)^2 = 2pt.$

Example 4

Suppose the drunken man’s house is at x=5. What’s the value of $E((X_t - 5)^2)$ at time t?

$E((X_t - 5)^2) = E(X_t^2 - 10X_t + 25) = E(X_t^2) - 10 E(X_t) + 25 = 25 + 2pt$.

Conclusion

As the reader can guess, successive differentiation gives us the higher moments $E(X_t^n)$ for various n. In fact, since

$P_{t+k}(x) = (p\cdot x + p\cdot x^{-1} + (1-2p))^k P_t(x),$

we can even consider relations across wider time steps.

Notice that we didn’t have to start with a fixed starting position for the drunken man. In particular, we could have any sequence of non-negative real u(m, 0), as long as the sum is 1. However, care would have to be taken to ensure that the higher moments are well-defined. E.g. if the probability distribution starts with

$u(m, 0) = \frac{6}{m^2\pi^2}$ for m>0 (u(m, 0) = 0 if m ≤ 0),

then the expected value is undefined since $\sum_{m=1}^\infty m\cdot u(m,0) = \frac{6}{\pi^2} \sum_{m=1}^\infty \frac 1 m$ which diverges since it is the renowned harmonic series. Roughly speaking, as m → ±∞, u(m, 0) should approach 0 exponentially fast in order for the higher moments to make sense.

## Higher-Dimensional Case

Let’s assume now that our drunken friend can walk in a two-dimensional plane. So the probability of him at location (mn) at time t is denoted u(mnt). At each time step, our friend can walk a unit distance in any of the four standard directions: north, south, east or west, each with probability p. The probability that he remains still for one time step is 1-4p.

The recurrence relation in t is easy to write:

u(mnt+1) = p(u(m-1, nt) + u(m+1, nt) + u(mn-1, t) + u(mn+1, t)) + (1-4p)u(mnt).

Proceeding as before, let’s define a two-parameter power series:

$P_t(x, y) = \sum_{m,n\in\mathbf{Z}} u(m,n,t) x^m y^n.$

The above recurrence relation then translates to:

$P_{t+1}(x,y) = (px + px^{-1} + py + py^{-1} +(1-4p))P_t(x,y).$

Let the random variable denoting his location at time t be (XtYt). In order to extract useful information, we’ll need to do partial differentiation. For convenience, we’ll denote $A(x,y) = px + px^{-1} + py + py^{-1} + (1-4p)$.

$\frac{\partial P_{t+1}}{\partial x} = (p - px^{-2})P_t + A\cdot \frac{\partial P_t}{\partial x},$

$\frac{\partial P_{t+1}}{\partial y} = (p - py^{-2})P_t + A\cdot\frac{\partial P_t}{\partial y}.$

The second derivatives are:

$\frac{\partial^2 P_{t+1}}{\partial x^2} = 2px^{-3} P_t + 2(p - px^{-2})\frac{\partial P_t}{\partial x} + A\frac{\partial^2 P_t}{\partial x^2},$

$\frac{\partial^2 P_{t+1}}{\partial y^2} = 2py^{-3} P_t + 2(p - py^{-2})\frac{\partial P_t}{\partial x} + A\frac{\partial^2 P_t}{\partial y^2},$

$\frac{\partial^2 P_{t+1}}{\partial x\partial y} = (p-py^{-2})\frac{\partial P_t}{\partial x} + (p-px^{-2})\frac{\partial P_t}{\partial y} + A\frac{\partial^2 P_t}{\partial x\partial y}.$

Substituting x=1, y=1 gives the following useful values:

$\left.\frac{\partial P_t}{\partial x}\right|_{1,1} = E(X_t), \ \left.\frac{\partial P_t}{\partial y}\right|_{1,1} = E(Y_t),$

$\left.\frac{\partial^2 P_t}{\partial x^2}\right|_{1,1} = E(X_t^2 - X_t),\ \left.\frac{\partial^2 P_t}{\partial y^2}\right|_{1,1} = E(Y_t^2 - Y_t),\ \left.\frac{\partial^2 P_t}{\partial x\partial y}\right|_{1,1} = E(X_t Y_t).$

In the next article, we’ll consider the limiting case, where the space and time intervals approach zero, and look at the corresponding differential equations.

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