Page 340 - Numerical Methods for Chemical Engineering

Important probability distributions 329

After N s steps (coin tosses) the net displacement that we have traveled is
n
1, if heads
x = l (2ζ j − 1) ζ j = (7.55)
0, if tails
j=1
The average (and most likely) displacement is zero, as we double back upon our path at
random,
7 8
n n n
1
x = l (2ζ j − 1) = l (2 ζ j − 1) = l 2 − 1 = 0 (7.56)
2
j=1 j=1 j=1
but, the mean-squared displacement is not zero,
7

8
n n n n
2 2
x = l (2ζ j − 1) l (2ζ k − 1) = l (2ζ j − 1)(2ζ k − 1)
j=1 k=1 j=1 k=1
n n 1, if heads
2

x = l 2 ζ ζ ζ = 2ζ j − 1 = (7.57)

j k j
j=1 k=1 −1, if tails
Using the independence of the coin tosses,

1, if j = k

ζ ζ = δ jk = (7.58)
j k
0, if j = k
we find that the mean-squared displacement is linear in the number of steps,
n n
n
n
n
2
2

x = l 2 ζ ζ = l 2 δ jk = l 2 (1) = l n (7.59)

j k
j=1 k=1 j=1 k=1 j=1
The binomial distribution
Now that we have seen some possible applications of this coin toss question, let us derive
the answer. First, let us say that we make ten coin tosses. We wish to compute the probability
of observing an exact ordered sequence of heads and tails, e.g.
H T T H H T H H T H
If the outcomes of each coin toss are independent, and if p H is the probability of a coin toss
returning heads and p T = 1 − p H is the probability that it returns tails, the probability of
observing the exact sequence above is
6
p H × p T × p T × p H × p H × p T × p H × p H × p T × p H = p p 4 (7.60)
H T
In general, the probability of observing an exact ordered sequence of n H heads and n T tails
n T
n H
is p p . Note that the sequence above is only one of the possible sequences containing
H T
n H heads and n T tails. Others include
H H T T H H T H T H
H H T H H T T H H T
H T H H T H T H H T
Since every sequence with the same number of heads and tails has the same probability of
being observed, the probability that we observe any sequence of n H heads and n T = n − n H