Limited Time Offer: Save 10% on all 2021 and 2022 Premium Study Packages with promo code: BLOG10    Select your Premium Package » Calculating Probabilities from Cumulative Distribution Function

A cumulative distribution function, $$F(x)$$, gives the probability that the random variable $$X$$ is less than or equal to $$x$$:

$$P(X ≤ x)$$

By analogy, this concept is very similar to the cumulative relative frequency.

A cumulative distribution is the sum of the probabilities of all values qualifying as “less than or equal” to the specified value. Perhaps an example will make this concept clearer.

Example: Cumulative Distribution

If we flipped a coin three times, we would end up with the following probability distribution of the number of heads obtained:

$$\begin{array}{c|c|c|c} \textbf{Heads (outcomes)} & \bf{0} & \bf{1} & \bf{2} & \bf{3} \\ \hline \text{Probability} & {1/8} & {3/8} & {3/8} & {1/8} \\ \end{array}$$

To come up with a cumulative distribution function, we have to calculate the cumulative probabilities:

The cumulative probability that $$X$$ is less than or equal to zero is 1/8.

To find the cumulative probability that $$X$$ is less than or equal to 1, we add $$P(X = 0)$$ and $$(P = 1)$$:

$$P(X \le 1) =\cfrac {1}{8} + \cfrac {3}{8} = \cfrac {1}{2}$$

Similarly,

$$P(X \le 2) = \cfrac {1}{8} + \cfrac {3}{8} + \cfrac {3}{8} = \cfrac {7}{8}$$

Lastly,

$$P(X \le 3) = \cfrac {1}{8} + \cfrac {3}{8} + \cfrac {3}{8} +\cfrac {1}{8} = 1$$

$$\begin{array}{c|c|c|c} \textbf{Heads (outcomes)} & \bf{0} & \bf{1} & \bf{2} & \bf{3} \\ \hline \text{Probability} & {1/8} & {3/8} & {3/8} & {1/8} \\ \hline \text{Cumulative prob.} & {1/8} & {4/8} & {7/8} & {8/8} \\ \end{array}$$ The CDF has two main properties:

1. All values in the CDF are between 0 and 1.
2. The CDF either increases or remains constant as the value of the specified outcome increases.

Interpreting the Cumulative Distribution Function

A cumulative distribution function can help us to come up with cumulative probabilities pretty easily. For example, we can use it to determine the probability of getting at least two heads, at most two heads, or even more than two heads. The probability of at most two heads from the cumulative distribution above is 0.875.

Example: Cumulative Distribution Function

Variable X can take the values 1, 2, 3, and 4. The probability of each outcome has been given below.

$$\begin{array}{c|c|c|c|c} \textbf{Outcome} & \bf{1} & \bf{2} & \bf{3} & \bf{4} \\ \hline \text{Probability} & {0.2} & {0.3} & {0.35} & {0.15} \\ \end{array}$$

Determine $$P(X ≤ 2)$$.

A. 0.5

B. 0.3

C. 0.85

Solution

You simply sum up the probabilities up to and including a given outcome and come up with a table similar to the one below:

$$\begin{array}{c|c|c|c|c} \textbf{Heads (outcomes)} & \bf{1} & \bf{2} &\bf {3} & \bf{4} \\ \hline \text{Probability} & {0.2} & {0.3} & {0.35} & {0.15} \\ \hline \text{Cumulative prob.} & {0.2} & {0.5} & {0.85} & {1} \\ \end{array}$$

From the table, it is clear that $$P(X \le 2) = 0.5$$.

Note to candidates: The standard notation for a cumulative distribution function is written in upper case $$F(x)$$. In contrast, that of a probability function is written in lower case $$f(x)$$.

Calculating Probabilities Given Cumulative Distribution Function

A cumulative distribution offers a convenient tool for determining probabilities for a given random variable. As seen  above, the cumulative distribution function, $$F(x)$$, gives the probability that the random variable $$X$$ is less than or equal to $$x$$ for every $$x$$ value. It is usually expressed as:

$$F(x) = P(X \le x)$$

Example: Cumulative Distribution Function

The random variable X has the following probability distribution function:

$$\begin{matrix} P(x) = \frac { x }{ 150 } & \text{ for x} = 10, 20, 30, 40, 50 \\ 0 & \text{otherwise} \end{matrix}$$

Calculate and interpret $$F(20)$$ and $$F(40)$$.

Solution

As you will recall, we can determine the probability of each outcome for a random variable given the probability distribution function (pdf).

\begin{align*} P(x) & =\cfrac {x}{150} \\ P(x) & = P(X = x) \\ \end{align*}

Therefore,

$$P(10) =\cfrac {10}{150}$$

Similarly,

\begin{align} P(20) &=\cfrac {20}{150} \\ P(30) &=\cfrac {30}{150}\\ P(40) &=\cfrac {40}{150}\end{align}

And lastly,

$$P(50) =\cfrac {50}{150}$$

Note to candidates: We can prove that our pdf is correct by testing the first rule of probability distribution functions by adding all the probabilities.

Now,

$$F(x) = P(X \le x)$$

Therefore,

\begin{align*} F(2) & = P(X \le 20) \\ & = P(X = 10) + P(X = 30) \\ & =\cfrac {10}{150} + \cfrac {20}{150} \\ &=\cfrac {30}{150} \text { or } \cfrac {1}{5} \\ \end{align*}

Interpretation: There is a 20% cumulative probability that outcomes 10 or 20 occur.

Similarly,

\begin{align*} F(40) & = P(X \le 40) \\ & = P(X = 10) + P(X = 20) + P(X = 30) + P(X = 40) \\ & =\cfrac {10}{150} + \cfrac {20}{150} + \cfrac {30}{150} + \cfrac {40}{150} \\ & = \cfrac {100}{150} \text{ or } 66.67\% \\ \end{align*}

Interpretation: there is a 66.67% cumulative probability that outcomes 10, 20, 30, or 40 occur.

Example: Calculating Probabilities Given Cumulative Distribution Function

Variable $$X$$ can take the values 1, 2, 3, and 4. The cumulative probability distribution has been given below. Use it to calculate:

(a) P(X = 2)

(b) P(X = 4)

$$\begin{array}{c|c|c|c|c} \textbf{Outcome} & \bf{1} & \bf{2} & \bf{3} & \bf{4} \\ \hline \text{Cumulative Probability Distribution} & {0.2} & {0.5} & {0.85} & {1} \\ \end{array}$$

Solution

$$F(x) = P(X \le x)$$

(a)

\begin{align*} F(2) & = P(X \le 2) = 0.5 \\ 0.5 & = P(X = 1) + P(X = 2) \\ & = 0.2 + P(X = 2)\\ P(X = 2)& = 0.5 – 0.2 = 0.3 \\ \end{align*}

Note to candidates: A simpler, more direct approach can be:

$$P(X = 2) = F(2) – F(3)$$

Therefore,

$$P(X = 2) = 0.5 – 0.2 = 0.3$$

(b)

\begin{align*} P(X = 4) & = F(4) – F(3) \\ &= 1 – 0.85 = 0.15 \\ \end{align*}

Question

Given the following cumulative probability distribution, determine $$P(X=2)$$.

$$\begin{array}{c|c|c|c|c} \textbf{Outcome} & \bf{0} & \bf{1} & \bf{2} & \bf{3} \\ \hline \text{Cumulative prob.} & {1/8} & {4/8} & {7/8} & {1} \\ \end{array}$$

1. $$\frac{7}{8}$$.
2. $$\frac{3}{8}$$.
3. $$\frac{1}{8}$$.

Solution

\begin{align*} P(X = 2)& = F(2) – F(1) \\ &=\cfrac {7}{8} – \cfrac {4}{8} \\ & =\cfrac {3}{8} \\ \end{align*}

Featured Study with Us CFA® Exam and FRM® Exam Prep Platform offered by AnalystPrep

Study Platform

Learn with Us

Subscribe to our newsletter and keep up with the latest and greatest tips for success
Online Tutoring Our videos feature professional educators presenting in-depth explanations of all topics introduced in the curriculum.

Video Lessons Sergio Torrico
2021-07-23
Excelente para el FRM 2 Escribo esta revisión en español para los hispanohablantes, soy de Bolivia, y utilicé AnalystPrep para dudas y consultas sobre mi preparación para el FRM nivel 2 (lo tomé una sola vez y aprobé muy bien), siempre tuve un soporte claro, directo y rápido, el material sale rápido cuando hay cambios en el temario de GARP, y los ejercicios y exámenes son muy útiles para practicar. diana
2021-07-17
So helpful. I have been using the videos to prepare for the CFA Level II exam. The videos signpost the reading contents, explain the concepts and provide additional context for specific concepts. The fun light-hearted analogies are also a welcome break to some very dry content. I usually watch the videos before going into more in-depth reading and they are a good way to avoid being overwhelmed by the sheer volume of content when you look at the readings. Kriti Dhawan
2021-07-16
A great curriculum provider. James sir explains the concept so well that rather than memorising it, you tend to intuitively understand and absorb them. Thank you ! Grateful I saw this at the right time for my CFA prep. nikhil kumar
2021-06-28
Very well explained and gives a great insight about topics in a very short time. Glad to have found Professor Forjan's lectures. Marwan
2021-06-22
Great support throughout the course by the team, did not feel neglected Benjamin anonymous
2021-05-10
I loved using AnalystPrep for FRM. QBank is huge, videos are great. Would recommend to a friend Daniel Glyn
2021-03-24
I have finished my FRM1 thanks to AnalystPrep. And now using AnalystPrep for my FRM2 preparation. Professor Forjan is brilliant. He gives such good explanations and analogies. And more than anything makes learning fun. A big thank you to Analystprep and Professor Forjan. 5 stars all the way! michael walshe
2021-03-18
Professor James' videos are excellent for understanding the underlying theories behind financial engineering / financial analysis. The AnalystPrep videos were better than any of the others that I searched through on YouTube for providing a clear explanation of some concepts, such as Portfolio theory, CAPM, and Arbitrage Pricing theory. Watching these cleared up many of the unclarities I had in my head. Highly recommended.