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Probability Mass Functions (PMF)



Key Terms
Definition & Physical Intuition
Mass vs. Density
PMF Notation
The Domain (Support)
The Two Fundamental Axioms
Representations: The Three Views
Building a PMF (From Story to Math)
Calculations Using the PMF
Derived Properties: Location Measures
Derived Properties: Spread Measures
Transformations
Connection to CDF
PMFs of Common Discrete Distributions
Common Mistakes
PMF at a Glance
PMF FAQ



Probability in the Discrete Case


When a random variable can take only isolated, countable values, probability is assigned directly to those values rather than spread over intervals. Each possible outcome carries a specific “mass” that reflects how likely it is to occur.

This page develops the Probability Mass Function (PMF) as the fundamental object governing discrete random variables. It shows how probability is distributed across individual points, how these masses must satisfy basic axioms, and how they support calculations such as event probabilities, averages, and variability. The emphasis is on understanding the structure and behavior of discrete probability models, not just listing particular distributions.

Key Terms

Probability Mass Function— pX(x)=P(X=x)p_X(x) = P(X = x) for discrete variables
Discrete Random Variable— the type of variable described by a PMF
Cumulative Distribution Function— FX(x)F_X(x), related to the PMF by summation
Expected Value— E[X]E[X], computed from the PMF
Variance— Var⁡(X)\operatorname{Var}(X), computed from the PMF

See All Probability Definitions →


Definition & Physical Intuition

Toss two coins and count the heads: X is 0, 1 or 2. How likely is each value?
The four equally likely results HHHH, HTHT, THTH and TTTT give the answer by counting.
Each value of X gets its own probability, and together the values use up all of it.
value k of Xresults that give itP(X = k)
0TT1/4
1HT, TH1/2
2HH1/4
totalall four results1
A rule that takes a value kk and returns P(X=k)P(X = k) holds everything there is to know about XX.
Probability sits on single points, like weights placed on a number line.
Think of it like placing weights on a number line. Each possible outcome gets a weight equal to its probability. Where you put more weight, that outcome is more likely.
Example: Roll a fair die. Put weight 16\frac{1}{6} at positions 1, 2, 3, 4, 5, and 6. Each spot has equal probability.
Example: Roll a loaded die favoring 6. Put weight 12\frac{1}{2} at position 6, and split the remaining 12\frac{1}{2} among the other five faces.
The definition
The probability mass function of a discrete random variable XX is the function p(k)=P(X=k)p(k) = P(X = k), giving the probability of each value kk.

What is a PMF?

A PMF gives the probability that a discrete random variable lands on each specific value.
So for two tosses, p(0)=14p(0) = \frac{1}{4}, p(1)=12p(1) = \frac{1}{2} and p(2)=14p(2) = \frac{1}{4}.
The weights must add up to 1.
The weights must sum to 1—that's your total probability budget.
PMF Weights Illustration Loaded Die: Probability as Weight 0.1 1 0.1 2 0.1 3 0.1 4 0.1 5 0.5 6 Outcome Value Probability 5× heavier!
0.0000.0500.1000.1500.2000.250012345678910Value (k)Probability P(X = k)
Binomial PMF, n = 10, p = 0.5

Each bar is a weight placed on a value of the discrete variable: the height at k is P(X = k), the weights add to 1 and the number line between the integers carries nothing. Here the heaviest weight sits at 5. Switch families and watch the weights redistribute on the PMF visualizer.

A PMF belongs to discrete variables only.
For a continuous variable every single value has probability 00, so a list of point weights says nothing; a density takes over, as compared in mass vs. density.
The set of values that carry weight is the support, defined next.
The rest of the page covers mass against density, notation, the support, the two axioms, the three views of a PMF, building one from a story, calculations, location and spread measures, transformations, the link with the CDF and common mistakes.

Mass vs. Density


The PMF is one of two ways to describe probability distributions. The key distinction is whether your variable takes discrete values or continuous values.

PMF (Mass): For discrete random variables. Probability concentrates at specific points. You can ask "What's P(X=5)P(X = 5)?" and get a number.

Example: Dice rolls, coin flips, number of customers.

PDF (Density): For continuous random variables. Probability spreads across intervals. P(X=5)=0P(X = 5) = 0 for any exact point—you need ranges like P(4.9<X<5.1)P(4.9 < X < 5.1).

Example: Height, weight, time measurements.

The PMF gives exact point probabilities. The PDF doesn't—it only gives probabilities over intervals.


PMF vs PDF PMF (Discrete) Exact Point Probabilities X P(X) 2 3 4 5 6 P(X=5) > 0 PDF (Continuous) Interval Probabilities X f(x) 4.9 5.1 5 P(X=5) = 0 P(4.9 < X < 5.1) > 0 Discrete: Exact probabilities Continuous: Area = probability

PMF Notation

Notation

PMF Notation

Three marks on this page are not about pp itself but about the set it is summed over, and each writes that set a different way.
The capital/lowercase split XX versus xx, the operator P(⋅)P(\cdot) and the mass/density pair pp versus ff all belong to the probability function. The cumulative FXF_X and the ≤\leq convention are at the CDF, and the signature X:Ω→RX: \Omega \to \mathbb{R} at random variables.
{x:pX(x)>0}\{x : p_X(x) > 0\}
the set of x such that p of x is greater than zero — the support of X
The support is defined by a condition, not listed. The strict >0> 0 is the whole definition: a value belongs exactly when it carries probability, so values the variable can never take are excluded rather than assigned zero.
CasesThe PMF itself is defined everywhere and returns 00 off the support, so the function's domain and its support are different sets. Only the support matters in a sum, since the rest contributes nothing.
Also writtenWritten supp⁡(X)\operatorname{supp}(X), or named SS or SXS_X and then used as a set. Concrete supports are given as rosters instead — {1,2,3,4,5,6}\{1,2,3,4,5,6\} for a die, {0,1,2,…}\{0,1,2,\ldots\} with an ellipsis when infinite.
Do not confuseReading the support as the list of values the variable is rather than the values it can be with positive probability. A value sitting inside the roster but carrying probability 00 is not in the support.
∑x∈SupportpX(x)=1\sum_{x \in \text{Support}} p_X(x) = 1
the sum of p of x, over all x in the support, equals one
The index under the sigma is a set membership, not a counter running from one bound to another. There is no start value and no end value — the sum ranges over whatever the support contains, which may be finite or infinite.
CasesBecause off-support values contribute 00, summing over the support and summing over every real xx give the same answer. The support form is written because it is the one that can actually be computed.
Also writtenCompressed to ∑xpX(x)=1\sum_x p_X(x) = 1 with the set left implicit, or spelled ∑k=0n\sum_{k=0}^{n} once the support is known to be a run of integers.
Do not confuseTreating it as an ordinary indexed sum and looking for the limits. There are none to find; the limits are replaced by a membership test.
pY(y)=∑x : g(x)=ypX(x)p_Y(y) = \sum_{x\,:\,g(x) = y} p_X(x)
p of y equals the sum of p of x over every x for which g of x equals y
Here the index is a condition on the answer: collect every input the transformation sends to this particular yy, and add their probabilities. The colon reads such that, the same role it plays in the support above.
CasesWhen gg is one-to-one the condition picks out a single xx and the sum has one term, which is why the linear case collapses to pX ⁣(y−ba)p_X\!\left(\frac{y-b}{a}\right) with no sigma at all. The sum is there for the many-to-one case, where probability from several inputs piles onto one output.
Also writtenWritten ∑x∈g−1(y)\sum_{x \in g^{-1}(y)} using the preimage, which says the same thing as a set rather than as a test.
Do not confuseExpecting pYp_Y to be pXp_X with yy substituted in. Transforming a variable moves mass and can merge it; the sigma is the merging.
DownloadPNG2×PDF
Notation · Cheat sheet
Random Variable Notation at a Glance
The marks from the notation cards across the random-variable pages: how each is read and what it means. Every group links to the full card on its page.
Random variables
X:Ω→RX : \Omega \to \mathbb{R}
X maps outcomes to numbers
A random variable is a function, not a variable.
{X≤x}\{X \le x\}
the event that X is at most x
Braces turn a condition into an event.
X∼Bin(n,p)X \sim \text{Bin}(n, p)
X is distributed as
Names the distribution and its parameters.
Probability functions
XX vs xx
the variable; a value
Capital for the random variable, lowercase for a number it takes.
P(⋅)P(\cdot)
the probability of
Takes an event, returns a number in [0,1][0, 1].
p(x)p(x) vs f(x)f(x)
mass; density
pp is a chance (discrete); ff is a rate (continuous).
PMFThis page
{x:pX(x)>0}\{x : p_X(x) > 0\}
the support of X
Every value that carries probability.
∑xpX(x)=1\sum_{x} p_X(x) = 1
the masses add to one
The sum runs over the support, not from a start to an end.
PDF
fX(x)f_X(x)
the density of X at x
A density, not a probability; it can exceed 11.
P(a≤X≤b)P(a \le X \le b)
X between a and b
The integral ∫abfX(x) dx\int_a^b f_X(x)\,dx.
P(X=a)=0P(X = a) = 0
a single point
For a continuous variable, always zero.
CDF
FX(x)F_X(x)
the CDF of X at x
Equals P(X≤x)P(X \le x): everything up to and including xx.
F(b)−F(a)F(b) - F(a)
F of b minus F of a
Equals P(a<X≤b)P(a < X \le b).
Indicators
IA,  1AI_A, \; \mathbf{1}_A
the indicator of A
11 if AA happens, 00 if not.
1{X≤x}\mathbf{1}\{X \le x\}
one if the condition holds
The brace form takes a statement instead of an event.
N=∑IAiN = \sum I_{A_i}
a count of events
How many of the AiA_i happened.
Same marks, different job
PP vs ppCapital PP takes an event; lowercase pp is a mass function of a value.
FF vs ffCapital FF accumulates (the CDF); lowercase ff is the density.
∼\simIs distributed as; in linear algebra, row-equivalent or similar.
{X≤x}\{X \le x\}An event, not a set of numbers.
Random variable symbols tableProbability
Random Variable Notation at a Glance·/probability/random-variablesLearn Math Class

The Domain (Support)


The support of a PMF is the set of values where the probability is nonzero—the places where probability mass actually sits.

Formally: The support of XX is {x:pX(x)>0}\{x : p_X(x) > 0\}.

Think of it as the active zone. Outside the support, pX(x)=0p_X(x) = 0. Nothing happens there.

Types of Support


Finite Support: The random variable can take only a limited number of values.

Example: A fair die. Support is {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}. Six values, finite.

Example: Flip a coin 10 times, count heads. Support is {0,1,2,…,10}\{0, 1, 2, \ldots, 10\}. Eleven values, finite.

Infinite Support: The random variable can take infinitely many values, but still countable.

Example: Flip a coin until you get heads. Count the number of flips. Support is {1,2,3,4,…}\{1, 2, 3, 4, \ldots\}. Could go on forever.

Example: Number of emails you receive in a day (Poisson model). Support is {0,1,2,3,…}\{0, 1, 2, 3, \ldots\}. No upper bound.

Why Support Matters


The support tells you:
• Where to look for probability
• What values are possible vs. impossible
• How to set up calculations (sums over the support)

If you're computing P(X∈A)P(X \in A) for some set AA, you only sum over values in AA that are also in the support. Outside the support contributes nothing.

The Two Fundamental Axioms


Every valid PMF must satisfy two non-negotiable rules. Break either one, and you don't have a PMF.

Axiom 1: Non-Negativity


PMF Non-negativity
pX(x)≥0 for all xp_X(x) \geq 0 \text{ for all } x
Learn more about this formula: PMF Non-negativity →


Probabilities cannot be negative. Ever.

Each point gets zero or positive mass. If pX(5)=−0.2p_X(5) = -0.2, you've made an error. Fix it.

Axiom 2: Normalization


PMF Normalization
∑x∈SupportpX(x)=1\sum_{x \in \text{Support}} p_X(x) = 1
Learn more about this formula: PMF Normalization →


All probabilities across the entire support must sum to exactly 1.

This is your total probability budget. You must allocate all of it, and you can't overspend.

Example
Fair die. pX(k)=16p_X(k) = \frac{1}{6} for k∈{1,2,3,4,5,6}k \in \{1,2,3,4,5,6\}.
Check: 6×16=16 \times \frac{1}{6} = 1. ✓

Example
Loaded die with pX(6)=12p_X(6) = \frac{1}{2} and pX(k)=110p_X(k) = \frac{1}{10} for k∈{1,2,3,4,5}k \in \{1,2,3,4,5\}.
Check: 12+5×110=12+12=1\frac{1}{2} + 5 \times \frac{1}{10} = \frac{1}{2} + \frac{1}{2} = 1. ✓

What These Axioms Guarantee


Together, these axioms ensure:
• No impossible probabilities (negative or greater than 1)
• Complete accounting (all outcomes covered)
• Consistency with basic probability theory

Violate either axiom, and your probability model breaks. These are the foundation everything else builds on.

Representations: The Three Views


    A PMF can be represented in three equivalent ways. Each offers different insights into how probability mass distributes.

    Graphical Representation


    The PMF appears as a series of vertical bars (lollipop diagram) or dots on the x-y plane. The x-axis shows values of the random variable, the y-axis shows probability.

    Key features:
  • Height of each bar represents probability at that point
  • Bars are discrete — isolated points, not continuous curves
  • All heights must be non-negative
  • Sum of all heights equals 1

    • Example: Fair die PMF shows six bars of equal height 16\frac{1}{6} at positions 1, 2, 3, 4, 5, 6.

    Example: Binomial distribution with n=10,p=0.5n=10, p=0.5 shows bars concentrated near the center (around 5) with symmetric decay on both sides.

    Functional Representation


    The PMF is defined by an explicit mathematical formula pX(k)p_X(k).

    Example: Fair die:

    Representations: The Three Views
    pX(k)={16if k∈{1,2,3,4,5,6}0otherwisep_X(k) = \begin{cases} \frac{1}{6} & \text{if } k \in \{1, 2, 3, 4, 5, 6\} \\ 0 & \text{otherwise} \end{cases}

    Example: Geometric distribution:

    Representations: The Three Views
    pX(k)=(1−p)k−1p for k∈{1,2,3,…}p_X(k) = (1-p)^{k-1}p \text{ for } k \in \{1, 2, 3, \ldots\}

    The functional form allows direct computation of probabilities and derivation of properties like expected value and variance.

    Tabular Representation


    The PMF can be displayed as a table listing each value and its probability.

    Example: Roll two dice, sum the results:



Sum (k) 2 3 4 5 6 7 8 9 10 11 12
pX(k) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

    Tables are especially useful when the PMF doesn't follow a simple formula or when presenting empirical data.

    Fact: Odd dice, the same table
    Dice numbered 1,2,2,3,3,41, 2, 2, 3, 3, 4 and 1,3,4,5,6,81, 3, 4, 5, 6, 8 produce this exact table: their 3636 face pairs hit each sum as often as ordinary dice do. Apart from ordinary dice, no other pair with positive whole-number faces does (Sicherman dice).

    The two-dice table comes from counting: of the 3636 equally likely ordered pairs of faces, one gives the sum 22, two give the sum 33, and six give the most common sum, 77. The table keeps only the sums and their chances. Which faces produced each sum is no longer recorded anywhere in it.

    Why Three Views?


  • Graphical: Best for intuition and pattern recognition
  • Functional: Best for calculation and theoretical analysis
  • Tabular: Best for lookup and concrete examples

    • All three describe the same probability distribution, just from different perspectives. Choose the representation that best serves your immediate purpose.

Representation Best for Form Example
Graphical intuition; seeing shape, peaks, and tails vertical bars (lollipops) or impulses at each value six bars of height 1/6 for a fair die
Functional calculation; deriving properties analytically explicit mathematical formula pX(k) pX(k) = (1−p)k−1·p (geometric)
Tabular lookup; concrete cases and empirical data value-probability listing the dice-sum table shown above
DownloadPNG2×PDF
Before you read
Three views of one PMF
A probability mass function can be drawn as bars, written as a formula, or laid out as a table.
Representing a PMF3 views
Graphical
Readsbars; height = probability
Functional
Readsa formula pX(k)p_X(k)
Examplesa die, the geometric distribution
Tabular
Readsa value-probability table
Examplethe sum of two dice
Takeaway
Every view sums to 1
Bar heights, formula values and table entries are all probabilities, and in every view they add up to 1.
Three views of a PMF·/probability/probability-function/pmfLearn Math Class

Building a PMF (From Story to Math)


Every PMF starts with a story—the situation, the process, the experiment. Your job is to translate that story into mathematical form.

The Construction Process


Step 1: Identify the Random Variable

What are you counting or measuring? Define it clearly.

Example: "Flip a coin 3 times, count heads" → Let XX = number of heads.

Step 2: Determine the Support

What values can the random variable take?

Example: XX can be 0, 1, 2, or 3. Support is {0,1,2,3}\{0, 1, 2, 3\}.

Step 3: Assign Probabilities

For each value in the support, determine its probability. Use counting, combinatorics, symmetry, or physical reasoning.

Example
Fair coin, 3 flips.
• P(X=0)=18P(X = 0) = \frac{1}{8} (only TTT)
• P(X=1)=38P(X = 1) = \frac{3}{8} (HTT, THT, TTH)
• P(X=2)=38P(X = 2) = \frac{3}{8} (HHT, HTH, THH)
• P(X=3)=18P(X = 3) = \frac{1}{8} (only HHH)

Step 4: Verify the Axioms

Check non-negativity: All probabilities ≥ 0? Yes.

Check normalization: Do they sum to 1?

18+38+38+18=1\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1 ✓

These are the two fundamental axioms every PMF must satisfy.

Step 5: Write the PMF

Express the pattern as a formula if possible, or as a table.

Formula: pX(k)=(3k)(12)3p_X(k) = \binom{3}{k} \left(\frac{1}{2}\right)^3 for k∈{0,1,2,3}k \in \{0,1,2,3\}

From Physical Setup to Mathematical Object


The key skill is moving from a concrete scenario to a precise probability function. Ask:
• What outcomes are possible?
• How likely is each one?
• Does the assignment satisfy the axioms?

Once built, the PMF becomes a complete mathematical description of the random variable's behavior.
Step Action Example: 3 coin flips, X = number of heads
1 Identify the random variable let X = number of heads in 3 fair flips
2 Determine the support support = {0, 1, 2, 3}
3 Assign a probability to each value P(X=0) = 1/8; P(X=1) = 3/8; P(X=2) = 3/8; P(X=3) = 1/8
4 Verify the axioms all probabilities ≥ 0 ✓; sum = 1/8 + 3/8 + 3/8 + 1/8 = 1 ✓
5 Write the PMF (formula or table) pX(k) = C(3, k) · (1/2)3 for k ∈ {0, 1, 2, 3}

Calculations Using the PMF


Once you have a PMF, you can compute probabilities for any event involving the random variable.

Single Point Probabilities


The PMF directly gives the probability of any specific value.

P(X=k)=pX(k)P(X = k) = p_X(k)

Example
Roll a fair die. P(X=4)=16P(X = 4) = \frac{1}{6}

Range Probabilities


To find P(X∈A)P(X \in A) for some set AA, sum the PMF over all values in AA.

P(X∈A)=∑k∈ApX(k)P(X \in A) = \sum_{k \in A} p_X(k)

Example: P(X≤3)=pX(1)+pX(2)+pX(3)=16+16+16=12P(X \leq 3) = p_X(1) + p_X(2) + p_X(3) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}

Example: P(2≤X≤5)=pX(2)+pX(3)+pX(4)+pX(5)=46=23P(2 \leq X \leq 5) = p_X(2) + p_X(3) + p_X(4) + p_X(5) = \frac{4}{6} = \frac{2}{3}

Complement Probabilities


To find P(X∉A)P(X \notin A), use the complement rule.

P(X∉A)=1−P(X∈A)P(X \notin A) = 1 - P(X \in A)

Example
P(X>3)=1−P(X≤3)=1−12=12P(X > 3) = 1 - P(X \leq 3) = 1 - \frac{1}{2} = \frac{1}{2}

Pitfall: Using the wrong complement
Writing P(X>3)=1−P(X<3)P(X > 3) = 1 - P(X < 3) leaves the face 33 in neither event and gives 1−26=231 - \frac{2}{6} = \frac{2}{3}. The complement of X>3X > 3 is X≤3X \leq 3, so P(X>3)=1−12=12P(X > 3) = 1 - \frac{1}{2} = \frac{1}{2}.

The complement rule needs the two events to split the support exactly: every value of XX lies in one of them and none lies in both. For the die, the faces 1,2,31, 2, 3 make up the event X≤3X \leq 3 and the faces 4,5,64, 5, 6 make up its complement, three faces of mass 16\frac{1}{6} on each side.

The Core Pattern


Every probability question reduces to:
1. Identify which values satisfy the condition
2. Sum the PMF over those values
3. Simplify

The PMF is your complete toolkit. If you know pX(k)p_X(k) for all kk, you can answer any probability question about XX.

Probability type Formula Fair-die example
Single point P(X = k) = pX(k) P(X = 4) = 1/6
Range / event P(X ∈ A) = Σk ∈ A pX(k) P(X ≤ 3) = 1/6 + 1/6 + 1/6 = 1/2
Complement P(X ∉ A) = 1 − P(X ∈ A) P(X > 3) = 1 − 1/2 = 1/2
Fair die PMF with the bars for 2, 3, 4 and 5 highlighted: P(2 to 5) = 4/6ky1234567P(2 ≤ X ≤ 5) = 1/6 + 1/6 + 1/6 + 1/6 = 4/6endpoints 2 and 5 includedA range probability: add the bars inside it
A range probability as bars

The fair-die PMF with the four bars for 2, 3, 4 and 5 picked out. A range probability is the sum of the bars inside the range: P(2 ≤ X ≤ 5) = 4 · 1/6 = 2/3. Both endpoints count because they are possible values; with a PMF, ≤ and < give different answers.

P(2 < X < 5) would drop both end bars: 2/6.

Derived Properties: Location Measures


The PMF contains all the information about a random variable. From it, we can compute summary measures that describe where the distribution is centered.

Expected Value (Mean)


The expected value is the probability-weighted average of all possible values.

Derived Properties: Location Measures
E[X]=∑k∈Supportk⋅pX(k)E[X] = \sum_{k \in \text{Support}} k \cdot p_X(k)

Think of it as the center of mass. Each value kk contributes based on its probability weight.

Example: Fair die.

Derived Properties: Location Measures
E[X]=1⋅16+2⋅16+3⋅16+4⋅16+5⋅16+6⋅16=216=3.5E[X] = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} = \frac{21}{6} = 3.5

The average outcome over many rolls is 3.5.

Pitfall: Averaging the values without weights
For the loaded die, 1+2+3+4+5+66=3.5\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5 treats every face as equally likely. Weighting each face by its probability, E[X]=110(1+2+3+4+5)+12⋅6=1.5+3=4.5E[X] = \frac{1}{10}(1 + 2 + 3 + 4 + 5) + \frac{1}{2} \cdot 6 = 1.5 + 3 = 4.5.

For the fair die every weight is the same, 16\frac{1}{6}, so each face pulls on the balance point equally and the mean lands exactly in the middle of the support. When the weights differ, as on the loaded die from the axioms section, the faces that carry more probability pull the center of mass toward themselves.

Median


The median is the value that splits the probability in half. It's the point where P(X≤m)≥0.5P(X \leq m) \geq 0.5 and P(X≥m)≥0.5P(X \geq m) \geq 0.5.

For discrete distributions, the median may not be unique if probability mass is concentrated at specific points.

Example
Fair die. The median is between 3 and 4, often reported as 3.5.

Mode


The mode is the value with the highest probability.

Derived Properties: Location Measures
mode=arg⁡max⁡kpX(k)\text{mode} = \arg\max_k p_X(k)

Example
Loaded die with pX(6)=12p_X(6) = \frac{1}{2} and pX(k)=110p_X(k) = \frac{1}{10} for k∈{1,2,3,4,5}k \in \{1,2,3,4,5\}. Mode is 6.

A distribution can have multiple modes if several values share the maximum probability.

Why These Matter


Fact: A valid PMF with no mean
The masses pX(k)=1k(k+1)p_X(k) = \frac{1}{k(k+1)} for k=1,2,3,…k = 1, 2, 3, \ldots add up to 11, since each equals 1k−1k+1\frac{1}{k} - \frac{1}{k+1} and the sum telescopes. Yet E[X]=12+13+14+⋯E[X] = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots is the harmonic series, which grows without bound.

Location measures summarize where the distribution sits on the number line. They answer: "What's a typical value?" Each measure captures a different notion of "typical."

On an infinite support the mean is no longer a finite list of products but an infinite sum. The masses must shrink toward zero for the probabilities to add up to 11, while the values kk they multiply keep growing, and the two pull in opposite directions. The median and the mode are not weighted sums of the values; they come from comparing masses, and every PMF has both.

Derived Properties: Spread Measures


While location tells you where the distribution is centered, spread tells you how much the values vary around that center.

Variance


Variance quantifies the average squared deviation from the mean. It's calculated directly from the PMF:

Derived Properties: Spread Measures
Var(X)=∑k∈Support(k−E[X])2⋅pX(k)\text{Var}(X) = \sum_{k \in \text{Support}} (k - E[X])^2 \cdot p_X(k)

Each value's squared distance from the mean is weighted by its probability. High-probability values far from the mean increase variance significantly.

Computational shortcut: Var(X)=E[X2]−(E[X])2\text{Var}(X) = E[X^2] - (E[X])^2

Pitfall: Subtracting the mean instead of its square
For the fair die, E[X2]−E[X]=916−3.5=353≈11.67E[X^2] - E[X] = \frac{91}{6} - 3.5 = \frac{35}{3} \approx 11.67. The shortcut subtracts the squared mean: 916−3.52=3512≈2.92\frac{91}{6} - 3.5^2 = \frac{35}{12} \approx 2.92.

For the fair die, the shortcut needs two averages taken with the same weights 16\frac{1}{6}: E[X2]E[X^2], the average of the squared faces 1,4,9,16,25,361, 4, 9, 16, 25, 36, which is 916\frac{91}{6}, and the mean E[X]=3.5E[X] = 3.5 found in the location section. The shortcut saves computing six separate deviations from the mean and squaring each one.

Standard Deviation


Standard deviation is simply Var(X)\sqrt{\text{Var}(X)}. It measures spread in the original units, making it more interpretable than variance.

A die with SD(X)≈1.71\text{SD}(X) \approx 1.71 means outcomes typically deviate about 1.7 units from the mean.

Range


The range is the span of the support: maximum value minus minimum value.

For a fair die: Range = 6−1=56 - 1 = 5

Range ignores probability—it treats rare and common outcomes equally. A single extreme value in the support sets the range, regardless of how unlikely it is.

What Spread Reveals


Two distributions can share the same mean but behave completely differently. One might cluster tightly around the center; another might scatter values widely. Spread measures capture this distinction. They tell you how predictable individual outcomes are.
The definition and the shortcut, side by side for the fair die.
DownloadPNG2×PDF
Variance of a fair die

Same variance, less work

The shortcut needs two averages instead of six separate deviations from the mean.
Before
Definition
∑(k−3.5)2⋅16=3512\displaystyle\sum (k - 3.5)^2 \cdot \dfrac{1}{6} = \dfrac{35}{12}
Six squared deviations: 6.25, 2.25, 0.25, 0.25, 2.25, 6.25
Use the shortcut
average the squares, subtract the squared mean
After
Shortcut
916−3.52=3512\dfrac{91}{6} - 3.5^2 = \dfrac{35}{12}
Two averages: 91/6 and 3.5
Subtracting the mean itself gives 916−3.5=353≈11.67\frac{91}{6} - 3.5 = \frac{35}{3} \approx 11.67, which is wrong. The shortcut subtracts the squared mean: 916−3.52=3512≈2.92\frac{91}{6} - 3.5^2 = \frac{35}{12} \approx 2.92.
What changed
Six squared deviations became two averages: the squared faces 1, 4, 9, 16, 25, 36 average to 91/6, and the mean is 3.5.
What stayed
The variance: 35/12 ≈ 2.92, so SD ≈ 1.71.
Variance of a fair die, two ways·/probability/probability-function/pmfLearn Math Class
Both routes give 35/12 ≈ 2.92; the shortcut needs only E[X²] and E[X].
Measure Category What it captures Formula from the PMF
Expected value (mean) Location center of mass — the probability-weighted average outcome E[X] = Σk k · pX(k)
Median Location value that splits the probability into halves smallest m with P(X ≤ m) ≥ 0.5
Mode Location the most likely value (highest PMF) argmaxk pX(k)
Variance Spread average squared deviation from the mean Var(X) = Σk (k − E[X])2 · pX(k)
Standard deviation Spread spread expressed in the same units as X √Var(X)
Range Spread span of the support (ignores probability) max(support) − min(support)

Transformations


If we take some random variable XX and apply function gg the result Y=g(X)Y=g(X) will be a new random variable.

The Transformation Process


Start with random variable XX with PMF pX(x)p_X(x).

Apply function gg to get Y=g(X)Y = g(X).

The new PMF pY(y)p_Y(y) is found by collecting all the probability mass from XX values that map to each yy value.

Transformations
pY(y)=∑x:g(x)=ypX(x)p_Y(y) = \sum_{x: g(x) = y} p_X(x)

Sum over all xx values that produce the same yy after transformation.

Linear Transformations


For Y=aX+bY = aX + b where a≠0a \neq 0:

Transformations
pY(y)=pX(y−ba)p_Y(y) = p_X\left(\frac{y - b}{a}\right)

The support shifts and scales accordingly.

Example: If XX is a fair die, then Y=2XY = 2X has support {2,4,6,8,10,12}\{2, 4, 6, 8, 10, 12\} with pY(k)=16p_Y(k) = \frac{1}{6} for each value.

Example: Y=X+3Y = X + 3 shifts the die to support {4,5,6,7,8,9}\{4, 5, 6, 7, 8, 9\} with probabilities unchanged.

Non-Injective Transformations


When multiple xx values map to the same yy, their probabilities combine.

Example
XX is a fair die, Y=(X−3.5)2Y = (X - 3.5)^2 (squared distance from mean).
Multiple values map to the same squared distance:
• pY(6.25)=pX(1)+pX(6)=16+16=13p_Y(6.25) = p_X(1) + p_X(6) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}
• pY(2.25)=pX(2)+pX(5)=13p_Y(2.25) = p_X(2) + p_X(5) = \frac{1}{3}
• pY(0.25)=pX(3)+pX(4)=13p_Y(0.25) = p_X(3) + p_X(4) = \frac{1}{3}

Effect on Expected Value


For any transformation gg:

LOTUS Discrete
E[g(X)]=∑xg(x)⋅pX(x)E[g(X)] = \sum_{x} g(x) \cdot p_X(x)
Learn more about this formula: LOTUS Discrete →


You don't need to find pY(y)p_Y(y) first. Work directly with the original PMF.

Linear transformations follow: E[aX+b]=aE[X]+bE[aX + b] = aE[X] + b

Pitfall: Applying the function to the mean
For Y=(X−3.5)2Y = (X - 3.5)^2, plugging in the mean gives (E[X]−3.5)2=0(E[X] - 3.5)^2 = 0, yet YY is never 00. Averaging g(x)g(x) over the PMF gives E[Y]=0.25+2.25+6.253=3512E[Y] = \frac{0.25 + 2.25 + 6.25}{3} = \frac{35}{12}.

The linear rule lets the mean pass straight through the transformation: scale and shift XX, and its mean is scaled and shifted in the same way. The squared distance Y=(X−3.5)2Y = (X - 3.5)^2 from the example above is not of this form. It bends the number line, folding faces from both sides of 3.53.5 onto the same value.

Key Insight


Transformations create new random variables, but the underlying probability structure stays connected to the original PMF. You're redistributing the same total probability mass, just arranging it differently.
Y = (X - 3.5) squared for a fair die: faces 3 and 4 give 0.25, faces 2 and 5 give 2.25, faces 1 and 6 give 6.25, each with probability 1/3yP(Y = y)1234567faces 3, 4faces 2, 5faces 1, 6two faces per value:1/6 + 1/6 = 1/3 eachSix faces, three values: the probabilities pool
A transformation that merges values

Y = (X − 3.5)² for a fair die. Faces 3 and 4 are both 0.5 from the mean, so both give Y = 0.25; faces 2 and 5 give 2.25; faces 1 and 6 give 6.25. Six equally likely faces become three values, each collecting two faces: p_Y(y) = 1/6 + 1/6 = 1/3.

E[Y] = (0.25 + 2.25 + 6.25)/3 = 35/12, the variance of the die.

Connection to CDF


The PMF and the CDF (Cumulative Distribution Function) are two ways to describe the same probability distribution. Each contains complete information about the random variable.

From PMF to CDF


The CDF accumulates probability from the PMF:

Connection to CDF
FX(x)=P(X≤x)=∑k≤xpX(k)F_X(x) = P(X \leq x) = \sum_{k \leq x} p_X(k)

The CDF at any point equals the sum of all PMF values up to and including that point.

Example
Fair die.
- FX(3)=pX(1)+pX(2)+pX(3)=16+16+16=12F_X(3) = p_X(1) + p_X(2) + p_X(3) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}
- FX(5.7)=pX(1)+⋯+pX(5)=56F_X(5.7) = p_X(1) + \cdots + p_X(5) = \frac{5}{6}

RULE
The CDF is a step function for discrete variables—it jumps at each value in the support, with jump height equal to the PMF at that point.

From CDF to PMF


Recover the PMF by taking differences in the CDF:

Connection to CDF
pX(k)=FX(k)−FX(k−)p_X(k) = F_X(k) - F_X(k^-)

where k−k^- represents the value just before kk in the support.

Example
For the fair die: pX(3)=FX(3)−FX(2)=36−26=16p_X(3) = F_X(3) - F_X(2) = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}

Why Both Exist


The PMF provides the probability of the random variable taking exactly a particular value and thus shows how probability distributes among all individual values of the domain.

The CDF answers the question: "What's the probability of at most this value?" and shows how probability accumulates across the domain.

Different questions favor different representations. The PMF is more direct for point probabilities. The CDF is better for cumulative and range probabilities. Both describe the same underlying distribution completely.
0.00.20.40.60.81.0012345678910Value (k)Cumulative Probability F(k)
Binomial CDF, n = 10, p = 0.5

Adding the PMF bars from the left produces a staircase: each riser at k has height P(X = k), so the PMF is recovered from the CDF by taking differences of consecutive steps, and the CDF from the PMF by running sums. The two descriptions carry the same information. Compare the risers with the bars on the discrete CDF explorer.

The common discrete families are catalogued next.

PMFs of Common Discrete Distributions


The standard discrete distributions each have distinct PMF shapes and behaviors.

Binomial Distribution


PMF: pX(k)=(nk)pk(1−p)n−kp_X(k) = \binom{n}{k} p^k (1-p)^{n-k} for k∈{0,1,…,n}k \in \{0, 1, \ldots, n\}

The PMF is symmetric when p=0.5p = 0.5, creating a bell-shaped pattern centered at n2\frac{n}{2}. For p<0.5p < 0.5, mass concentrates toward 0, creating right skew. For p>0.5p > 0.5, mass concentrates toward nn, creating left skew. The peak occurs near npnp, which is also the expected value. As nn increases, the PMF spreads wider but maintains its shape relative to pp. The binomial coefficient (nk)\binom{n}{k} creates the characteristic rise and fall pattern, with probabilities increasing until the mode and then decreasing. For extreme values of pp near 0 or 1, the distribution becomes highly skewed with most mass concentrated at one end.

Read More About Binomial Distribution

Geometric Distribution


PMF: pX(k)=(1−p)k−1pp_X(k) = (1-p)^{k-1} p for k∈{1,2,3,…}k \in \{1, 2, 3, \ldots\}

The PMF decreases monotonically—it never increases. Maximum mass always sits at k=1k=1, the first possible value. Each subsequent value has probability (1−p)(1-p) times the previous, creating exponential decay. For pp close to 1, the PMF drops sharply, concentrating almost all mass at small values. For pp close to 0, the PMF decays slowly, spreading mass across many values. The geometric PMF has the memoryless property: the shape of the tail remains proportional regardless of where you start. Unlike bounded distributions, the tail extends infinitely but thins out rapidly. The ratio between consecutive probabilities is constant: pX(k+1)pX(k)=(1−p)\frac{p_X(k+1)}{p_X(k)} = (1-p) always.

Read More About Geometric Distribution

Poisson Distribution


PMF: pX(k)=λke−λk!p_X(k) = \frac{\lambda^k e^{-\lambda}}{k!} for k∈{0,1,2,…}k \in \{0, 1, 2, \ldots\}

The PMF starts at pX(0)=e−λp_X(0) = e^{-\lambda}, rises to a peak near λ\lambda, then decays toward zero. For small λ\lambda (less than 1), the mode is at 0 and the PMF shows strong right skew. For moderate λ\lambda (around 5-10), the PMF becomes roughly symmetric and bell-shaped. For large λ\lambda (greater than 20), the PMF looks nearly normal. The rise-and-fall pattern is governed by the competition between λk\lambda^k (which grows) and k!k! (which grows faster). The peak shifts rightward as λ\lambda increases, and the variance increases in proportion, with the standard deviation growing like λ\sqrt{\lambda}—both mean and variance equal λ\lambda. The Poisson PMF never reaches zero for any finite kk, though it becomes negligibly small far from λ\lambda.

Read More About Poisson Distribution

Negative Binomial Distribution


PMF: pX(k)=(k−1r−1)pr(1−p)k−rp_X(k) = \binom{k-1}{r-1} p^r (1-p)^{k-r} for k∈{r,r+1,r+2,…}k \in \{r, r+1, r+2, \ldots\}

The PMF starts at k=rk=r (the minimum possible value). For r=1r = 1 it decreases from there; for r≥2r \geq 2 it first rises whenever p<1−1rp < 1 - \frac{1}{r}, peaks, and then declines. This is essentially a sum of rr independent geometric random variables, so its shape resembles a widened, shifted geometric distribution. As rr increases, the mode shifts rightward and the distribution spreads out. The parameter pp controls the decay rate: high pp produces fast decay and concentration near rr, while low pp produces slow decay and a long tail. When r=1r=1, this reduces exactly to the geometric distribution. The PMF has heavier tails than the Poisson, making it useful for overdispersed count data. The ratio test shows probabilities eventually decrease monotonically, though the initial behavior depends on both parameters.

Read More About Negative Binomial Distribution

Hypergeometric Distribution


PMF: pX(k)=(Kk)(N−Kn−k)(Nn)p_X(k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}

The PMF is unimodal with a peak near n⋅KNn \cdot \frac{K}{N}, the expected proportion of successes. The shape closely resembles the binomial distribution but is slightly more concentrated—sampling without replacement reduces variance. Unlike binomial, the hypergeometric support is finite and bounded by the population constraints: kk cannot exceed either KK (total successes available) or nn (sample size). As the population size NN grows relative to sample size nn, the hypergeometric PMF converges to binomial with p=KNp = \frac{K}{N}. For small populations or large samples, the distinction from binomial becomes significant. The PMF is symmetric when K=N2K = \frac{N}{2}, left-skewed when K>N2K > \frac{N}{2}, and right-skewed when K<N2K < \frac{N}{2}. The finite population creates a correction factor that shrinks variance compared to binomial sampling with replacement.

Read More About Hypergeometric Distribution

Discrete Uniform Distribution


PMF: pX(k)=1np_X(k) = \frac{1}{n} for k∈{a,a+1,…,b}k \in \{a, a+1, \ldots, b\} where n=b−a+1n = b - a + 1

The PMF is completely flat—constant probability across the entire support. No value is more likely than any other, so there is no single peak: every value in the support is a mode. This represents maximum uncertainty given only knowledge of the support boundaries. The PMF has zero skewness and is perfectly symmetric around the midpoint a+b2\frac{a+b}{2}. Adding or removing even one value from the support changes all probabilities—with nn values, each gets exactly 1n\frac{1}{n} of the total probability mass. The uniform PMF serves as a reference distribution: deviations from uniformity indicate structure or bias in the data. Despite its simplicity, the discrete uniform appears frequently in practice: fair dice, random selection, lottery draws. The PMF provides no information about which specific value will occur—all are equally plausible.

Read More About Discrete Uniform Distribution

PMF Patterns


Each PMF has characteristic behavior: where mass concentrates, how it spreads, whether it's symmetric or skewed, how parameters affect shape. These patterns reflect the underlying random process and determine the distribution's expected value and variance. Recognizing PMF shapes helps identify which distribution models your data.
Distribution PMF Support Shape & key behavior
Binomial(n, p) C(n, k) · pk(1−p)n−k {0, 1, …, n} bell-shaped at p = 0.5; peak near np; skewed when p ≠ 0.5
Geometric(p) (1−p)k−1 · p {1, 2, 3, …} monotonically decreasing; peak at k = 1; exponential decay
Poisson(λ) λk · e−λ / k! {0, 1, 2, …} rises to a peak near λ then decays; approximately normal for large λ
Negative Binomial(r, p) C(k−1, r−1) · pr(1−p)k−r {r, r+1, r+2, …} sum of r geometrics; heavier tail than Poisson; reduces to geometric for r = 1
Hypergeometric(N, K, n) C(K, k) · C(N−K, n−k) / C(N, n) bounded by N, K, n (max(0, n−N+K) to min(n, K)) unimodal; peak near n · K/N; more concentrated than binomial (finite population)
Discrete Uniform(a, b) 1 / (b − a + 1) {a, a+1, …, b} flat — every value equally likely; no peak

Interactive visualization of six fundamental discrete distributions

Discrete Uniform

Equal probability for finite outcomes

Explanation

A discrete uniform distribution assigns equal probability to each value in a finite range. The probability mass function is P(X=k)=1b−a+1P(X = k) = \frac{1}{b - a + 1} for a≤k≤ba \leq k \leq b. The expected value is E[X]=a+b2E[X] = \frac{a + b}{2}, and the variance is Var(X)=n2−112\text{Var}(X) = \frac{n^2 - 1}{12}, where n=b−a+1n = b - a + 1. Common examples include rolling a fair die, selecting a random card from a deck, or generating a random number from a finite range.

0.0000.0400.0800.1200.1600.200123456Value (k)Probability P(X = k)E[X]=3.50
Discrete uniform PMF with its mean marked

The catalogue of discrete mass functions ranges from the flat uniform pattern shown here through the peaked binomial and Poisson shapes to the decaying geometric tail; each is a different rule for splitting unit mass across integers. The explorer marks the mean of each. Compare the six families side by side on the distribution explorer.

Common mistakes with these functions are collected next.

Common Mistakes


    Confusing PMF with PDF


    The PMF applies only to discrete random variables. For continuous variables, you need a PDF (Probability Density Function). The PMF gives exact point probabilities: P(X=k)P(X = k). The PDF does not—for continuous variables, P(X=x)=0P(X = x) = 0 for any specific point.

    Don't apply PMF formulas to continuous data, and don't try to use PDF methods on discrete outcomes.

    Forgetting to Check the Axioms


    Not every function is a valid PMF. Before using any assignment as a PMF, verify:

  • Non-negativity: pX(k)≥0p_X(k) \geq 0 for all kk
  • Normalization: ∑kpX(k)=1\sum_k p_X(k) = 1

    • If either fails, you don't have a valid probability model. See axioms.

    Summing Over the Wrong Set


    When computing P(X∈A)P(X \in A), sum only over values in both AA and the support. Don't include values outside the support—they contribute zero anyway. Don't skip values inside the support that belong to AA.

    Example: For a die, P(X>4)=pX(5)+pX(6)P(X > 4) = p_X(5) + p_X(6), not pX(4)+pX(5)+pX(6)p_X(4) + p_X(5) + p_X(6).

    Treating Probabilities as Fixed


    The PMF depends on parameters. Change the parameters, and the entire PMF changes. A binomial PMF with p=0.3p = 0.3 looks completely different from one with p=0.7p = 0.7, even with the same nn.

    Don't assume a distribution has one "standard" PMF—parameters matter.

    Confusing Support with Domain


    The support is where pX(k)>0p_X(k) > 0. Outside the support, pX(k)=0p_X(k) = 0. The PMF is technically defined everywhere, but only the support matters for calculations.

    Don't waste time summing over values where the PMF is zero.

    Ignoring Discrete Structure


    Discrete random variables take isolated values, not continuous ranges. P(2<X<3)P(2 < X < 3) might be zero if nothing sits between 2 and 3. Don't treat discrete variables as if they vary smoothly.

    Always check whether your variable is discrete or continuous before choosing methods.

    Using the wrong complement

    The probability of an event is found as one minus the probability of the opposite event, but the opposite event is written with the wrong inequality, so a boundary value falls into neither event.

    Wrong working: for a fair die,
    P(X>3)=1−P(X<3)=1−26=23P(X > 3) = 1 - P(X < 3) = 1 - \frac{2}{6} = \frac{2}{3}


    Right working: the complement of X>3X > 3 is X≤3X \leq 3, the faces 1,2,31, 2, 3.
    P(X>3)=1−P(X≤3)=1−36=12P(X > 3) = 1 - P(X \leq 3) = 1 - \frac{3}{6} = \frac{1}{2}


    Why it happens: the opposite of "greater than" is read as "less than", and for continuous variables that slip costs nothing, since a single point has probability zero. For a discrete variable the boundary value carries its own mass.

    How to check: list the values of the event and of its complement. Together they must cover the support exactly once: here {4,5,6}\{4, 5, 6\} and {1,2,3}\{1, 2, 3\}. With X<3X < 3 the face 33 is in neither list, and the wrong answer exceeds the direct count pX(4)+pX(5)+pX(6)=12p_X(4) + p_X(5) + p_X(6) = \frac{1}{2} by exactly pX(3)=16p_X(3) = \frac{1}{6}.

    Averaging the values without weights

    The expected value is taken as the plain average of the values in the support, as if each were equally likely, and the PMF is not used.

    Wrong working: for the loaded die with pX(6)=12p_X(6) = \frac{1}{2} and pX(k)=110p_X(k) = \frac{1}{10} for k∈{1,2,3,4,5}k \in \{1, 2, 3, 4, 5\},
    E[X]=1+2+3+4+5+66=3.5E[X] = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5


    Right working: each value is multiplied by its probability.
    E[X]=110(1+2+3+4+5)+12⋅6=1.5+3=4.5E[X] = \frac{1}{10}(1 + 2 + 3 + 4 + 5) + \frac{1}{2} \cdot 6 = 1.5 + 3 = 4.5


    Why it happens: for a fair die the plain average and the weighted average agree, and the fair die is usually the first example met.

    How to check: compare the answer with where the mass sits. Half the probability is on the face 66, so the mean must lie well above the midpoint 3.53.5 of the faces.

    Subtracting the mean instead of its square

    In the shortcut Var(X)=E[X2]−(E[X])2\text{Var}(X) = E[X^2] - (E[X])^2, the mean is subtracted as it is, without being squared.

    Wrong working: for a fair die, E[X2]=1+4+9+16+25+366=916E[X^2] = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} and E[X]=3.5E[X] = 3.5, so
    Var(X)=916−3.5=353≈11.67\text{Var}(X) = \frac{91}{6} - 3.5 = \frac{35}{3} \approx 11.67


    Right working: the mean is squared before it is subtracted.
    Var(X)=916−3.52=18212−14712=3512≈2.92\text{Var}(X) = \frac{91}{6} - 3.5^2 = \frac{182}{12} - \frac{147}{12} = \frac{35}{12} \approx 2.92


    Why it happens: the formula holds two quantities that look alike, E[X2]E[X^2] and (E[X])2(E[X])^2, and after the work of computing E[X2]E[X^2] the second square is easy to drop.

    How to check: take the square root. The wrong variance gives a standard deviation of about 3.423.42, more than the largest distance of any face from the mean, 6−3.5=2.56 - 3.5 = 2.5. The right one gives 35/12≈1.71\sqrt{35/12} \approx 1.71, the value quoted in the standard deviation section.

    Applying the function to the mean

    The expected value of g(X)g(X) is needed, and gg is applied to E[X]E[X] instead of to each value of XX before averaging.

    Wrong working: for a fair die and Y=(X−3.5)2Y = (X - 3.5)^2,
    E[Y]=(E[X]−3.5)2=(3.5−3.5)2=0E[Y] = (E[X] - 3.5)^2 = (3.5 - 3.5)^2 = 0


    Right working: by LOTUS, each value g(x)g(x) is weighted by pX(x)=16p_X(x) = \frac{1}{6}.
    E[Y]=6.25+2.25+0.25+0.25+2.25+6.256=17.56=3512E[Y] = \frac{6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25}{6} = \frac{17.5}{6} = \frac{35}{12}


    Why it happens: for a linear function the shortcut is correct, E[aX+b]=aE[X]+bE[aX + b] = aE[X] + b, and it is easy to carry it over to functions that are not linear.

    How to check: look at the values YY can take. YY is always 0.250.25, 2.252.25 or 6.256.25, and an expected value lies between the smallest and the largest value of the variable, so E[Y]E[Y] cannot be 00.
Mistake Wrong thinking Correct view
Confusing PMF with PDF applying PMF formulas to continuous data the PMF is discrete-only; for continuous variables use a PDF, and P(X = x) = 0 for any single point
Skipping axiom verification assuming any function is a valid PMF check pX(k) ≥ 0 everywhere and Σ pX(k) = 1 before using
Summing over the wrong set including non-support values, or skipping support values that belong to A sum only over values in both A and the support — outside the support contributes zero
Treating probabilities as fixed assuming a distribution has one “standard” PMF parameters change the PMF entirely — binomial with p = 0.3 looks nothing like p = 0.7
Confusing support with domain summing over the whole real line, or over values where pX is zero only the support matters; outside it the PMF is zero and contributes nothing
Ignoring discrete structure treating discrete variables as smooth or continuous discrete values are isolated; P(2 < X < 3) may be exactly 0 if no value sits between 2 and 3

PMF at a Glance

The page has unpacked the probability mass function from many angles — from its defining axioms and three modes of representation through summary measures, transformations, and the family of common discrete distributions. The table below collects the essentials into a single reference card, pairing each aspect of the PMF with its concise statement and a concrete formula or example.
Aspect Statement Example or formula
What a PMF is a function pX(x) = P(X = x) for a discrete random variable fair die: pX(k) = 1/6 for k ∈ {1, …, 6}
Support the set of values where pX(x) > 0 finite ({1, …, 6}) or infinite countable ({1, 2, 3, …})
Two axioms non-negative everywhere; total over the support equals 1 pX(k) ≥ 0 for all k; Σ pX(k) = 1
Three representations graphical, functional, tabular — the same PMF viewed three ways bars at each value; a formula; a value-probability table
Computing event probabilities sum the PMF over the values that make up the event P(X ≤ 3) for a die = pX(1) + pX(2) + pX(3) = 1/2
Summary measures location (mean, median, mode) and spread (variance, SD, range) — all computed from the PMF E[X] = Σ k · pX(k); Var(X) = Σ (k − E[X])2 · pX(k)
Connection to the CDF accumulate the PMF to get the CDF; differ to recover the PMF FX(x) = Σk ≤ x pX(k); pX(k) = FX(k) − FX(k−)
Common forms named discrete distributions, each with a characteristic PMF shape binomial, geometric, Poisson, negative binomial, hypergeometric, discrete uniform

PMF FAQ

Does P(X > 4) include 4?

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No — a strict inequality excludes the endpoint. For a die, P(X > 4) is p(5) + p(6), not p(4) + p(5) + p(6). This matters far more for discrete variables than continuous ones: dropping or adding a single value changes the answer by that value's whole probability, whereas for a continuous variable an endpoint contributes nothing.Read more →

Why can P(2 < X < 3) be zero for a discrete variable?

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Because a discrete variable takes isolated values, and there may be nothing between 2 and 3 for it to land on. An interval containing no point of the support carries no probability at all. Continuous variables never behave this way — any interval of positive length inside their support carries positive probability.Read more →

What is the difference between the support and the domain of a PMF?

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The PMF is technically defined everywhere — it returns 0 outside the values the variable can take. The support is the smaller set where p_X(k) > 0. Only the support matters for calculations, since summing over anything else adds zeros. The two words are often used loosely as synonyms, but the distinction saves wasted work on infinite domains.Read more →

When do you need the subscript in p_X(x)?

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Only when more than one variable is in play. The subscript names which variable the function belongs to, so p_X(x) and p_Y(y) stay distinct on the same page. With a single variable in context you can drop it and write p(x). It becomes mandatory once joint and marginal functions appear together.Read more →

Can different dice give the same distribution of sums as ordinary dice?

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Yes, and exactly one other pair with positive whole-number faces does: the Sicherman dice, numbered 1,2,2,3,3,41, 2, 2, 3, 3, 4 and 1,3,4,5,6,81, 3, 4, 5, 6, 8. Their 3636 face pairs hit each sum from 22 to 1212 as often as ordinary dice do, so the PMF table of the sum is identical. The table keeps only the sums and their chances, so it cannot tell the two pairs apart.Read more →

Does every PMF have an expected value?

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No. The masses pX(k)=1k(k+1)p_X(k) = \frac{1}{k(k+1)} for k=1,2,3,…k = 1, 2, 3, \ldots form a valid PMF: each equals 1k−1k+1\frac{1}{k} - \frac{1}{k+1}, so the sum telescopes to 11. But E[X]=12+13+14+⋯E[X] = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots is the harmonic series, which grows without bound. The median and the mode still exist, because they come from comparing masses rather than from a weighted sum.Read more →