Chapter 3 (part b): Discrete random variables
STAT2001/STAT2013/STAT6013/STAT6039 - Introductory
Mathematical Statistics (for Actuarial Studies)/Principles of
Mathematical Statistics (for Actuarial Studies)
Lecturer: Associate Professor Janice Scealy 1
1Research School of Finance, Actuarial Studies and Statistics, ANU
• Calculate summary measures for discrete distributions (e.g.
distribution mean/average).
• Describe the shape, location and spread of distributions.
• Properties of discrete probability distributions.

HAMBURGER
2 / 19
Expectation
Two coins are tossed. How many heads can we expect to come up?
Let Y = number of heads. Then
p(y) =
8><>:
1/4 y = 0
1/2 y = 1
1/4 y = 2
The answer seems to be 1 (the middle value).
But what exactly do we mean by “expect”?
A mental experiment: Suppose we toss the two coins 1000 times, and
each time record the number of heads, y.
The result would be something like 1,1,2,0,1,2,0,1,1,1,...,1,0.
We’d get about 250 zero’s, 500 one’s and 250 two’s.
3 / 19
Expectation continued
So the average of the 1000 values of Y would be approximately
250(0)+500(1)+250(2)
1000 = 0(1/4) + 1(1/2) + 2(1/4) = 1.
This agrees with our intuitive answer.
Observe that
0(1/4) + 1(1/2) + 2(1/4) = 0p(0) + 1p(1) + 2p(2) =
2X
y=0
yp(y).
This leads to the following definition.
Suppose Y is a discrete random variable with pmf p(y). Then the
expected value (or mean) of Y is
E(Y) =
X
y
yp(y).
(The sum is over all possible values y of the rv Y .)
We may also write Y’s mean as µY or µ.
µ is a measure of central tendency, in the sense that it represents the
average of a hypothetically infinite number of independent
realisations of Y .
4 / 19
Example 10
Suppose that Y is a random variable which equals 5 with probability
0.2 and 7 with probability 0.8. Find the expected value of Y .
E(Y) =
X
y
yp(y) = 5(0.2) + 7(0.8) = 6.6.
This means that if we were to generate many independent realisations
of Y , so as to get a sequence like 7, 7, 5, 7, 5, 7, ..., the average of
these number would be close to 6.6.
As the sequence got longer, the average would converge to 6.6. More
on this later.
5 / 19
Example 11
Find the mean of the Bernoulli distribution.
Let Y v Bern(p). Then
p(y) =
(
p y = 1
1 p y = 0.
So Y has mean
µ =
1X
y=0
yp(y) = 0p(0) + 1p(1) = 0(1 p) + 1p = p.
Thus for example, if we toss a fair coin thousands of times, and each
time write 1 when a head comes up and 0 otherwise, we will get a
sequence like 0,0,1,0,1,1,1,0,... The average of these 1’s and 0’s will
be about 1/2, corresponding to the fact that each such number has a
Bernoulli distribution with parameter 1/2 and thus a mean of 1/2.
6 / 19
Example 12
Find the mean of the binomial distribution.
Let Y v Bin(n, p). Then Y has mean
µ =
nX
y=0
y
✓
n
y
◆
py(1 p)ny
=
nX
y=1
y
n!
y!(n y)!p
y(1 p)ny (the first term is zero)
= np
nX
y=1
(n 1)!
(y 1)!(n 1 (y 1))!p
y1(1 p)n1(y1)
= np
mX
x=0
m!
x!(m x)!p
x(1 p)mx (x = y 1 and m = n 1)
= np (since the sum equals 1, by the binomial theorem)
This makes sense. For example, if we roll a die 60 times, we can
expect 60(1/6) = 10 sixes.
y
It 1 0
7 / 19
Example 12 limits
sum y 1 to y n
I
y 1 0 to y i
n t
I
2 0 to a n I
µ
8 0 to a M
8 / 19
Expectations of functions of random variables
Suppose that Y is a discrete random variable with pmf p(y), and g(t) is
a function. Then the expected value (or mean) of g(Y) is defined to be
E (g(Y)) =
X
y
g(y)p(y).
The text presents this equation as Theorem 3.2 and provides a proof
for it. We have instead defined the expected value of a function of a
rv, with no need for a proof.
Example 13 Suppose that Y v Bern(p). Find E(Y2).
E(Y2) =
X
y
y2p(y) = 02(1 p) + 12p = p.
(same as E(Y); in fact, E(Yk) = p for all k). Ok
O
I k I
9 / 19
Laws of expectation
1. If c is a constant, then E(c) = c.
2. E {cg(Y)} = cE {g(Y)}.
3.
E {g1(Y) + g2(Y) + ...+ gk(Y)} = E {g1(Y)}+E {g2(Y)}+...+E {gk(Y)} .
Proof of 1st law: E(c) =
P
y cp(y) = c
P
y p(y) = c(1) = c.
Example 14 Suppose that Y v Bern(p). Find E(3Y2 + Y 2).
E(3Y2 + Y 2) = 3E(Y2) + E(Y) 2
= 3p+ p 2
= 4p 2.
(recall from Example 13 that E(Yk) = p for all k)
10 / 19
Special expectations
1. The kth raw moment of Y is µ0k = E(Yk).
2. The kth central moment of Y is µk = E
⇣
(Y µ)k
⌘
.
3. The variance of Y is Var(Y) = 2 = µ2 = E
⇣
(Y µ)2
⌘
.
4. The standard deviation of Y is SD(Y) = =
p
Var(Y).
We can also write Var(Y) as V(Y) or 2Y .
Note that µ01 = µ.
Also, µ1 = E
⇣
(Y µ)1
⌘
= E(Y) µ = µ µ = 0.
11 / 19
Example 15
Suppose that p(y) = y/3, y = 1, 2. Find µ03 and .
µ03 = E(Y3) =
P
y y
3p(y) = 13
1
3

+ 23
2
3

= 173 .
µ = E(Y) =
P
y yp(y) = 1(1/3) + 2(2/3) = 5/3.
2 = µ2 = E
⇣
(Y µ)2
⌘
=
P
y(y µ)2p(y) =
1 53
2 1
3 +

2 53
2 2
3 =
2
9 .
Hence =
p
2/3 = 0.4714.
The various moments provide information about the nature of a
distribution.
We have already seen that the mean provides a measure of central
tendency.
The variance and standard deviation provide measures of dispersion.
Distributions that are highly disperse have a large variance.
12 / 19
Variance example
Example: Suppose X has pmf p(x) = 1/2, x = 1, 3 and Y has pmf
p(y) = 1/2, y = 0, 4. Find Var(X) and Var(Y). Which distribution is
the more disperse?
Both distributions have a mean of 2 (= average of 1 and 3 = average of
0 and 4).
Var(X) = (1 2)20.5+ (3 2)20.5 = 1 and
Var(Y) = (0 2)20.5+ (4 2)20.5 = 4.
We see that Var(Y) > Var(X). This corresponds to the fact that Y’s
distribution is the more disperse of the two.
pint Ply
112 112
o I i n o d 4 Y
13 / 19
Binomial Pmf
ply
111
Mylarge Mysmall
ipeaned flat
T T
Mi su mean 1st raw moment
Ma measure spread dispersioncentral
moments My measures shewner binomial
abode
shewed to right
measures kurtosis MITumnetricY
how flat or peatied
First 4 moments tell you a lot about the distribution
14 / 19
Two important results (for computing variances)
1. Var(Y) = E(Y2) (E(Y))2, or equivalently, 2 = µ02 µ2.
2. Var(a+ bY) = b2Var(Y).
proof of 1:
LHS = E

(Y µ)2 = E(Y22Yµ+µ2) = E(Y2)2µE(Y)+µ2 = RHS.
proof of 2:
LHS = E
⇣
(a+ bY E(a+ bY))2
⌘
= E
⇣
(a+ bY a bµ)2
⌘
=
b2E
⇣
(Y µ)2
⌘
= RHS.
Example 16 Find the variance of the Bernoulli distribution.
Recall that if Y v Bern(p), then E(Y) = E(Y2) = p. Therefore Y has
variance Var(Y) = p p2 = p(1 p) .
What is the variance of X = 2 5Y?
Var(X) = (5)2Var(Y) = 25p(1 p).
15 / 19
Moment generating functions
The moment generating function (mgf) of a random variable Y is
defined to be
m(t) = E(etY).
Mgf’s have two important uses:
1. To compute raw moments, according to the formula:
µ0k = m
(k)(0)
(See Thm 3.12 in the text for a general proof, and below for the case
k = 1.)
2. To identify distributions according to the result:
If the mgf of a rv Y is the same as that of another rv X, we can
conclude that Y has the same distribution as X.
(This follows from “the uniqueness theorem”, a result in pure
mathematics.)
Rtn derivative of mgf
evaluated at t o
16 / 19
Moment generating functions continued
Note: m(k)(0) denotes the kth derivative of m(t), evaluated at t = 0,
and may also be written as
dkm(t)
dtk

t=0
.
We may also write m(1)(t) as m0(t), and m(2)(t) as m00(t), etc.
For all mgf’s, m(t), it is true that
m(0)(0) = m(0) = E(YY0) = E(1) = 1 .
Proof that µ = µ01 = m0(0):
m0(t) = ddtm(t) =
d
dtE(e
Yt) = ddt
P
y e
ytp(y) =
P
y
d
dt e
ytp(y) =P
y ye
ytp(y).
So m0(0) =
P
y ye
y0p(y) =
P
y yp(y) = E(Y) = µ.
17 / 19
Example 17
Use the mgf technique to find the mean and variance of the binomial
distribution.
Let Y v Bin(n, p). Then Y has mgf
m(t) = E(eYt) =
Pn
y=0 e
ytn
y

py(1 p)ny =Pn
y=0
n
y

(pet)y(1 p)ny = {(pet) + (1 p)}n by the binomial
theorem.
Thus m(t) = (1 p+ pet)n.
Then, m0(t) = dm(t)dt = n (1 p+ pet)n1 pet, by the chain rule for
differentiation.
(This rule is: dudt =
du
dv
dv
dt , where here v = 1 p+ pet and
u = m(t) = vn.)
So µ = µ01 = m0(0) = n

1 p+ pe0n1 pe0 = np (as before).
18 / 19
Example 17 continued
m00(t) = d
2m(t)
dt2 =
dm0(t)
dt =
np
n
(1 p+ pet)n1 et + et(n 1) (1 p+ pet)n2 pet
o
,
by the product rule for differentiation.
(This rule is: d(uv)dt = u
dv
dt + v
du
dt , where here u = (1 p+ pet)n1 and
v = et.)
µ02 = m00(0) =
np
n
1 p+ pe0n1 e0 + e0(n 1) 1 p+ pe0n2 pe0o =
np {1+ (n 1)p}.
Therefore 2 = µ02 µ2 = np {1+ (n 1)p} (np)2 = np(1 p).
19 / 19
Example 18
A random variable Y has the mgf m(t) = 18(1+ e
t)3. Find the
probability that Y equals three.
m(t) = (1 12 + 12et)3 = (1 p+ pet)n, where n = 3 and p = 1/2.
Thus m(t) is the mgf of a random variable whose distribution is
binomial with parameters 3 and 1/2. Therefore Y v Bin(3, 1/2), and
so P(Y = 3) = 1/8.