Discrete uniform and binomial distribution functions

CFA level I / Quantitative Methods: Application / Common Probability Distributions / Discrete uniform and binomial distribution functions

A discrete uniform distribution is the simplest of all probability distributions. The outcomes in the distribution are discrete i.e. distinct, separable, and clearly differentiated values. The number of outcomes is countable. A uniform distribution means that all the outcomes have an equal probability of occurrence.

A discrete uniform random variable is a variable for which the probability of all possible outcomes is equal. For a discrete uniform distribution, F(xn) = np(x) where p(x) = 1/N given that x is one of the outcomes of the variable X and N is the total number of possible outcomes. For example- tossing a fair coin, throwing a single dice.

A binomial distribution is a distribution where only two results are possible at each node. The results are seen as either a success or a failure. The probability of success and failure remains constant at each node.

If the outcomes of a variable have only two possible outcomes, then it is referred to as Bernoulli random variable. If the probability of success is p, then the probability of failure will be 1-p as only two outcomes are possible.

A binomial random variable is defined as the number of successes in n Bernoulli trials (a trial where only two outcomes are possible). A binomial random variable is the sum of Bernoulli random variables Yi where i=1 to n.

X = Y1 + Y2 +....+ Yn

The probability of success, p, is constant for all trials and all trials are independent.

There is a total of n number of trials.

The expected value of a binomial random variable X = E(X) = np
The probability of exact x successes in n trials is given by p(x) = nCx px(1-p)n-x
Variance of the binomial random variable = np(1-p)

Example 1: Calculating probabilities for discrete uniform distribution

The throwing of a dice follows a discrete uniform distribution with the outcomes ranging from 1 to 6. Calculate the following for the random variable:

(a) p(4)
(b) F(5)
(c) F(9)
(d) p(9)

Solution:

There are only six outcomes possible - 1, 2, 3, 4, 5, and 6.
Since the distribution is discrete and uniform, the probability of each outcome = 1/6.

(a) p(4) = p(1) = p(2) = p(3) = p(5) = p(6) = 1/6.

(b) F(5) represents the cumulative probability that the random variable has a value less than or equal to 5. It is the fifth possible outcome if we arrange the outcomes in the ascending order. Therefore, F(5) = 5*1/6 = 5/6.

(c) F(9) is the cumulative probability that the random variable has a value less than or equal to 9. Since all the values of the random variable are less than 9, the F(9) = 1.

(d) p(9) is the probability that the random variable has a value of 9. But it can take values ranging from 1 to 6 only. So, p(9) = 0.



Example 2: Calculating probabilities for binomial distribution

The returns on a stock are following a binomial distribution. The results will either be positive or negative. The probability of positive result in each period is 0.60. Calculate the following for the binomial random variable for a total of five trials:

(a) The probability that the result will be positive for at least three times
(b) The probability that the result will be positive for at most four times
(c) The probability that the result will be positive for exactly two times
(d) The expected value of positive result
(e) The variance of the random variable

Solution:

The probability that the result will be negative = 1-0.6 = 0.4.

(a) The probability that the result will be positive for at least 3 times = Exact probability for three positive results + Exact probability for four positive results + Exact probability for five positive results = 1 - (Exact probability of zero positive result + Exact probability of one positive result - Exact probability of two positive results ) = 5C3(0.6)3(0.4)2 + 5C4(0.6)4(0.4)1 + 5C5(0.6)5(0.4)0 = 0.68256.

(b) The probability that the result will be positive for at most 4 times = Exact probability of zero positive result + Exact probability for one positive result + Exact probability for two positive results + Exact probability for three positive results + Exact probability for four positive results = 1 - Exact probability of five positive results = 1 - 5C5(0.6)5(0.4)0 = 1 - 0.07776 = 0.92224.

(c) The probability that the result will be positive for exactly two times = 5C2(0.6)2(0.4)3 = 0.2304.

(d) The expected value of positive result = np = 5*0.6 = 3.

(e) The variance of the random variable = np(1-p) = 5(0.6)(0.4) = 1.2.


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