Where the term in brackets = n! / {(n - x)!x!}, and p = the probability of "success" on each trial.
i: Calculate the expected value and variance of a binomial random variable.
Expected value of X:E(X) = n * p
Variance of X:V(X) = n * p * (1 - p)
Example:Using empirical probabilities, for any given day the probability that the DJIA will increase is 0.67. We will assume that the only other outcome is that it decreases in a day. Hence, p(UP) = 0.67 and p(DOWN) = 0.33. We will assume that whether the DJIA increases in one day is independent of whether it decreases in anther day. What is the probability that the DJIA will increase three out of five days? What is the expected number of up days in a five-day period?
Here we define success as UP, so p = 0.67. The definition of success is critical to any binomial problem. The n items are the five days: n = 5. The number of successes we are computing the probability for is x = 3. The formula is:
p(3) = P(X = 3) = 5! / [(5 - 3)! * 3!] * (0.673) * (0.335 - 3)
p(3) = 10 * 0.301 * 0.109
p(3) = 0.328
Expected value of X: E(X I n = 5, p
= 0.67) = 5 * 0.67 =3.35
Variance of X: V(X) = n * p * (1 - p) = 5 * 0.67 * 0.33 = 1.106
j: Construct a binomial tree to describe stock price movement and calculate the expected terminal stock price.
Example:Assume the DJIA is at 10,000. It moves an average of 2 percent a day. Using the probabilities from our previous example in LOS 1.D.i,what is the expected value of the DJIA at the end of two days?
 
Day 1 prob.
Day 1 values
Day 2 prob.
Day 2 values
p(x) * x
 
 
 
P(UU) = 0.4489
DJIA = 10,404
4,670.36
 
P(U) = 0.67
DJIA = 10,200
 
 
 
 
 
 
P(UD) = 0.2211
DJIA = 9,996
2,210.12
10,000
 
 
 
 
 
 
 
 
P(DU) = 0.2211
DJIA = 9,996
2,210.12
 
P(D) = 0.33
DJIA = 9,800
 
 
 
 
 
 
P(DD) = 0.1089
DJIA = 9,604
1,045.88
 
 
 
 
Expected value =
10,136.46