## A

site i + 2 in Figure 2.17, we could interpret this to mean that the walker is within D/2 to the left or D/2 to the right of site i + 2. Thus, perhaps a more precise way to think about px, t) is that px, t) D is the probability that the walker is within ±D/2 of the site i = x/D. This may seem like a lot of pedantry, and to some extent it is for just now, but thinking carefully about what these things mean will be enormously helpful later on, when we let the distance between sites shrink towards 0.

We will now derive an equation for px, t). How can the walker be at spatial point x (site i) at time t + r? There are many ways, but they all boil down to this argument: the walker had to be at spatial point y at time t and take a jump of size |x — y|. We thus define j by y = x — Ds. Assuming that the walker's location at time t and the size of the jump are independent of each other allows us to multiply probabilities (you will be reminded of these rules in Chapter 3) so that we have px, t + r) = p(s)px — Ds, t) (2.48)

We Taylor expand this equation around the point (x, t) to obtain px o+p^+o(r) = ^2 p(s)

where subscripts denote partial derivatives. We simplify this expression by thinking about the sums. For example, we know that XL P(s) = 1 because a jump of some size (including 0) must occur. Furthermore, since px, t) does not depend upon j, we know that XL p(s)px, t) = px, t) s p(s). This takes care of the first term in square brackets on the right hand side of Eq. (2.49). The fourth term will similarly simplify to o(D2). The second and third terms require a bit more thought. Factoring the things that do not depend upon s out of the second term allows us to rewrite it as —Dpx^s p(s)s. We recognize the sum as the average jump size: that is s is the size of the jump from site j to site i and p(s) is the chance of making this jump. We will denote this average by the symbol m1 (for first moment). Exactly the same kind of argument will apply to the third term on the right hand side of Eq. (2.49), except that we will have the square of the jump size; hence we will use the symbol m2 for that summation. We put all this together by subtracting px, t) from both sides of Eq. (2.49), divide by r and obtain o(D) D D2 o(D2)

0 0
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