Probability Mass, Density, and Cumulative Probability

Separate discrete probability mass from continuous density and interval area. Use the cumulative distribution function to describe accumulated probability and recover interval probabilities by subtraction.

The word “probability” refers to different numerical objects in discrete and continuous models. A discrete value can carry probability mass. A continuous point has no interval width and normally carries zero probability; a density describes how probability is concentrated nearby. The cumulative distribution function provides one language that works in both settings.

Discrete Values Carry Probability Mass

For a discrete random variable $X$ , the probability mass function (PMF) is:

p_X(x)=P(X=x).

Every mass is nonnegative, and the masses across the support sum to one:

p_X(x)\ge0, \qquad \sum_{x\in\operatorname{supp}(X)}p_X(x)=1.

Consider a variable with the following PMF:

$x$	$0$	$1$	$2$
$p_X(x)$	$0.2$	$0.5$	$0.3$

The event $\{X\ge1\}$ contains the values $1$ and $2$ , so:

P(X\ge1)=p_X(1)+p_X(2)=0.5+0.3=0.8.

A PMF value is already a probability. Its horizontal location identifies one possible value, and its height gives the mass placed exactly there.

Exercise: Add discrete probability mass

Using the table above, what is $P(X\ge1)$ ?

Check the event probability

Compute it first, then check your number.

HintIdentify the included values

The event contains $X=1$ and $X=2$ .

SolutionSum the masses at one and two

The two included values carry masses $0.5$ and $0.3$ . Therefore:

P(X\ge1)=P(X=1)+P(X=2)=0.5+0.3=0.8.

The mass at $X=0$ is excluded because that value does not satisfy the event.

Continuous Values Use Probability Density

For a continuous random variable, a probability density function (PDF), written $f_X(x)$ , describes probability per unit of $x$ . A valid density is nonnegative and has total area one:

f_X(x)\ge0, \qquad \int_{-\infty}^{\infty}f_X(x)\,dx=1.

Probability over an interval is area under the density:

P(a\le X\le b)=\int_a^b f_X(x)\,dx.

“The probability that $X$ lies from $a$ through $b$ equals the integral of its density from $a$ to $b$ .” For a continuous variable, probability is the area over the interval, not the density height at one point.

Suppose $f_X(x)=0.2$ on $[0,5]$ and is zero elsewhere. The density is constant, and its total rectangular area is $5\times0.2=1$ . The probability between $1$ and $3$ is:

P(1\le X\le3)=\int_1^3 0.2\,dx=0.2(3-1)=0.4.

The density height $0.2$ is not the probability of one point. For a continuous random variable:

P(X=x)=0

even where $f_X(x)$ is positive. A single point has zero width, so its area is zero.

Exercise: Calculate area under a uniform density

A density equals $0.2$ on $[0,5]$ and zero elsewhere. What is $P(1\le X\le3)$ ?

Check the interval probability

Compute it first, then check your number.

HintMultiply height by interval width

The interval width is $3-1=2$ .

SolutionIntegrate the constant density

Since the density is constant on the interval:

P(1\le X\le3)=\int_1^3 0.2\,dx=0.2\times2=0.4.

The result is an interval area and is therefore a probability between zero and one.

A Density Can Be Greater Than One

A probability must not exceed one, but a density is not a probability. If a variable is uniform on the short interval $[0,0.5]$ , its density must have height $2$ so that its total area is one:

\int_0^{0.5}2\,dx=1.

The density value $2$ is valid because probability comes from height multiplied by width. On the subinterval $[0.1,0.2]$ , the probability is $2(0.2-0.1)=0.2$ .

Exercise: Interpret a density above one

A random variable is uniform on $[0,0.25]$ , with density $f_X(x)=4$ on that interval. Is this a valid density?

Choose the valid interpretation

Yes, because its total area is one.

No, because a density may never exceed one.

Select one choice, then check.

HintCalculate the rectangle's area

Multiply the support width $0.25$ by the height $4$ .

SolutionNormalization depends on area

The density is nonnegative and its total area is:

0.25\times4=1.

It is therefore valid. The bound $0\le P(A)\le1$ applies to event probabilities, not to density values measured per unit of $X$ .

The CDF Accumulates Probability

The cumulative distribution function (CDF) of any real-valued random variable is:

F_X(x)=P(X\le x).

Every CDF has four structural properties:

$0\le F_X(x)\le1$ ;
it never decreases as $x$ increases;
it approaches $0$ far to the left;
it approaches $1$ far to the right.

A CDF is also right-continuous: its value at $x$ includes any mass placed exactly at $x$ .

For a discrete variable, the CDF is a step function. At a value $x$ , its jump size equals the point mass:

P(X=x)=F_X(x)-F_X(x^-),

where $F_X(x^-)$ is the accumulated probability immediately to the left of $x$ .

For a continuous variable with density $f_X$ :

F_X(x)=\int_{-\infty}^{x}f_X(t)\,dt,

and wherever the CDF is differentiable:

F_X'(x)=f_X(x).

Thus the CDF is accumulated probability, while the density is its local rate of accumulation.

Exercise: Read point mass from a CDF jump

A discrete CDF has $F_X(2^-)=0.35$ and $F_X(2)=0.60$ . What is $P(X=2)$ ?

Check the point mass

Compute it first, then check your number.

HintMeasure the jump at two

Subtract the accumulated probability immediately before $2$ from the value that includes $2$ .

SolutionThe CDF jump equals the point mass

The jump size is:

P(X=2)=F_X(2)-F_X(2^-)=0.60-0.35=0.25.

This jump is possible because the variable is discrete. A continuous CDF has no point-mass jumps.

CDF Differences Give Interval Probability

Because $F_X(b)$ includes all mass at or below $b$ , while $F_X(a)$ includes all mass at or below $a$ :

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

If $F_X(2)=0.35$ and $F_X(5)=0.80$ , then:

P(2<X\le5)=0.80-0.35=0.45.

For a continuous random variable, individual endpoints have probability zero, so $P(a<X\le b)$ , $P(a\le X\le b)$ , and the other endpoint variants are equal. For a discrete variable, endpoint inclusion can change the answer and must be read carefully.

Exercise: Calculate an interval from a CDF

If $F_X(2)=0.35$ and $F_X(5)=0.80$ , what is $P(2<X\le5)$ ?

Check the interval probability

Compute it first, then check your number.

HintRemove the earlier accumulation

The probability through $2$ is already included in the probability through $5$ .

SolutionSubtract the two cumulative probabilities

The required interval excludes values at or below $2$ and includes values up to $5$ . Therefore:

P(2<X\le5)=F_X(5)-F_X(2)=0.80-0.35=0.45.

Compare PMFs, PDFs, and CDFs Visually

The following plots place a discrete PMF beside its step-shaped CDF and a continuous uniform PDF beside its continuous CDF. Change the discrete masses or the uniform interval and inspect what changes locally and cumulatively.

Plot discrete mass, continuous density, and their CDFs

Edit either distribution. In the left column, height means mass or density; in the right column, height always means accumulated probability.

Ready to run.

The CDF always has probability on its vertical axis. By contrast, the PMF has probability mass on its vertical axis and the PDF has probability per unit of $X$ . This difference is also visible in their units.

Units Distinguish Mass from Density

Probability mass and cumulative probability are unitless. A density has inverse units. If $X$ is measured in seconds, $f_X(x)$ has units of probability per second. Integrating over an interval measured in seconds cancels those units and produces a unitless probability.

Before using a distribution, identify whether a displayed function is a PMF, PDF, or CDF. For a PMF, add heights. For a PDF, integrate area. For a CDF, subtract accumulated values. Check endpoint inclusion for discrete variables, and never interpret a continuous density height as exact-point probability.

The next lesson studies expectation, which uses a distribution to form a probability-weighted average of a random variable.