- CFA Exams
- 2023 Level I
- Topic 1. Quantitative Methods
- Learning Module 4. Common Probability Distributions
- Subject 2. Probability Function

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##### Subject 2. Probability Function PDF Download

Every random variable is associated with a probability distribution that describes the variable completely. A

**probability function**is one way to view a probability distribution. It specifies the probability that the random variable takes on a specific value; P(X = x) is the probability that a random variable X takes on the value x.A probability function has two key properties:

- 0 ≤ P(X=x) ≤ 1, because probability is a number between 0 and 1.
- ΣP(X=x) = 1. The sum of the probabilities P(X=x) over all values of X equals 1. If there is an exhaustive list of the distinct possible outcomes of a random variable and the probabilities of each are added up, the probabilities must sum to 1.

The following examples will utilize these two properties in order to examine whether they are probability functions.

*Example 1*p(x) = x/6 for X = 1, 2, 3, and p(x) = 0 otherwise

- Substituting into p(x): p(1) = 1/6, p(2) = 2/6 and p(3) = 3/6

Note that it is not necessary to substitute in any other values, as p(x) is only non-zero for X values 1, 2 and 3.

In all 3 cases, p(x) lies between 0 and 1, as 1/6, 2/6 and 3/6 are all values in the range 0 to 1 inclusive.

So, the first property is satisfied. - Summing the probabilities gives 1/6 + 2/6 + 3/6 = 1, showing the second property is also satisfied.

*Example 2*p(x) = (2x - 3)/16 for X = 1, 2, 3, 4 and p(x) = 0 otherwise

Substituting into p(x): p(1) = -1/16

STOP HERE!

It is impossible for any probability to be negative, so it's not necessary to continue. Property 1 is violated, so it can be said straightaway that p(x) is not a probability function.

Note that individual probabilities in a continuous case cannot occur, so P(X = 5), say, is 0 if X is continuous.

In a continuous case, only a range of values can be considered (that is, 0 < X < 10), whereas in a discrete case, individual values have positive probabilities associated with them.

For a

**discrete random variable**, the shorthand notation is p(x) = P(X = x). For**continuous random variables**, the probability function is denoted f(x) and called**probability density function (pdf)**, or just the density. This function is effectively the continuous analogue of the discrete probability function p(x).

- The probability density function, which has the symbol f(x), does not give probabilities, despite its name. Instead, it is the area between the graph and the horizontal axis that gives probabilities. Because of this, the height of f(x) is not restricted to the range 0 to 1, and the graph, which in itself is not a probability, is unrestricted as far as its height is concerned.
- From this information, it follows that the area under the entire graph (i.e., between the graph and the x-axis) must equal 1, because this area encapsulates all the probability contained in the random variable. Recall that for discrete distributions, the probabilities add up to 1.
- Because continuous random variables are concerned with a range of values, individual values have no probabilities, because there is no area associated with individual values. Rather, probabilities are calculated over a range of values. Another way of saying this is that p(x) = 0 for every individual X.
- If a discrete random variable has many possible outcomes, then it can be treated as a continuous random variable for conciseness, and ranges of values can be considered in determining probabilities.

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**User Contributed Comments**
7

User |
Comment |
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wms3 |
f(x) is a probability density function pdf and it is the area under the function that determines probability so this mus be equal to 1. The line can go above one! p(x) is for discrete random variables and f(x) is for continuous. |

stitcher |
The probability density function measures an area under the curve. An area cannot be negative space |

bidisha |
Loving the stop here |

tomalot |
STOP HERE! |

BossMan |
p(x) = x/6 for x = 1, 2, 3 and p(x) = 0 otherwise. Does that mean that the outcome can ONLY be a 1, a 2 or a 3? Nothing else is possible? |

Olesya_CFA |
@Bossman, right. The problem specifies that everything beyond 1, 2, and 3 gives us zero. |

MathLoser |
Remember to read that 4 dots at the end carefully, guys. |

I was very pleased with your notes and question bank. I especially like the mock exams because it helped to pull everything together.