Cumulative Distribution Functions and Their Meaning

What a Cumulative Distribution Function Measures

A probability model can answer questions about many different cutoffs. For example, it can tell us the chance that a claim amount is no more than a specified threshold, or that a continuous measurement falls below a limit. The cumulative distribution function, abbreviated CDF, organizes all such threshold probabilities into one function.

The key word is cumulative: a CDF value includes probability from every value at or below the chosen threshold. It is not generally the probability of the random variable being exactly equal to that threshold. This distinction is especially important for discrete variables, which can assign positive probability to individual values.

Definition: For a random variable $X$, its cumulative distribution function is the function $F_X$ defined for every real number $x$ by $$ F_X(x)=P(X\le x). $$ Thus $F_X(x)$ is the probability that $X$ takes a value less than or equal to the threshold $x$, under the specified probability model.

The argument $x$ is a threshold, not an observed value that $X$ must take. If $X$ is a claim count that can only be 0, 1, 2, or 3, then $F_X(1.6)$ still has a clear meaning: it is the probability that the count is at most 1.6, which is the probability that the count is either 0 or 1. If $X$ is continuous, the same definition applies; the probability is accumulated over all values at or below the threshold.

The CDF’s value is a probability, so it lies between 0 and 1. To calculate it, use the form of the probability model: add the probabilities of eligible values for a discrete random variable, or integrate the density over the relevant range for a continuous random variable. The examples below show both methods.

From a CDF to an Interval Probability

A CDF also provides a convenient way to calculate the probability that $X$ lies between two thresholds. For $a<b$, subtract the cumulative probability up to $a$ from the cumulative probability up to $b$:

P(a<X\le b)=F_X(b)-F_X(a).

This formula follows from separating the event $X\le b$ into two disjoint events: $X\le a$, and $a<X\le b$. Their probabilities add, so rearranging gives the result:

P(X\le b)=P(X\le a)+P(a<X\le b), $$ $$ P(a<X\le b)=P(X\le b)-P(X\le a)=F_X(b)-F_X(a).

The endpoint convention in this formula matters. It describes values strictly greater than $a$ and less than or equal to $b$. For a discrete variable, changing an endpoint can include or exclude a value with positive probability. For a continuous variable with a density, the probability at any one exact point is zero, so including or excluding a single endpoint does not change the interval probability. The formula above, however, has the stated endpoint convention in either setting.

Worked Example: A Discrete Service-Request Count

A hypothetical help desk models the number $X$ of urgent service requests received during one shift. Its probability model is $P(X=0)=0.10$, $P(X=1)=0.25$, $P(X=2)=0.40$, and $P(X=3)=0.25$, with no other possible values. Find $F_X(1)$, $F_X(2.4)$, and $P(1<X\le3)$. There is no interest or mortality component in this one-shift count model. The method is to add the probabilities of all possible values at or below a threshold; this avoids treating a CDF value as a point probability.

Step 1: Evaluate $F_X(1)$. By definition, this is $P(X\le1)$. The possible values satisfying the condition are 0 and 1:

$$ F_X(1)=P(X=0)+P(X=1)=0.10+0.25=0.35. $$

Step 2: Evaluate $F_X(2.4)$. The possible values no greater than 2.4 are 0, 1, and 2. There is no possible value equal to 2.4, and the value 3 is above the threshold:

$$ F_X(2.4)=0.10+0.25+0.40=0.75. $$

Step 3: Find $P(1<X\le3)$. Applying the interval formula gives:

P(1<X\le3)=F_X(3)-F_X(1)=1.00-0.35=0.65.

As a check, the values strictly greater than 1 and at most 3 are 2 and 3, whose probabilities sum to $0.40+0.25=0.65$. Therefore the three requested results are $0.35$, $0.75$, and $0.65$.

Reading CDF Values for a Continuous Variable

For a continuous random variable with density $f_X$, probabilities over ranges are found by integrating the density. In particular, the CDF at a threshold $x$ is the probability accumulated up to that threshold. If the variable is supported on a finite interval and has no probability outside it, this means integrating the density over the portion of that interval at or below $x$.

The definition remains $F_X(x)=P(X\le x)$; it does not change when we move from a discrete model to a continuous one. What changes is the calculation: probabilities are added across discrete values, whereas a density is integrated across a continuous range. A density value itself is not the probability of an exact point. That distinction is essential when interpreting a CDF for a continuous model.

Worked Example: Cumulative Probability From a Continuous Density

Suppose a hypothetical continuous measurement $X$, in units, has density $f_X(x)=x/8$ for $0<x<4$, and density 0 outside this interval. There is no interest or mortality component; this is a continuous probability model with the stated density and support. Find $F_X(3)$ and $P(1<X\le3)$. The method is integration followed by the CDF interval formula; this avoids using a density value as though it were a probability.

Step 1: Check the density’s total probability. The density integrates over its support to:

\int_0^4 \frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_0^4 =\frac{16}{16}-0 =1.

Thus the stated density assigns total probability 1 to the interval from 0 to 4.

Step 2: Calculate the cumulative probability at 3. The event $X\le3$ covers the part of the support from 0 through 3. Its probability is the integral of the density over that range:

F_X(3)=P(X\le3)=\int_0^3\frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_0^3 =\frac{9}{16} =0.5625.

So the probability that $X$ is at most 3 is $0.5625$.

Step 3: Calculate the interval probability. First evaluate the CDF at 1:

F_X(1)=\int_0^1\frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_0^1 =\frac{1}{16} =0.0625.

Now subtract the cumulative probability through 1 from the cumulative probability through 3:

P(1<X\le3)=F_X(3)-F_X(1) =\frac{9}{16}-\frac{1}{16} =\frac{8}{16} =\frac{1}{2} =0.5000.

As a direct check, integration over the interval gives the same result: $\int_1^3 (x/8)\,dx=(9-1)/16=0.5000$. The results are $F_X(3)=0.5625$ and $P(1<X\le3)=0.5000$.

A CDF Value Is Not Necessarily a Point Probability

A common exam error is to read $F_X(x)$ as $P(X=x)$. The definition shows why this is incorrect: $F_X(x)$ concerns the whole event $X\le x$. For example, in the service-request model, $F_X(1)=0.35$, but the point probability $P(X=1)$ is only $0.25$. The CDF value includes both $X=0$ and $X=1$.

A second error is to use the wrong inequality at a threshold. The definition includes equality: $F_X(x)=P(X\le x)$, not $P(X<x)$. In a discrete model, this can change the answer. At a threshold equal to a possible value, that value’s probability must be included. If the threshold falls between two possible values, the CDF includes all possible values below it and none above it.

For continuous models, remember the roles of the two quantities: a density describes probability per unit of measurement, while the CDF describes accumulated probability up to a threshold. For example, in the density model above, $f_X(3)=3/8$ is a density value, whereas $F_X(3)=9/16$ is a probability. They are not interchangeable.

Key takeaway: Start with the event $X\le x$. Then use the specified model to calculate its probability: sum eligible point probabilities for a discrete variable or integrate the density over the relevant range for a continuous variable. For $a<b$, $F_X(b)-F_X(a)$ gives $P(a<X\le b)$.

Check Your Understanding

For each problem, first identify the event represented by the requested CDF value or interval. Answers are provided for checking.

A hypothetical equipment monitor records a discrete alert count $Z$, with $P(Z=0)=0.15$, $P(Z=1)=0.45$, and $P(Z=4)=0.40$. Find $F_Z(1)$, $F_Z(2)$, and $P(1<Z\le4)$. (Answer: $0.60,\ 0.60,\ 0.40$)
A hypothetical claim amount $X$, in dollars, has probabilities $P(X=0)=0.10$, $P(X=1000)=0.40$, $P(X=2500)=0.30$, and $P(X=5000)=0.20$. Find $F_X(1000)$ and $P(X>1000)$. (Answer: $0.50,\ 0.50$)
A continuous variable $Y$ has density $f_Y(y)=2y/25$ for $0<y<5$, and zero elsewhere. Find $F_Y(2.5)$. (Answer: $0.2500$)
For the density in Question 3, find $P(1<Y\le4)$. (Answer: $0.6000$)
A hypothetical measurement $W$ is uniform on the interval from 0 to 12, meaning its density is $1/12$ on that interval and zero elsewhere. Find $F_W(7)$ and $P(4<W\le9)$. (Answer: $7/12,\ 5/12$)

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