The Central Idea
The cumulative distribution function, or CDF, of a random variable \(X\) is defined by
\[ \boxed{F_X(x)=P(X\le x)}. \]The CDF tells us how much probability has accumulated at or below the threshold \(x\).
Unlike a PMF or PDF, the CDF is defined for every random variable, regardless of whether the distribution is discrete, continuous, or mixed.
What Cumulative Means
The word cumulative means that probability is collected from the left up to a specified point.
If
\[ F_X(4)=0.72, \]then
\[ \boxed{P(X\le4)=0.72}. \]Seventy-two percent of the total probability lies at or below \(4\).
The CDF Is a Function of the Threshold
In
\[ F_X(x)=P(X\le x), \]the input \(x\) is a threshold. As the threshold moves to the right, more possible values of \(X\) are included in the event.
Therefore the accumulated probability can stay the same or increase, but it cannot decrease.
A CDF Is Always Between 0 and 1
Since a CDF value is itself a probability,
\[ \boxed{0\le F_X(x)\le1} \]for every real number \(x\).
Values below \(0\) or above \(1\) cannot occur in a valid CDF.
A CDF Is Nondecreasing
If
\[ ais contained in
\[ \{X\le b\}. \]Therefore,
\[ \boxed{F_X(a)\le F_X(b)}. \]A valid CDF may remain flat over an interval, but it can never move downward.
The Extreme Limits of a CDF
Far to the left of the distribution, essentially no probability has yet accumulated:
\[ \boxed{\lim_{x\to-\infty}F_X(x)=0}. \]Far to the right, all probability has accumulated:
\[ \boxed{\lim_{x\to\infty}F_X(x)=1}. \]These limits are fundamental requirements for every cumulative distribution function.
Right-Continuity
Every CDF is right-continuous. At any point \(x\),
\[ \boxed{\lim_{t\to x^+}F_X(t)=F_X(x)}. \]This property matters especially for discrete distributions, where the CDF may jump at values having positive probability.
The Four Core CDF Properties
A valid cumulative distribution function satisfies these structural conditions:
- \(0\le F_X(x)\le1\) for all \(x\).
- \(F_X(x)\) is nondecreasing.
- \(F_X(x)\) is right-continuous.
- \(\lim_{x\to-\infty}F_X(x)=0\) and \(\lim_{x\to\infty}F_X(x)=1\).
These properties apply to discrete, continuous, and mixed probability distributions.
Building a CDF from a PMF
If \(X\) is discrete with PMF \(p_X(x)\), then
\[ \boxed{F_X(x)=\sum_{t\le x}p_X(t)}. \]The CDF is obtained by adding all probability masses at values less than or equal to the chosen threshold.
Example: A Simple Discrete Distribution
Suppose \(X\) has the PMF
| \(x\) | \(p_X(x)\) |
|---|---|
| \(0\) | \(0.10\) |
| \(1\) | \(0.25\) |
| \(2\) | \(0.40\) |
| \(3\) | \(0.25\) |
The probabilities sum to \(1\), so this is a valid discrete distribution.
Accumulating the Discrete Probabilities
The CDF values at the support points are
\[ F_X(0)=0.10, \] \[ F_X(1)=0.10+0.25=0.35, \] \[ F_X(2)=0.10+0.25+0.40=0.75, \]and
\[ F_X(3)=1. \]Each new support point adds its probability mass to the running total.
The Complete Discrete CDF
For the preceding distribution,
\[ F_X(x)= \begin{cases} 0, & x<0,\\ 0.10, & 0\le x<1,\\ 0.35, & 1\le x<2,\\ 0.75, & 2\le x<3,\\ 1, & x\ge3. \end{cases} \]Notice that the CDF is constant between possible values and jumps when a support point is reached.
Discrete CDFs Look Like Step Functions
A discrete CDF is typically a step function.
Between support points, no additional probability mass is encountered, so the cumulative probability remains constant. At a support point, the CDF jumps upward by the probability assigned to that value.
The Size of a Jump Equals Point Probability
For any random variable, the probability concentrated at a point \(x\) is the size of the CDF jump there:
\[ \boxed{P(X=x)=F_X(x)-F_X(x^-)} \]where
\[ F_X(x^-)=\lim_{t\to x^-}F_X(t). \]For a discrete distribution, these jumps recover the PMF.
Example: Recovering a PMF from a CDF
Suppose a CDF jumps from \(0.35\) immediately before \(x=2\) to \(0.75\) at \(x=2\).
Then
\[ P(X=2)=0.75-0.35. \]Therefore,
\[ \boxed{P(X=2)=0.40}. \]The jump size is exactly the probability mass at \(2\).
Building a CDF from a PDF
If \(X\) is continuous with probability density function \(f_X(x)\), then
\[ \boxed{F_X(x)=\int_{-\infty}^{x}f_X(t)\,dt}. \]The CDF is the total area under the density curve to the left of the threshold \(x\).
Why a Different Variable Appears Inside the Integral
In
\[ F_X(x)=\int_{-\infty}^{x}f_X(t)\,dt, \]the symbol \(t\) is a temporary integration variable. The input of the CDF remains \(x\).
Using a different symbol inside the integral helps distinguish the moving threshold \(x\) from the variable being integrated.
Example: Uniform Distribution on \([0,4]\)
Suppose
\[ f_X(x)= \begin{cases} \frac14, & 0\le x\le4,\\ 0, & \text{otherwise}. \end{cases} \]For \(x<0\), no probability has accumulated:
\[ F_X(x)=0. \]For \(0\le x\le4\),
\[ F_X(x)=\int_0^x\frac14\,dt=\frac{x}{4}. \]For \(x\ge4\), all probability has accumulated:
\[ F_X(x)=1. \]The Uniform CDF
Therefore,
\[ \boxed{ F_X(x)= \begin{cases} 0, & x<0,\\ \frac{x}{4}, & 0\le x\le4,\\ 1, & x>4. \end{cases}} \]Unlike the discrete step-function example, this CDF rises continuously across its support.
Finding Probabilities Directly from a CDF
By definition,
\[ \boxed{P(X\le b)=F_X(b)}. \]The probability above a threshold follows from the complement rule:
\[ \boxed{P(X>b)=1-F_X(b)}. \]These formulas work for both discrete and continuous random variables.
A General Interval Formula
For any random variable,
\[ \boxed{P(aWhy Endpoint Details Matter for Discrete Variables
If \(X\) is discrete, then
\[ P(amay differ because \(P(X=a)\) may be positive.
Therefore the exact inequality symbols must be respected when using a CDF with a discrete distribution.
Including the Left Endpoint
For a general random variable,
\[ P(a\le X\le b) = F_X(b)-F_X(a^-), \]where
\[ F_X(a^-)=\lim_{t\to a^-}F_X(t). \]This subtracts only the probability strictly below \(a\), leaving any mass at \(a\) included.
Continuous Endpoint Simplification
For a continuous random variable, individual points have probability zero. Therefore,
\[ F_X(a^-)=F_X(a). \]Consequently,
\[ P(aExample: Interval Probability from a Continuous CDF
Use the uniform CDF
\[ F_X(x)=\frac{x}{4} \]for \(0\le x\le4\). Then
\[ P(1Recovering a PDF from a CDF
If a continuous CDF is differentiable at \(x\), then its density is
\[ \boxed{f_X(x)=F_X'(x)}. \]This follows from the Fundamental Theorem of Calculus because
\[ F_X(x)=\int_{-\infty}^{x}f_X(t)\,dt. \]The CDF accumulates density, and differentiation recovers the local rate of accumulation.
Example: Differentiate a CDF
Suppose
\[ F_X(x)=x^2 \]for \(0\le x\le1\), with \(F_X(x)=0\) for \(x<0\) and \(F_X(x)=1\) for \(x>1\).
On \(0
Therefore,
PMFs and PDFs require different accumulation operations:
The CDF packages the accumulated result into the same object:
This is one reason the CDF is so fundamental in probability theory.
If a CDF is constant over an interval, then no probability is accumulating
there.
For example, if
for every \(x\) between \(2\) and \(5\), then
The CDF Unifies Discrete and Continuous Probability
Flat Portions of a CDF
Steep Portions of a Continuous CDF
Where a differentiable continuous CDF rises rapidly, its derivative is large.
Since
\[ f_X(x)=F_X'(x), \]a steep CDF corresponds to a region of relatively high probability density.
Jumps Signal Point Mass
A jump in a CDF indicates positive probability concentrated at a single point.
If the CDF jumps by \(0.12\) at \(x=7\), then
\[ \boxed{P(X=7)=0.12}. \]Purely continuous distributions have no jumps.
Continuous CDF Does Not Mean Constant Density
A continuous CDF may curve, flatten, or become steep depending on how the density changes.
Continuity of the CDF means it has no jumps. It does not mean the CDF is a straight line or that the underlying density is uniform.
A CDF Can Be Continuous Without Being Differentiable Everywhere
Some continuous distributions have CDFs that are not differentiable at every point.
Therefore the relationship
\[ f_X(x)=F_X'(x) \]should be used where an ordinary density exists and the derivative is defined. The CDF itself remains the more general object.
Example: A Three-Point Distribution
Suppose
\[ P(X=1)=0.2,\qquad P(X=3)=0.5,\qquad P(X=5)=0.3. \]Then
\[ F_X(x)= \begin{cases} 0, & x<1,\\ 0.2, & 1\le x<3,\\ 0.7, & 3\le x<5,\\ 1, & x\ge5. \end{cases} \]Using the Three-Point CDF
From the CDF,
\[ P(X\le3)=F_X(3)=0.7. \]Also,
\[ P(1Example: A Quadratic Continuous CDF
Let
\[ F_X(x)= \begin{cases} 0, & x<0,\\ \frac{x^2}{9}, & 0\le x\le3,\\ 1, & x>3. \end{cases} \]Then
\[ P(X\le2)=F_X(2)=\frac49. \]Also,
\[ P(1Recovering the Density in the Quadratic Example
For \(0
Therefore,
So the corresponding density is
Quantiles from a CDF
A CDF can also be used to locate thresholds associated with specified cumulative probabilities.
A value \(q\) satisfying
\[ F_X(q)\approx p \]identifies a threshold near the \(p\)-level of the distribution.
This idea leads to medians, percentiles, and more formal definitions of quantiles.
The Median and the CDF
Informally, a median is a value near the point where half the probability lies at or below it and half lies at or above it.
For a continuous distribution with strictly increasing CDF, the median \(m\) satisfies
\[ \boxed{F_X(m)=0.5}. \]Discrete distributions require slightly more care because the CDF may jump across \(0.5\).
Common Error: Confusing a CDF with a PDF
A CDF and a PDF are different functions.
The CDF gives cumulative probability:
\[ F_X(x)=P(X\le x). \]A PDF gives density, and probabilities arise by integration.
For a continuous distribution,
\[ F_X(x)=\int_{-\infty}^{x}f_X(t)\,dt. \]Common Error: Assuming a CDF Must Be Strictly Increasing
A CDF must be nondecreasing, but it may remain constant.
Flat regions simply indicate that no probability is accumulating over those ranges.
Common Error: Allowing a CDF to Decrease
A function that decreases somewhere cannot be a valid CDF.
If \(a
so
Common Error: Forgetting the Right Endpoint in the Definition
The standard definition is
\[ F_X(x)=P(X\le x), \]not
\[ P(XCommon Error: Ignoring Jumps in a Discrete CDF
A discrete CDF does not normally rise smoothly.
If the random variable has positive probability at \(x\), the CDF jumps there by
\[ P(X=x). \]Those jumps are essential information, not graphing artifacts.
Common Error: Using \(F(b)-F(a)\) for the Wrong Endpoint Pattern
For a general random variable,
\[ F(b)-F(a) \]gives
\[ P(aA Reliable Procedure for Reading a CDF
- Identify the threshold or interval in the probability question.
- Use \(F(x)=P(X\le x)\) for cumulative probabilities.
- Use \(1-F(x)\) for probabilities strictly above \(x\).
- Use \(F(b)-F(a)\) for \(P(a
- For discrete variables, check carefully whether endpoints are included.
- Read jump sizes as point probabilities.
- For continuous variables, endpoint inclusion does not affect probability.
A Reliable Procedure for Constructing a Discrete CDF
- List the possible values in increasing order.
- Record the PMF value at each support point.
- Add probabilities cumulatively from left to right.
- Keep the CDF constant between support points.
- At each support point, jump upward by its probability mass.
- Use \(0\) below the smallest support value and \(1\) at and beyond the upper end when the support is finite.
A Reliable Procedure for Constructing a Continuous CDF
- Identify the support of the density.
- Set the CDF equal to \(0\) below the support.
- Integrate the density from the left end of the support to \(x\).
- Simplify the resulting expression.
- Set the CDF equal to \(1\) after all probability has accumulated.
- Check that the result is nondecreasing and approaches \(0\) and \(1\) at the extremes.
Practice 1: Read a CDF Value
Suppose
\[ F_X(6)=0.82. \]Interpret this statement as a probability.
Practice 2: Use a Complement
Suppose
\[ F_X(4)=0.63. \]Find
\[ P(X>4). \]Practice 3: Interval Probability
Suppose
\[ F_X(2)=0.30 \]and
\[ F_X(7)=0.88. \]Find
\[ P(2Practice 4: Jump Size
A CDF has
\[ F_X(3^-)=0.45 \]and
\[ F_X(3)=0.70. \]Find \(P(X=3)\).
Practice 5: Construct a Discrete CDF
Suppose
\[ P(X=0)=0.2,\qquad P(X=1)=0.5,\qquad P(X=2)=0.3. \]Write the CDF \(F_X(x)\).
Practice 6: Uniform CDF
Suppose
\[ f_X(x)= \begin{cases} \frac15, & 0\le x\le5,\\ 0, & \text{otherwise}. \end{cases} \]Find \(F_X(x)\) for \(0\le x\le5\).
Practice 7: Probability from the Uniform CDF
Using Practice 6, find
\[ P(1Practice 8: Recover a Density
Suppose
\[ F_X(x)=x^3 \]for \(0\le x\le1\), with \(F_X(x)=0\) for \(x<0\) and \(F_X(x)=1\) for \(x>1\).
Find the density on \(0
Practice 9: Valid CDF?
A proposed function decreases from \(0.7\) to \(0.6\) as \(x\) increases.
Can it be a valid CDF? Explain.
Practice 10: Endpoint Awareness
For a discrete random variable, explain why
\[ P(2Practice Answers
Practice 1
By definition,
\[ \boxed{P(X\le6)=0.82}. \]Practice 2
\[ P(X>4)=1-F_X(4)=1-0.63. \]Therefore,
\[ \boxed{P(X>4)=0.37}. \]Practice 3
\[ P(2Therefore,
\[ \boxed{P(X=3)=0.25}. \]Practice 5
The cumulative probabilities are \(0.2\), \(0.7\), and \(1\). Hence
\[ \boxed{ F_X(x)= \begin{cases} 0, & x<0,\\ 0.2, & 0\le x<1,\\ 0.7, & 1\le x<2,\\ 1, & x\ge2. \end{cases}} \]Practice 6
For \(0\le x\le5\),
\[ F_X(x)=\int_0^x\frac15\,dt. \]Therefore,
\[ \boxed{F_X(x)=\frac{x}{5}}. \]Practice 7
\[ P(1Practice 8
\[ f_X(x)=F_X'(x)=3x^2. \]Therefore,
\[ \boxed{f_X(x)=3x^2\quad\text{for }0No. A CDF must be nondecreasing. Once cumulative probability has reached \(0.7\), it cannot fall to \(0.6\) as the threshold increases.
Practice 10
The formula
\[ F_X(5)-F_X(2) \]removes all probability at or below \(2\), so it excludes any probability mass at \(X=2\). The event \(2\le X\le5\) includes that mass.
What to Remember
The cumulative distribution function is
\[ \boxed{F_X(x)=P(X\le x)}. \]For a discrete random variable,
\[ \boxed{F_X(x)=\sum_{t\le x}p_X(t)}. \]For a continuous random variable with density,
\[ \boxed{F_X(x)=\int_{-\infty}^{x}f_X(t)\,dt}. \]The CDF is nondecreasing, right-continuous, approaches \(0\) on the far left, and approaches \(1\) on the far right.
For any random variable,
\[ \boxed{P(aWhere We Go Next
A distribution tells us how probability is allocated and accumulated. Often, however, we want a single numerical summary of where the distribution is centered.
The next lesson, Expected Value, develops the probability-weighted average of a random variable and shows how expectation summarizes long-run average behavior.