Business Statistics
Discrete Random Variables
David Lander
Peking University
HSBC Business School
UK Campus
Module 1, 2026–27
Roadmap for “probability”
1. Fundamentals
2. Discrete random variables
3. Continuous random variables
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Last class
• Independence
◦ Dependence (does not imply causation)
• Conditional probability
◦ Probabilities given that another event occurs
◦ Reduced sample space
• Bayes’ theorem
◦ Prior and posterior probabilities
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This class
• Introduce discrete random variables
◦ Define a discrete random variable
◦ Determine and plot the distribution of a random variable
◦ Become familiar with the binomial distribution
• Determine the expected value of a random variable
◦ Compute mean, variance, and standard deviation
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Discrete random variables
Random variables
• A random variable assigns a numerical value to each possible
outcome of a random process (or experiment)
• We usually denote these using capital letters, e.g., X , Y , Z
• And the values that these variables may take, we denote by the
corresponding lower case letters
◦ e.g., x (for X ), y (for Y ), z (for Z )
• Examples:
◦ X = 1 if a coin lands heads, and X = 0 if it lands tails
◦ The proportion of heads in a fixed number n of coin tosses
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Random variables: example
• Consider an experiment of tossing two coins
• Suppose we are interested in the number of heads that occur
• We can define a (random) variable to study this specific event
◦ Let X represent the number of heads that occurs
▷ X can only take on the values 0, 1, 2
▷ We write X ∈ {0, 1, 2} to describe its possible values
◦ Therefore, we are interested in the events X = 0, X = 1, and X = 2
• Since the outcome (i.e., number of heads) entails uncertainty, the
value of X is uncertain; hence, we call X a random variable
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Discrete Random Variables
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Discrete random variables
• A random variable is called a discrete random variable if it
assumes only a finite number of values
◦ Or if its values can be placed in one-to-one correspondence with
the positive integers
• We will study continuous random variables later
• Examples:
1. Suppose a die is rolled and X is the number that turns up. Then X
is a random variable, and X ∈ {1, 2, 3, 4, 5, 6}
2. Suppose a coin is successively tossed until a head appears. If Y is
the number of such tosses, then Y is a random variable and
Y ∈ {1, 2, 3, . . .}
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Discrete Random Variables
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Distribution of a discrete random variable
• If X is a random variable, the probability of the event that X
assumes the value x is denoted P(X = x)
◦ Similarly, we can consider the probabilities of events, such as X ≤ x
or X > x
• If X is discrete, then the function f (·) that assigns the number
P(X = x) to each possible value of X is called the probability
mass function (or the pmf) of the random variable X :
f (x) = P(X = x)
• We can describe this by saying that “f (x) is the probability that X
takes the value x”
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Discrete Random Variables
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Distribution of a discrete random variable
• Suppose X counts the number of heads when two fair coins are
tossed independently; determine the distribution of X
• Goal: find the probabilities of the events X = 0, X = 1, and X = 2
• Our usual equiprobable sample space is S = {HH, HT , TH, TT }
• Hence, we have the following events:
◦ The event X = 0 is {TT }, with probability 14
◦ The event X = 1 is {HT , TH}, with probability 12
◦ The event X = 2 is {HH}, with probability 41
• Hence, if f (·) is the distribution of X , i.e., f (x) = P(X = x), then
f (0) =
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1
4
f (1) =
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1
2
f (2) =
1
4
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Distribution of a discrete random variable
• We can plot the distribution of X :
Figure: Graph of the distribution of X
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Distribution of a discrete random variable
• We can plot the histogram of X :
Figure: Probability histogram of X
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Distribution of a discrete random variable
• Note, in this example, if we sum the probabilities:
2
X
f (x) = f (0) + f (1) + f (2) =
x=0
1 1 1
+ + =1
4 2 4
• This property holds for every probability mass function
• For any probability mass function f (·), we have 0 ≤ f (x) ≤ 1 for all
P
x, and x f (x) = 1
• This also means that the sum of the areas of the rectangles in a
probability histogram equals 1
◦ This is important, and it will appear again when we consider
continuous random variables (later)
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Discrete Random Variables
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Expectations
• The distribution for a random variable X gives the relative
frequencies of the values of X in the long run
• It is often very useful to determine the “average” value of X in the
long run
• We call this the expected value of X (or taking expectations of X ):
X
E(X ) =
xf (x)
x
• Note that this is the mean of X , which we often refer to by the
Greek letter µ (“mu”)
◦ We write µ = E(X ) for the mean of X
◦ This is interpreted as the average value of X in the long run
• We consider discrete random variables whose means and
variances are finite
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Discrete Random Variables
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Expectations
• If X takes the n distinct values x1 , x2 , . . . , xn with equal probability,
f (xi ) = n1 for i = 1, 2, . . . , n, then:
µ=
X
xf (x) =
x
n
X
i=1
n
xi
1
1X
=
xi
n
n
i=1
• Which takes the familiar form of a simple average
• In general, when we are not dealing with equiprobable outcomes,
P
the sum, x xf (x), is in fact a weighted average, where the
weights are the probabilities, f (x)
◦ Recall:
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P
x f (x) = 1, hence the “weights” indeed sum to 1
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Expectations: example
• Going back to our example:
X ∈ {0, 1, 2}, with f (0) =
1
1
1
, f (1) = , f (2) =
4
2
4
• The expected value of X is:
E(X ) =
2
X
x=0
1
1
1
xf (x) = (0) + (1) + (2) = 1
4
2
4
• If we repeat this two-coin experiment independently many times,
the average number of heads per trial (two coin tosses)
approaches 1
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Discrete Random Variables
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Expectations: example
• The mean E(X ) = 1 is a property of the distribution; the observed
average in a finite sample need not equal 1
• In n independent repetitions of the two-coin experiment, let ki be
the number of trials with X = i. The observed average is:
(0)k0 + (1)k1 + (2)k2
k0
k1
k2
= (0) + (1) + (2)
n
n
n
n
• The ratios kn0 , kn1 , kn2 are the relative frequencies of the events
X = 0, X = 1, and X = 2, respectively
• As the number of trials (n), increases, we expect:
k0
1
→ ,
n
4
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k1
1
→ ,
n
2
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k2
1
→
n
4
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Measures of central tendency
• Since E(X ) is the average value of X in the long run, it is a
commonly-used measure of the central tendency of X
• In simple terms, it provides us with a value, µ, which lies near the
“center” of the distribution of X
◦ To be precise, the mean µ summarises the distribution using a
probability-weighted average of its possible values; it need not be a
value that X can actually take
• It, however, does not tell us about the dispersion (or spread) of X
◦ This refers to how “stretched out” (or “wide”) the distribution of X is
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Measures of central tendency
• Consider two random variables, X and Y , with distributions f (x)
and g(y), respectively:
Figure: Probability distributions for X and Y
• These distributions are clearly different, particularly in terms of
dispersion
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Measures of dispersion: variance
• There are various ways to measure dispersion for a random
variable X
• The most commonly-used measures of dispersion are the
standard deviation and variance of X
• The variance of X is defined as:
Var(X ) = E((X − µ)2 ) =
X
(x − µ)2 f (x)
x
• It is the long-run average of the squares of the deviations of X
from µ
• In other words, for the variance we can consider a different
random variable, Y = (X − µ)2 (and take expectations of that)
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Business Statistics
Discrete Random Variables
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Measures of dispersion: variance
• We can write a simpler expression for the variance:
X
Var(X ) =
(x − µ)2 f (x)
x
X
=
(x 2 − 2xµ + µ2 )f (x)
x
X
=
(x 2 f (x) − 2xµf (x) + µ2 f (x))
=
=
x
X
x
X
x 2 f (x) − 2µ
X
xf (x) + µ2
x
X
f (x)
x
x f (x) − 2µ(µ) + µ (1)
2
2
x
=
X
x 2 f (x) − µ2
x
= E(X 2 ) − [E(X )]2
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Business Statistics
Discrete Random Variables
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Measures of dispersion: standard deviation
• Since (X − µ)2 is involved in Var(X ), and both X and µ have the
same units of measurement, the units of Var(X ) are those of X 2
◦ e.g., if X is measured in dollars, then Var(X ) is measured in dollars
squared
• It is often convenient to have a measure of dispersion in the same
units as X
• Therefore, the standard deviation of X is often used:
σ = σ(X ) =
p
Var(X )
• Which is usually denoted by the Greek letter σ (“sigma”)
• It immediately follows that σ2 = Var(X ) (“sigma squared”)
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Discrete Random Variables
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Chebyshev’s inequality
• Suppose X has a finite mean µ and a finite, positive standard
deviation σ
• Chebyshev’s inequality states that, for k > 1:
P(X ∈ (µ − k σ, µ + k σ)) ≥
k2 − 1
1
=1− 2
k2
k
• For example, taking k = 2 gives:
P(µ − 2σ < X < µ + 2σ) ≥
3
4
• Thus, the probability of falling within two standard deviations of the
mean is at least 34
• This bound applies regardless of the shape of the distribution
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Discrete Random Variables
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Binomial distribution
• We will begin our study of specific probability distributions by
considering the binomial distribution
• A Bernoulli trial is a single trial with two possible outcomes:
◦ Success (S), with probability p
◦ Failure (F ), with probability 1 − p
• A binomial experiment consists of:
◦ A fixed number n of Bernoulli trials
◦ Independent trials
◦ The same success probability p on every trial
• This applies to a wide range of random processes characterised
by binary outcomes
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Business Statistics
Discrete Random Variables
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Binomial distribution
• A binomial random variable counts the number of successes in
n independent Bernoulli trials with the same success probability p
• Examples:
◦ The number of heads in n independent tosses of a coin with a fixed
probability of heads
◦ The number of sixes in n independent rolls of a fair die, with
success probability p = 61
• The latter example reveals that random processes with several
(i.e., more than 2) outcomes may nonetheless generate (binary)
events that follow a Binomial distribution
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Discrete Random Variables
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Binomial distribution
• Consider n = 3 trials; let R denote the number of successful
outcomes
• What is the probability of three successes? (SSS)
◦ P(R = 3) = (S ∩ S ∩ S) = P(S)P(S)P(S) = p · p · p = p3
• What is the probability of two successes and one failure? (SSF )
◦ P(R = 2) = P(S)P(S)P(F ) = p · p · (1 − p) = p2 (1 − p)1
• What is the probability of one success and two failures? (SFF )
◦ P(R = 1) = P(S)P(F )P(F ) = p · (1 − p) · (1 − p) = p(1 − p)2
• A pattern emerges: for n trials, with r successes
◦ P(R = r ) = pr (1 − p)n−r
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Discrete Random Variables
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Binomial distribution
• But we’re missing something; consider n = 2 trials
• For 1 success and 1 failure, there are two possible outcomes:
◦ SF and FS, each occurring with probability p · (1 − p)
• Therefore, the probability of exactly one successful outcome is
2p(1 − p)
• In the previous slide we did not take into account all the possible
combinations
◦ Recall the example of a coin toss, where the outcomes are
{HH, TH, HT , TT }
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Discrete Random Variables
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Permutations with repeated objects
• We will state the result and then demonstrate it with an example
• The number of distinguishable permutations of n objects such that
n1 are of one type, n2 are of a second type, . . . , and nk are of a
k th type, where n1 + n2 + . . . + nk = n, is
n!
n1 !n2 ! . . . nk !
• Note: each “type” is a unique outcome
• In the special case where we have a binary outcome, then the
number of permutations becomes n1n!
!n2 ! , where n1 + n2 = n
• So, if we substitute n1 = r into the above equation:
n!
n!
=
= n Cr
n1 !n2 !
r !(n − r )!
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Discrete Random Variables
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Permutations with repeated objects: example
• Therefore, for a Bernoulli trial, we can determine the number of
permutations with repeated objects simply by determining the
combinations (n Cr )
• If we want to know how many combinations of a specific event can
arise in an experiment, e.g., two successes and one failure, we
3!
calculate 3 C2 = 2!1!
=3
◦ Here n is the number of trials, and r is the number of successes
◦ This exercise is simply re-ordering SSF in every possible
combination: SSF , SFS, FSS
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Business Statistics
Discrete Random Variables
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Binomial distribution
• If X is the number of successes in n independent trials of a
binomial experiment with probability p of success and (1 − p) of
failure on any trial, then the distribution f for X is given by:
f (x) = P(X = x) = n Cx px (1 − p)n−x
◦ Where x is an integer such that 0 ≤ x ≤ n
• Any random variable with this distribution is called a binomial
random variable and is said to have a binomial distribution
• The mean and standard deviation of X are given by:
µ = np,
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σ=
Business Statistics
p
np(1 − p)
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Binomial distribution: example
• Suppose X is a binomial random variable with n = 4 and p = 13
• Find the distribution for X :
4!
16
16
P(X = 0) = 4 C0 ( 31 )0 ( 23 )4 = 0!4!
· 1 · 16
81 = 1 · 1 · 81 = 81
4!
8
8
· 13 · 27
= 4 · 13 · 27
= 32
P(X = 1) = 4 C1 ( 31 )1 ( 23 )3 = 1!3!
81
4!
8
P(X = 2) = 4 C2 ( 31 )2 ( 23 )2 = 2!2!
· 19 · 49 = 6 · 91 · 49 = 27
4!
1
1
8
P(X = 3) = 4 C3 ( 31 )3 ( 23 )1 = 3!1!
· 27
· 23 = 4 · 27
· 23 = 81
4!
1
1
1
P(X = 4) = 4 C4 ( 13 )4 ( 23 )0 = 4!0!
· 81
· 1 = 1 · 81
· 1 = 81
• µ = np = 4( 13 ) = 43 , and
q
q
√
p
σ = np(1 − p) = 4 · 31 · 23 = 89 = 2 3 2
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Discrete Random Variables
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Binomial distribution: example
• We can plot the probability histogram:
Figure: Binomial distribution, with n = 4 and p = 31 .
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Discrete Random Variables
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Recap
• Discrete random variable, X
• Probability mass function (pmf): f (x) = P(X = x)
• Expectations, mean, variance, and standard deviation
P
◦ µ = E(X ) = x xf (x)
P
◦ σ2 = Var(X ) = E((X − µ)2 ) = x (x − µ)2 f (x)
• Chebyshev’s inequality
• Bernoulli trial, binomial distribution
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