Page 342 - Numerical Methods for Chemical Engineering

Important probability distributions 331

The Gaussian (normal) distribution

Let us return to the example of the random walk, in which the net displacement after n steps
of length l is
n
1, if heads
x = l (2ζ j − 1) ζ j = (7.65)
0, if tails
j=1
If we perform n coin tosses, and n H are heads, the net displacement is
x
= n H − n T = n H − (n − n H ) = 2n H − n (7.66)
l
Hence for a given displacement, the numbers of heads and tails are
1 $ x % 1 $ x %
n H = n + n T = n − n H = n − (7.67)
2 l 2 l
For a fair coin, we obtain from the binomial distribution,
n n

n 1 n! 1
P(n, n H ) = = (7.68)
n H 2 n H !(n − n H )! 2
the probability of observing a net displacement x after n steps of length l,
n

n! 1
P(x; n,l) = (7.69)
[(n + x/l)/2]![(n − x/l)/2]! 2
We evaluate this equation in the limit n →∞, taking the natural logarithm,
ln[P(x; n,l)] = ln(n!) − ln{[(n + x/l)/2]!}− ln{[(n − x/l)/2]!}− n ln 2 (7.70)

We remove the factorials by using Stirling’s approximation
) *
N N ' N
0
ln(N!) = ln m = ln(m) ≈ ln xdx = N ln N − N (7.71)
m=1 m=1 1
Therefore, for large n,wehave
n + x/l n + x/l n + x/l

ln[P(x; n,l)] ≈ n ln n − n − ln +
2 2 2

n − x/l n − x/l n − x/l
− ln + − n ln 2 (7.72)
2 2 2
After cancelling out terms, this simplifies to
n + x/l n + x/l

ln[P(x; n,l)] ≈− ln
2 2
n

n − x/l n − x/l $ %
− ln − n ln (7.73)
2 2 2
We further simplify the expression by noting
n n + x/l n − x/l n
$ % 1 $ %
n ln = + ln (7.74)
2 2 2 2