Cumulative Distribution Functions
A cumulative distribution function or CDF, expresses a probability's function in terms of lowest to highest value, by giving the probability that a random variable X is less than or equal to a particular value x. Expressed in shorthand, the cumulative distribution function is P(X < x). A cumulative distribution function is constructed by summing up, or cumulating all values in the probability function that are less than or equal to x. The concept is similar to the cumulative relative frequency covered earlier in this study guide, which computed values below a certain point in a frequency distribution.
Example: Cumulative Distribution Function
For example, the following probability distribution includes the cumulative function.
X = x  P(X = x)  P(X < x) or cdf 
< 12  0.15  0.15 
12 to 3  0.15  0.30 
3 to 4  0.25  0.55 
4 to 10  0.25  0.80 
> 10  0.2  1.0 
From the table, we find that the probability that x is less than or equal to 4 is 0.55, the summed probabilities of the first three P(X) terms, or the number found in the cdf column for the third row, where x < 4. Sometimes a question might ask for the probability of x being greater than 4, for which this problem is 1  P(x < 4) = 1  0.55 = 0.45. This is a question most people should get  but one that will still have too many people answering 0.55 because they weren't paying attention to the "greater than".
Discrete Uniform Random Variable
A discrete uniform random variable is one that fulfills the definition of "discrete", where there are a finite and countable number of terms, along with the definition of "uniform", where there is an equally likely probability that the random variable X will take any of its possible values x. If there are n possible values for a discrete uniform random variable, the probability of a specific outcome is 1/n.
Example: Discrete Uniform Random Variable
Earlier we provided an example of a discrete uniform random variable: a random day is oneseventh likely to fall on a Sunday. To illustrate some examples on how probabilities are calculated, take the following discrete uniform distribution with n = 5.
X = x  P(X = x)  P(X < x) 
2 
0.2 
0.2 
4 
0.2 
0.4 
6 
0.2 
0.6 
8 
0.2 
0.8 
10 
0.2 
1.0 
According to the distribution above, we have the probability of x = 8 as 0.2. The probability of x = 2 is the same, 0.2.
Suppose that the question called for P(4 < X < 8). The answer would be the sum of P(4) + P(6) + P(8) = 0.2 + 0.2 + 0.2 = 0.6.
Suppose the question called for P(4 < X < 8). In this case, the answer would omit P(4) and P(8) since it's less than, NOT less than or equal to, and the correct answer would be P(6) = 0.2. The CFA exam writers love to test whether you are paying attention to details and will try to trick you  the probability of such tactics is pretty much a 1.0!
Binomial Random Variable
Binomial probability distributions are used when the context calls for assessing two outcomes, such as "success/failure", or "price moved up/price moved down". In such situations where the possible outcomes are binary, we can develop an estimate of a binomial random variable by holding a number of repeating trials (also known as "Bernoulli trials"). In a Bernoulli trial, p is the probability of success, (1  p) is the probability of failure. Suppose that a number of Bernoulli trails are held, with the number denoted by n. A binomial random variable X is defined as the numberof successes in n Bernoulli trials, given two simplifying assumptions: (1) the probabilityp of success is the same for all trials and (2) the trials are independent of each other.
Thus, a binomial random variable is described by two parameters: p (the probability of success of one trial) and n (the number of trials). A binomial probability distribution with p = 0.50 (equal chance of success or failure) and n = 4 would appear as:
x (# of successes) 
p(x) 
cdf, P(X < x) 
0 
0.0625 
0.0625 
1 
0.25 
0.3125 
2 
0.375 
0.6875 
3 
0.25 
0.9325 
4 
0.0625 
1.0000 
The reference text demonstrates how to construct a binomial probability distribution by using the formula p(x) = (n!/(n  x)!x!)*(p^{x})*(1  p)^{nx}. We used this formula to assemble the above data, though the exam would probably not expect you to create each p(x); it would probably provide you with the table, and ask for an interpretation. For this table, the probability of exactly one success is 0.25; the probability of three or fewer successes is 0.9325 (the cdf value in the row where x = 3); the probability of at least one is 0.9325 (1  P(0)) = (1  0.0625) = 0.9325.
Calculations
The expected value of a binomial random variable is given by the formula n*p. In the example above, with n = 4 and p = 0.5, the expected value would be 4*0.5, or 2.
The variance of a binomial random variable is calculated by the formula n*p*(1  p). Using the same example, we have variance of 4*0.5*0.5 = 1.
If our binomial random variable still had n = 4 but with a greater predictability in the trial, say p = 9, our variance would reduce to 4*0.9*0.1 = 0.36. For successive trials (i.e. for higher n), both mean and variance increase but variance increases at a lower rate  thus the higher the n, the better the model works at predicting probability.
Creating a Binomial Tree
The binomial tree is essentially a diagram showing that the future value of a stock is the product of a series of up or down movements leading to a growing number of possible outcomes. Each possible value is called a node.
Figure 2.9: Binomial Tree 
Continuous Uniform Distribution
A continuous uniform distribution describes a range of outcomes, usually bound with an upper and lower limit, where any point in the range is a possibility. Since it is a range, there are infinite possibilities within the range. In addition, all outcomes are all equally likely (i.e. they are spread uniformly throughout the range).
To calculate probabilities, find the area under a pdf curve such as the one graphed here. In this example, what is the probability that the random variable will be between 1 and 3? The area would be a rectangle with a width of 2 (the distance between 1 and 3), and height of 0.2, 2*0.2 = 0.4.
What is the probability that x is less than 3? The rectangle would have a width of 3 and the same height: 0.2. 3*0.2 = 0.6
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