Chapter 3: Random Variables and Discrete Distributions  
MATH/STAT 394: Probability I  
Arman Jahangiri  
University of Washington Department of Mathematics  
Summer 2026  
Arman Jahangiri  
Chapter 3: Random Variables and Discrete Distributions  
Summer 2026  
1
 
Learning Goals  
By the end of this lecture, students should be able to:  
▶ define a random variable;  
▶ compute and interpret PMFs and CDFs;  
▶ recognize and use common discrete distributions;  
▶ work with functions of random variables;  
Arman Jahangiri  
Chapter 3: Random Variables and Discrete Distributions  
Summer 2026  
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Random Variables  
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Chapter 3: Random Variables and Discrete Distributions  
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Why Random Variables?  
In many experiments, the sample space itself is complicated.  
Examples:  
▶ sequences of coin tosses;  
▶ poker hands;  
▶ paths of a random walk;  
▶ survey responses.  
Usually we are not interested in the full outcome itself.  
Instead, we care about a numerical summary:  
▶ number of Heads (i.e., 3 instead of HHTHTTT);  
▶ total score;  
▶ duration of a game;  
▶ number of successes.  
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Chapter 3: Random Variables and Discrete Distributions  
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Definition of a Random Variable  
Definition  
Let (S, F, P) be a probability space. A random variable (r.v.) is a (measurable) function  
X : S → R.  
Measurability means that for every (Borel) set B ∈ B,  
X
−1(B) = { s ∈ S : X(s) ∈ B } ∈ F.  
Important  
The randomness comes from the outcome  
s ∈ S.  
The function X itself is deterministic; it simply converts outcomes into numerical values.  
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Chapter 3: Random Variables and Discrete Distributions  
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5
Coin Toss Example  
Flip a fair coin twice.  
S = {HH, HT, TH, TT}.  
Define:  
Then,  
X = number of Heads.  
X(HH) = 2,  
X(HT) = 1,  
X(TH) = 1,  
X(TT) = 0.  
Thus,  
X : S → {0, 1, 2}.  
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Chapter 3: Random Variables and Discrete Distributions  
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Indicator Random Variables  
Indicator Random Variable  
For an event A, define  
(
1, if A occurs,  
IA  
=
0, otherwise.  
Indicator random variables are extremely useful.  
Example.  
I
.
{first toss is Heads}  
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Chapter 3: Random Variables and Discrete Distributions  
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Discrete Random Variables  
Definition  
A random variable is discrete if it takes values in a finite or countably infinite set.  
Examples:  
▶ number of Heads in 10 tosses;  
▶ number of emails received today;  
▶ number of customers entering a store.  
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Chapter 3: Random Variables and Discrete Distributions  
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PMFs and Distributions  
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Chapter 3: Random Variables and Discrete Distributions  
Summer 2026  
9
Probability Mass Functions (PMFs)  
Definition  
The probability mass function (PMF) of a discrete r.v. X is  
pX (x) = P(X = x).  
The PMF assigns probability mass to each possible value of the random variable.  
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Chapter 3: Random Variables and Discrete Distributions  
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PMF Example: Number of Heads  
Suppose:  
X = number of Heads in two fair tosses.  
Then,  
1
4
1
2
1
4
pX (0) =  
,
pX (1) =  
,
pX (2) =  
.
All other values have probability 0. We will see in the future that this is called a Binomial random variable with  
parameters n = 2 and p = 0.5.  
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Chapter 3: Random Variables and Discrete Distributions  
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Properties of a Valid PMF  
A valid PMF must satisfy:  
1
Nonnegativity:  
pX (x) ≥ 0.  
2
Total probability equals 1:  
X
pX (x) = 1.  
x
Interpretation  
A PMF distributes total probability mass 1 across the support of the random variable.  
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Chapter 3: Random Variables and Discrete Distributions  
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Example: Sum of Two Dice  
Roll two fair dice. Define T :  
Define  
T = sum of the two die rolls  
The possible values of T:  
Support(T) = {2, 3, . . . , 12}.  
and the PMF,  
t
2
3
4
5
6
7
8
9
10 11 12  
1
36  
2
36  
3
36  
4
36  
5
36  
6
36  
5
36  
4
36  
3
36  
2
36  
1
36  
pT (t) = P(T = t)  
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Chapter 3: Random Variables and Discrete Distributions  
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Named Distributions  
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Chapter 3: Random Variables and Discrete Distributions  
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Bernoulli Distribution  
A Bernoulli random variable models the outcome of a binary experiment: success ↔ 1,  
failure ↔ 0.  
Bernoulli PMF  
A random variable X has the Bernoulli distribution with parameter 0 ≤ p ≤ 1, written X ∼ Bern(p), if its PMF is  
given by  
p,  
x = 1,  
pX (x) =  
1 − p, x = 0,  
0,  
otherwise,  
Interpretation: P(success) = P(X = 1) = p,  
P(failure) = P(X = 0) = 1 − p. PMF verification: Since 0 ≤ p ≤ 1,, we  
have pX (x) ≥ 0. Also,  
X
pX (x) = p + (1 − p) = 1.  
x
Hence pX (x) satisfies the properties of a PMF.  
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Chapter 3: Random Variables and Discrete Distributions  
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Bernoulli Distribution: PMF  
X ∼ Bern(p),  
P(X = 1) = p,  
P(X = 0) = 1 − p.  
A Bernoulli random variable models a single success/failure  
experiment.  
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Chapter 3: Random Variables and Discrete Distributions  
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Binomial Distribution  
Binomial Distribution  
Suppose an experiment consists of  
▶ n independent Bernoulli trials;  
▶ where each trial share the same success probability p.  
If X is the number of successes, then  
X ∼ Bin(n, p).  
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Chapter 3: Random Variables and Discrete Distributions  
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Binomial PMF  
If  
X ∼ Bin(n, p),  
then  
ꢀ ꢁ  
n
pX (k) = P(X = k) =  
pk (1 − p)n−k  
,
k
for  
k = 0, 1, . . . , n.  
Interpretation:  
ꢂ ꢃ  
n
k
▶
: choose which k trials are successes;  
▶ pk : probability of the k successes;  
▶ (1 − p)n−k : probability of the n − k failures.  
PMF verification: Since 0 ≤ p ≤ 1, we have pX (k) ≥ 0. Also,  
n
n
ꢀ ꢁ  
X
X
n
pX (k) =  
pk (1 − p)n−k = (p + (1 − p))n = 1,  
k
k=0  
k=0  
by the Binomial Theorem.  
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Chapter 3: Random Variables and Discrete Distributions  
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Binomial Distribution: PMF  
X ∼ Bin(n, p),  
  !  
n
P(X = k) =  
pk (1 − p)n−k  
.
k
Binomial distribution counts the number of successes in n independent Bernoulli trials.  
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Chapter 3: Random Variables and Discrete Distributions  
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Example: Coin Tosses  
Flip a fair coin 10 times.  
Let:  
X = number of Heads.  
Then,  
X ∼ Bin(10, 12 ).  
For example:  
 
!
ꢀ ꢁ  
10  
10  
3
1
P(X = 3) =  
.
2
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Chapter 3: Random Variables and Discrete Distributions  
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Geometric Distribution  
Geometric Distribution  
Suppose independent Bernoulli trials are repeated until the first success occurs.  
Let  
X = number of trials until the first success.  
If each trial share the same success probability p, then  
X ∼ Geom(p).  
Possible values:  
1, 2, 3, . . . (i.e., all positive integers)  
Interpretation:  
▶ X = 1 means success on the first trial;  
▶ X = 2 means failure then success;  
▶ X = 3 means two failures then success;  
▶ etc.  
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Chapter 3: Random Variables and Discrete Distributions  
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Geometric PMF  
If  
X ∼ Geom(p),  
P(X = k) = (1 − p)k−1p,  
k = 1, 2, 3, . . .  
then  
for  
Interpretation:  
▶ the first k − 1 trials must fail;  
▶ the k-th trial must succeed.  
Hence,  
P(X = k) = (1 − p)k−1  
failures  
p
.
|{z}  
|
{z  
}
success  
PMF verification: Since 0 ≤ p ≤ 1, we have pX (x) ≥ 0. Also,  
∞
∞
∞
X
X
X
1
1
pX (k) =  
(1 − p)k−1p = p  
(1 − p)k = p ×  
= p × = 1,  
1 − (1 − p)  
p
k=1  
k=1  
k=0  
using the geometric sequence formula.  
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Chapter 3: Random Variables and Discrete Distributions  
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Geometric Distribution: PMF  
X ∼ Geom(p),  
pX (x) = P(X = k) = (1 − p)k−1p,  
k = 1, 2, 3, . . .  
The Geometric distribution models the waiting time until the first success.  
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Chapter 3: Random Variables and Discrete Distributions  
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Memoryless Property of Geom(p)  
The Geometric distribution is the only discrete distribution that satisfies the memoryless property. We will see  
the continuous counterpart (exponential distribution) in Chapter 5b.  
P(X > m + n | X > m) = P(X > n).  
Interpretation:  
If the first m trials were all failures, the process probabilistically restarts from scratch.  
Example:  
P(X > 10 | X > 7) = P(X > 3).  
The past failures do not affect future waiting times.  
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Chapter 3: Random Variables and Discrete Distributions  
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Negative Binomial Distribution  
Negative Binomial Distribution  
Suppose independent Bernoulli trials are repeated until the rth success occurs.  
Let  
X = number of trials needed to obtain r successes.  
If each trial has success probability p, then  
X ∼ NegBin(r, p).  
Possible values:  
r, r + 1, r + 2, . . .  
Interpretation:  
▶ the experiment stops once the rth success appears;  
▶ Geometric distribution is the special case  
r = 1.  
Arman Jahangiri  
Chapter 3: Random Variables and Discrete Distributions  
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Negative Binomial PMF  
If  
X ∼ NegBin(r, p),  
then  
ꢀ
ꢁ
k − 1  
pX (k) = P(X = k) =  
pr (1 − p)k−r  
,
r − 1  
for  
k = r, r + 1, r + 2, . . .  
Interpretation:  
▶ among the first k − 1 trials, there must be exactly r − 1 successes;  
▶ the k-th trial must be a success.  
Thus,  
ꢀ
|
ꢁ
}
k − 1  
P(X = k) =  
pr  
(1 − p)k−r  
.
|{z}  
|
{z  
}
r − 1  
{z  
r successes  
k−r failures  
arrange first r−1 successes  
PMF verification: Since 0 ≤ p ≤ 1, we have pX (x) ≥ 0.  
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Chapter 3: Random Variables and Discrete Distributions  
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PMF verification (conitnue)  
 
!
∞
∞
X
X
k − 1  
pX (k) =  
pr (1 − p)k−r  
.
r − 1  
k=r  
k=r  
Let  
j = k − r.  
Then k = j + r, so  
 
!
∞
∞
X
X
j + r − 1  
pX (k) = pr  
(1 − p)j .  
r − 1  
k=r  
j=0  
Using the negative binomial series  
identity,  
 
!
∞
X
j + r − 1  
1
xj =  
,
|x| < 1,  
r − 1  
(1 − x)r  
j=0  
with x = 1 − p, we get  
∞
X
1
1
pr  
pX (k) = pr  
= pr  
= 1.  
(1 − (1 − p))r  
k=r  
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Chapter 3: Random Variables and Discrete Distributions  
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Understanding the Formula  
To have  
X = k,  
the following must happen:  
1
In the first k − 1 trials, r − 1 successes must occur.  
2
The final trial must be a success.  
Number of ways:  
 
!
k − 1  
.
r − 1  
Probability of each arrangement:  
Multiplying gives:  
p r−1(1 − p)k−r · p = pr (1 − p)k−r  
.
 
!
k − 1  
P(X = k) =  
pr (1 − p)k−r  
.
r − 1  
Note that both the formula and the definition implies that the Negative Binomial distribution is a generalization of  
the Geometric distribution (when r = 1, Negative Binomial reduces to Geometric).  
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Chapter 3: Random Variables and Discrete Distributions  
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Example: Third Heads  
Flip a fair coin repeatedly.  
Let  
X = number of tosses needed to obtain 3 Heads.  
Then,  
X ∼ NegBin(3, 12 ).  
For example,  
  !  
ꢀ ꢁ ꢀ ꢁ  
3
2
4
1
2
1
2
P(X = 5) =  
.
2
Interpretation:  
▶ among the first 4 tosses, exactly 2 must be Heads;  
▶ the 5th toss must also be Heads.  
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Chapter 3: Random Variables and Discrete Distributions  
Summer 2026 29  
Geometric vs. Negative Binomial  
Distribution  
Stops After  
Support  
Geometric  
first success  
1, 2, 3, . . .  
Negative Binomial  
rth success  
r, r + 1, . . .  
Key Relationship  
The Geometric distribution is a special case of the Negative Binomial distribution:  
Geom(p) = NegBin(1, p).  
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Chapter 3: Random Variables and Discrete Distributions  
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Hypergeometric Distribution  
Hypergeometric Story  
An urn contains:  
▶ w white balls,  
▶ b black balls.  
Sample n balls without replacement.  
If X is the number of white balls selected, then  
X ∼ HGeom(w, b, n).  
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Chapter 3: Random Variables and Discrete Distributions  
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Hypergeometric PMF  
If  
X ∼ HGeom(w, b, n),  
then  
ꢂ ꢃꢂ  
ꢃ
w
k
b
n−k  
P(X = k) =  
.
ꢂ
ꢃ
w+b  
n
The possible values are  
max(0, n − b) ≤ k ≤ min(w, n).  
Interpretation:  
▶ choose k white balls from the w white balls;  
▶ choose n − k black balls from the b black balls;  
▶ divide by the totral number of samples of size n.  
Thus,  
ꢀ
ꢁ
ꢀ
|
}
ꢁ
}
w
k
b
n − k  
| {z }  
choose white balls  
ꢀ
{z  
choose black balls  
P(X = k) =  
.
ꢁ
w + b  
n
|
{z  
all samples of size n  
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Chapter 3: Random Variables and Discrete Distributions  
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Hypergeometric Distribution: PMF  
X ∼ HGeom(w, b, n),  
ꢂ ꢃꢂ  
ꢃ
w
k
b
n−k  
P(X = k) =  
ꢂ
ꢃ
.
w+b  
n
The Hypergeometric distribution arises from  
sampling without replacement.  
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Chapter 3: Random Variables and Discrete Distributions  
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Binomial vs. Hypergeometric  
Distribution  
Binomial  
Replacement? Independent?  
Yes  
No  
Yes  
No  
Hypergeometric  
Key Difference  
Hypergeometric sampling is without replacement.  
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Chapter 3: Random Variables and Discrete Distributions  
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Discrete Uniform Distribution  
Discrete Uniform  
Choose one value uniformly from a finite set C.  
Then,  
X ∼ DUnif(C).  
All values are equally likely.  
Example:  
X ∼ DUnif({1, 2, . . . , 6}).  
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Chapter 3: Random Variables and Discrete Distributions  
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Poisson Distribution  
Poisson Distribution  
A random variable X has the Poisson distribution with parameter λ > 0 if  
−λ λk  
P(X = k) = e  
,
k = 0, 1, 2, . . . .  
k!  
Notation:  
X ∼ Pois(λ).  
Poisson random variables often model counts of rare events.  
PMF verification: Since  
λ > 0,  
we have  
Also,  
P(X = k) ≥ 0.  
∞
∞
∞
−λ λk  
λ
k
X
X
X
pX (k) =  
e
= e−λ  
.
k!  
k!  
k=0  
k=0  
k=0  
Using the exponential series,  
∞
k
X
λ
= eλ,  
k!  
k=0  
so  
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∞
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Chapter 3:
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Summer 2026 36  
Poisson Distribution: PMF  
X ∼ Pois(λ),  
−λ λk  
P(X = k) = e  
,
k = 0, 1, 2, . . .  
k!  
The Poisson distribution models the number  
of events occurring in a fixed interval of time  
or space.  
Typical examples include:  
▶ phone calls arriving,  
▶ radioactive decays,  
▶ website visits,  
▶ accidents in a day.  
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Chapter 3: Random Variables and Discrete Distributions  
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CDFs  
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Chapter 3: Random Variables and Discrete Distributions  
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Cumulative Distribution Functions  
Definition  
The cumulative distribution function (CDF) of X is  
FX (x) = P(X ≤ x).  
▶ The CDF accumulates all the probability mass that exists from −∞ to (inclusive) x.  
▶ In other words, FX (x) = P(X ∈ (−∞, x]) = P(−∞ < X ≤ x) are three different ways of introducing the same  
thing, the CDF of X at point x.  
▶ Note that X ≤ x (or X ∈ (−∞, x], or −∞ < X ≤ x), actually refer to an EVENT:  
{X ≤ x} = {s ∈ S : X(s) ≤ x} = X−1((−∞, x]))  
▶ In general, {X ∈ A} = {s ∈ S : X(s) ∈ A} = X−1(A)  
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Chapter 3: Random Variables and Discrete Distributions  
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PMF vs. CDF  
For discrete random variables:  
▶ PMF:  
pX (x) = P(X = x)  
gives probability at one point;  
▶ CDF:  
FX (x) = P(X ≤ x)  
gives cumulative probability.  
The CDF is increasing and ranges from 0 to 1.  
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Chapter 3: Random Variables and Discrete Distributions  
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Properties of a CDF  
Key properties  
Every CDF satisfies:  
▶ 0 ≤ F(x) ≤ 1,  
▶ F(x) is increasing,  
▶
x→lim−∞ F(x) = 0,  
▶
lim F(x) = 1.  
x→∞  
Interactive visualization of PMF/PDF and CDF accumulation.  
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Chapter 3: Random Variables and Discrete Distributions  
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Functions of Random Variables  
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Chapter 3: Random Variables and Discrete Distributions  
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Functions of Random Variables  
If X is a random variable and  
g : R → R,  
then  
Y = g(X)  
is also a random variable.  
Examples:  
√
Y = X2,  
Y = X,  
Y = |X|,  
Y = 2X + 1.  
Question: How to find FY (y)? We will see later in Transformations of r.v.s.  
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Chapter 3: Random Variables and Discrete Distributions  
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Several Random Variables on One Space  
Different random variables may be defined on the same experiment.  
Example:  
▶ X = first die ∈ { 1, 2, 3, 4, 5, 6 };  
▶ Y = second die ∈ { 1, 2, 3, 4, 5, 6 };  
▶ X + Y = total ∈ { 1, 2, . . . , 11, 12 };  
▶ max(X, Y ) = maximum ∈ { 1, 2, 3, 4, 5, 6 };  
▶ |X − Y | = difference ∈ { 0, 1, 2, 3, 4, 5 };  
Each gives a different numerical summary of the same experiment.  
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Chapter 3: Random Variables and Discrete Distributions  
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Summary  
▶ A random variable is a function from the sample space to R.  
▶ Discrete random variables are described by PMFs.  
▶ Important distributions:  
Bernoulli, Binomial, Hypergeometric, Discrete Uniform.  
▶ CDFs accumulate probability:  
FX (x) = P(X ≤ x).  
▶ Functions of random variables are themselves random variables.  
▶ Random variables help convert complicated experiments into manageable numerical objects.  
Next lecture  
Expectation  
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Chapter 3: Random Variables and Discrete Distributions  
Summer 2026 45