Page 343 - Numerical Methods for Chemical Engineering

332 7 Probability theory and stochastic simulation

hence,
1
n + x/l n + x/l $ %
n
ln[P(x; n,l)] ≈− ln + ln
2 2 2
1
n
n − x/l n − x/l $ %
− ln + ln (7.75)
2 2 2
The rule ln(ab) = ln(a) + ln(b) yields
1
n + x/l n + x/l 2
ln[P(x; n,l)] ≈− ln
2 2 n
n − x/l n − x/l 2
1
− ln
2 2 n

n + x/l 9 . x /: n − x/l 9 . x /:
≈− ln 1 + − ln 1 − (7.76)
2 nl 2 nl
In the limit of large n, the high degree of backtracking in the random walk means that |x|
nl, hence we use the Taylor expansions around x/(nl) = 0,
x x 1 x x x 1 x
. / $ % 2 . / $ % 2
ln 1 + ≈ − ln 1 − ≈− − (7.77)
nl nl 2 nl nl nl 2 nl
to obtain
n + x/l x 1 $ x % n − x/l x 1 $ x %
2 1 2 1
ln[P(x; n,l)] ≈− − − − −
2 nl 2 nl 2 nl 2 nl
(7.78)
Collecting terms yields the particularly simple result
x 2
ln[P(x; n,l)] ≈− (7.79)
2nl 2
- +∞
Taking the exponential and renormalizing to P(x; n,l)dx = 1 yields
−∞
1 x 2
P(x; n,l) = √ exp − 2 (7.80)
2πnl 2 2nl
2
2
Defining σ = nl produces the Gaussian (normal) distribution
1 x 2
P(x; σ) = √ exp − 2 (7.81)
σ 2π 2σ
We note by symmetry of the exponential argument that the average is zero:
x x
+∞ +∞
' ' 2
x = E(x) = xP(x; σ)dx = √ exp − 2 dx = 0 (7.82)
−∞ σ 2π 2σ
−∞
The variance is
' ' 2 2
+∞ +∞ x x
2 2 2
var(x) = E[(x − E(x)) ] = x P(x; σ)dx = √ exp − 2 dx = σ
−∞ σ 2π 2σ
−∞
(7.83)
Therefore, σ is the standard deviation of the Gaussian distribution. The Gaussian distribu-
tion is plotted in Figure 7.9 for various values of σ.