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Discrete Random Variables
Chia-Ping Chen
Professor
Department of Computer Science and Engineering
National Sun Yat-sen University
Probability
Random Variables
A random variable is a function whose domain is a sample
space and whose codomain is the set of real numbers
X : Ω 7→ R
That is, an outcome ω of a random experiment is mapped to
X(ω), which is a real number.
Prof. C. Chen
Discrete Random Variables
Example
Flip a coin until the first head shows up.
The number of flips is a random variable
X(ωn ) = n
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Discrete Random Variables
Discrete Random Variable
A random variable is discrete if its image is finite or
countable.
For a discrete random variable X, the image can be represented
by
X = {x1 , x2 , . . . }
For example, in the previous example
X = {1, 2, . . . }
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Discrete Random Variables
Probability of a Discrete Random Variable
Let (Ω, F, P(·)) be a probability model, and X be a discrete
random variable defined on Ω.
For the probability of X = x to be well-defined, it is required
that
{ω | X(ω) = x}
is an event in F.
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Discrete Random Variables
Probability Mass Function
For every xi ∈ X , the set of outcomes mapped by X to xi
{X = xi } = {ω ∈ Ω X(ω) = xi }
is an event in F.
The probability of the event
{X = x}
as a function of x is called the probability mass function
(PMF) of X, denoted by
pX (x) = P(X = x)
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Discrete Random Variables
Abstraction
It can be tedious to start with a random experiment to derive
the PMF of a random variable. More often than not, we begin
with PMF.
a discrete random variable is specified by a PMF
leaving the details of random experiments behind
applicable in different scenarios
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Discrete Random Variables
Examples
What do the following items have in common?
head or tail
brother or sister
catch-up or fall-behind
win or lose
pass or fail
yes or no
busy or idle
strike or ball
They can all be analyzed by a random variable X taking a
value in {0, 1}.
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Discrete Random Variables
Common Discrete Random Variables
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Discrete Random Variables
Experiment: Flip a Coin Once
Flip a coin once. The number of head
X(head) = 1
X(tail) = 0
is a random variable.
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Discrete Random Variables
Abstraction: Bernoulli Random Variable
A Bernoulli random variable takes a value of either 1 or 0.
The PMF of a Bernoulli random variable is
(
pX (x) =
p,
if x = 1
1 − p, if x = 0
It is denoted by
X ∼ Bernoulli(p)
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Discrete Random Variables
Experiment: Flip a Coin n Times
Flip a coin n times. The number of times X that the head
shows up is a random variable.
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Discrete Random Variables
Abstraction: Binomial Random Variables
A binomial random variable takes a value in
{0, 1, . . . , n}
The PMF of a binomial random variable is
!
pX (x) =
n x
p (1 − p)n−x
x
It is denoted by
X ∼ binomial(n, p)
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Discrete Random Variables
Experiment: Just Flip
Flip a coin until a head shows up, then stop. The number of
flips X is a random variable.
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Discrete Random Variables
Abstraction: Geometric Random Variables
A geometric random variable takes a value in
N = {1, 2, . . . }
The PMF of a geometric random variable is
pX (x) = (1 − p)x−1 p
where 0 ≤ p ≤ 1 is a parameter. It is denoted by
X ∼ geometric(p)
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Discrete Random Variables
Limit of a Binomial Random Variable
Suppose
Y ∼ binomial(n, p)
with
n 1,
Then
p1
!
pY (x) =
(np)x −np
n x
e
p (1 − p)n−x ≈
x
x!
since
(1 − p)n−x ≈ (1 − p)n = (1 − p)
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−np
−p
= (1 − p)
− p1 −np
Discrete Random Variables
≈ e−np
Poisson Random Variables
A Poisson random variable takes a value in
{0} ∪ N
The PMF of a Poisson random variable is
pX (x) = e−λ
λx
x!
where λ > 0 is a parameter. It is denoted by
X ∼ Poisson(λ)
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Discrete Random Variables
Binomial and Poisson
A binomial random variable can be approximated by a Poisson
random variable, and vice versa.
Specifically, if
n 1,
p1
then
binomial(n, p) ≈ Poisson(np)
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Discrete Random Variables
Coin and Random Variables
random experiment
flip a coin once
flip a coin n times
flip it until first head shows up
flip it many many times, rare heads
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abstraction
Bernoulli
binomial
geometric
Poisson
Discrete Random Variables
PMFs
Note the similarity between binomial(20, 0.5) and Poisson(10)
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Discrete Random Variables
Functions of a Random Variable
A function of a random variable is a random variable.
Let
Y = g(X)
Then
ω −→ X(ω) −→ Y (ω) = g(X(ω))
So Y is a random variable.
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Discrete Random Variables
PMF of Y
For y in the image of Y , define set
Sy = {x|g(x) = y}
⇒ {ω | Y (ω) = y} =
[
{ω | X(ω) = x}
x∈Sy
⇒ pY (y) = P(Y = y) = P({ω | Y (ω) = y})


[
= P
{X = x}
x∈Sy
=
X
P (X = x)
x∈Sy
=
X
pX (x)
x∈Sy
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Discrete Random Variables
Example 2.1 Function of a Uniform Random Variable
Suppose the PMF of random variable X is
(
pX (x) =
1
9,
0,
if x is an integer in [−4, 4]
otherwise
Find pY (y) for
Y = |X|
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Discrete Random Variables
Expectation and Variance
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Discrete Random Variables
Expectation
The expectation (or mean or expected value) of a discrete
random variable X is
X
E[X] =
x pX (x)
x∈X
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Discrete Random Variables
Example 2.2
Consider 2 independent coin tosses, each with a probability of
3/4 for a head, and let X be the number of heads obtained.
Decide the PMF and the mean of X.
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Discrete Random Variables
Variance and Standard Deviation
variance
h
var(X) = E (X − E[X])2
i
standard deviation
σX =
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q
var(X)
Discrete Random Variables
Example 2.3
Suppose the PMF of X is
(
pX (x) =
1
9,
0,
if x is an integer in [−4, 4]
otherwise
Find the variance of X through the PMF of
Y = (X − E[X])2
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Discrete Random Variables
Expectation of a Function of Random Variable
Let X be a random variable and
Y = g(X)
Then
E[Y ] =
X
g(x) pX (x)
x∈X
That is
X
E[g(X)] =
g(x) pX (x)
x∈X
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Discrete Random Variables
Proof
E[Y ] =
X
y pY (y)
y∈Y
=
=
X
y
X
y∈Y
x∈Sy
X
X
pX (x)
y pX (x)
y∈Y x∈Sy
=
X
X
g(x) pX (x)
y∈Y x∈Sy
=
X
g(x) pX (x)
x∈X
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Discrete Random Variables
Example 2.3 (continued)
Suppose the PMF of X is
(
pX (x) =
1
9,
0,
if x is an integer in [−4, 4]
otherwise
Find the variance of X.
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Discrete Random Variables
Linear Function of a Random Variable
Y = aX + b
⇒ E[Y ] =
X
(ax + b)pX (x) = aE[X] + b
x∈X
⇒ var(Y ) = E[(aX + b − (aE[X] + b))2 ]
= E[(aX − aE[X])2 ]
= a2 E[(X − E[X])2 ]
= a2 var(X)
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Discrete Random Variables
An Equality for Variance
Variance is the difference of second moment and mean square.
var(X) = E[(X − E[X])2 ]
=
X
(x − E[X])2 pX (x)
x∈X
=
X
x2 − 2xE[X] + E 2 [X] pX (x)
x∈X
= E[X 2 ] − 2E 2 [X] + E 2 [X]
= E[X 2 ] − E 2 [X]
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Discrete Random Variables
Example 2.4 Time to School
If the weather is good (with probability 0.6), Alice walks the 2
miles to school at a speed of 5 miles per hour. Otherwise, she
rides her motorcycle at a speed of 30 miles per hour. What is
the mean time T for Alice to get to school?
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Discrete Random Variables
Example 2.5 Bernoulli Random Variable
Find the mean and variance of
X ∼ Bernoulli(p)
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Discrete Random Variables
Example 2.6
What is the mean and the variance of a roll of 6-sided die?
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Discrete Random Variables
Discrete Uniform Random Variable
A discrete uniform random variable has a constant PMF.
For a uniform random variable X taking an integer in [a, b],
denoted by
X ∼ uniform[a, b]
the PMF is
pX (x) =
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1
b−a+1
Discrete Random Variables
Mean and Variance
E[X] =
b
X
x pX (x) =
x=a
b
X
a+b
1
x=
b − a + 1 x=a
2
Y =X −a+1
⇒ Y ∼ uniform[1, b − a + 1]
⇒ var(X) = var(Y ) = E[Y 2 ] − E 2 [Y ]
b−a+1
X
1
=
y2 −
b − a + 1 y=1
=
b−a+2
2
(b − a)(b − a + 2)
12
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Discrete Random Variables
2
Example 2.7 Poisson Random Variable
Find the mean and variance of
Z ∼ Poisson(λ)
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Discrete Random Variables
Mean
pZ (z) = e−λ
λz
z!
∞
X
z e−λ
⇒ E[Z] =
z=0
= e−λ
=e
−λ
= λe
λz
z!
∞
X
λz
(z − 1)!
z=1
0
∞
X
λz +1
z 0 =0
∞
X
−λ
z0!
0
λz
z0!
z 0 =0
= λe−λ eλ
=λ
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Discrete Random Variables
Variance
2
E[Z ] =
=
∞
X
z=0
∞
X
z
2
e
z
−λ λ
z!
z(z − 1)e−λ
z=1
= e−λ
−λ
=e
∞
X
λz
+λ
(z − 2)!
z=2
0
∞
X
λz +2
z 0 =0
∞
X
2 −λ
=λ e
∞
λz X
λz
+
ze−λ
z! z=1
z!
z0!
+λ
0
λz
+λ
z0!
z 0 =0
= λ2 e−λ eλ + λ = λ2 + λ
⇒ var(Z) = E[Z 2 ] − E 2 [Z] = (λ2 + λ) − λ2 = λ
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Discrete Random Variables
Example: Geometric Random Variable
For
G ∼ geometric(p)
the mean is
E[G] =
and the variance is
var(G) =
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1
p
1−p
p2
Discrete Random Variables
Mean (Expectation)
E[G] =
∞
X
kpG (k)
k=1
=p+
=p+
=p+
∞
X
k=2
∞
X
kpG (k)
(k 0 + 1)pG (k 0 + 1)
k0 =1
∞
X
(k 0 + 1)(1 − p)pG (k 0 )
k0 =1
= p + (1 − p) (E[G] + 1)
⇒ pE[G] = 1
⇒ E[G] =
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1
p
Discrete Random Variables
Variance
E[G2 ] =
∞
X
k 2 pG (k) = p +
k=1
=p+
∞
X
k=2
∞
X
k 2 pG (k)
(k 0 + 1)2 pG (k 0 + 1)
k0 =1
=p+
∞
X
(k 02 + 2k 0 + 1)(1 − p)pG (k 0 )
k0 =1
= p + (1 − p) E[G2 ] + 2E[G] + 1
⇒ E[G2 ] =
2−p
p2
⇒ var(G) = E[G2 ] − E 2 [G] =
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1−p
p2
Discrete Random Variables
Example 2.8 Game
Consider a game where a player is given 2 questions.
Question A is answered correctly with probability 0.8, and
the prize money is 100.
Question B is answered correctly with probability 0.5, and
the prize money is 200.
If the first question attempted is answered incorrectly, the quiz
terminates. If it is answered correctly, the player attempts the
second question to earn more money.
Which question should be answered first to maximize the
expected total prize money?
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Discrete Random Variables
Multiple Random Variables
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Discrete Random Variables
2 Random Variables
Let X and Y be discrete random variables.
For any x ∈ X and y ∈ Y, since the sets
{ω|X(ω) = x}, {ω|Y (ω) = y}
are events, the set
{ω|X(ω) = x, Y (ω) = y} = {ω|X(ω) = x} ∩ {ω|Y (ω) = y}
is also an event.
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Discrete Random Variables
Joint Probability Mass Function
The probability of the event
{X = x, Y = y}
as a function of x and y is called the joint probability mass
function (joint PMF) of X and Y , denoted by
pXY (x, y) = P(X = x, Y = y)
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Discrete Random Variables
Marginal Probability
The marginal probability mass function (marginal PMF)
of a random variable can be computed by a joint PMF
pX (x) =
X
pXY (x, y)
y∈Y
This follows from
P(X = x) =
X
P(X = x, Y = y)
y∈Y
Similarly
pY (y) =
X
pXY (x, y)
x∈X
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Discrete Random Variables
Example 2.9 Joint Probability Table
Consider two random variables X and Y , described by the joint
probability shown in Fig. 2.10. Find the marginal PMFs.
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Discrete Random Variables
Function of Multiple Random Variables
A function of multiple random variables is a random variable.
For
Z = g(X, Y )
surely
Z(ω) = g(X(ω), Y (ω))
is a mapping from Ω to R.
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Discrete Random Variables
PMF of Z
For z in the image of Z, define
Sz = {(x, y) | g(x, y) = z}
[
⇒ {Z = z} =
{X = x, Y = y}
(x,y)∈Sz
⇒ pZ (z) = P(Z = z) =
X
pXY (x, y)
(x,y)∈Sz
⇒ E[Z] =
X
zpZ (z) =
z∈Z
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X X
g(x, y) pXY (x, y)
x∈X y∈Y
Discrete Random Variables
Example 2.9 (continued)
Find the PMF and the expectation of
Z = X + 2Y
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Discrete Random Variables
Linear Function
Z = a1 X1 + a2 X2 + · · · + an Xn
⇒ E[Z] = E[a1 X1 + a2 X2 + · · · + an Xn ]
=
X
pX1 ...Xn (x1 , . . . , xn )[a1 x1 + · · · + an xn ]
x1 ,...,xn
=
X
pX1 ...Xn (x1 , . . . , xn )a1 x1 + · · · +
x1 ,...,xn
X
pX1 ...Xn (x1 , . . . , xn )an xn
x1 ,...,xn
=
X
pX1 (x1 )a1 x1 + · · · +
X
pXn (xn )an xn
xn
x1
= a1 E[X1 ] + · · · + an E[Xn ]
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Discrete Random Variables
Example 2.10 Grade A
Each student in a 300-student class has probability of 1/3 of
getting an A, independent of any other student. What is the
mean of X, the number of students that get an A?
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Discrete Random Variables
Example 2.11
Suppose n people throw their hats in a box and then each picks
one hat at random. What is the expected value of H, the
number of people retrieving their own hats?
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Discrete Random Variables
Let Xi indicate whether the ith person picks his own hat.
⇒ H = X1 + · · · + Xn
⇒ E[H] =
=
n
X
i=1
n
X
E[Xi ]
P(Xi = 1)
i=1
hat still in the box hat retrieved
=
n
X
z
i=1
=
}|
{
n−i+1
n
z
}|
{
1
n−i+1
n
X
1
i=1
n
=1
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Discrete Random Variables
Conditional Probability and Independence
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Discrete Random Variables
Conditional Probability Mass Function
The conditional probability mass function (conditional
PMF) of a random variable X conditioned on an event A of
non-zero probability (a non-null event) is defined by
pX|A (x) =
P({X = x} ∩ A)
P(A)
Probability is re-distributed to A and re-normalized by P(A).
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Discrete Random Variables
Normalization
X
pX|A (x) = 1
x∈X
A=
[
({X = x} ∩ A)
x∈X
⇒ P(A) =
X
P({X = x} ∩ A)
x∈X
X
⇒ 1=
pX|A (x)
x∈X
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Discrete Random Variables
Example 2.12
Let X be the roll of a fair 6-sided die. Let A be the event that
the roll is an even number. What is the conditional PMF of X
given A?
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Discrete Random Variables
Example 2.13
A student will take a certain test repeatedly until he passes the
test, each time with a probability p of passing, independent of
the previous attempts, up to a maximum number of n times.
What is the conditional PMF of the number of attempts K,
given that the student passes the test?
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Discrete Random Variables
A = {the student eventually passes the test}
⇒ Ac = {the student fails the test n times}
⇒ P(Ac ) = (1 − p)n
⇒ P(A) = 1 − (1 − p)n
⇒ pK|A (k) = P(K = k|A)
P({K = k} ∩ A)
P(A)
(1 − p)k−1 p
=
, k = 1, 2, . . . , n
1 − (1 − p)n
=
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Discrete Random Variables
Conditional Probability Mass Function
The conditional probability mass function of X conditioned on
a random variable Y is defined by
pX|Y (x|y) = P(X = x|Y = y)
That is
pX|Y (x|y) =
pXY (x, y)
pY (y)
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Discrete Random Variables
Chain Rule
Marginal PMF, conditional PMF, and joint PMF are related by
pXY (x, y) = pX (x)pY |X (y|x)
= pY (y)pX|Y (x|y)
This is known as the chain rule.
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Discrete Random Variables
Example 2.14
Professor May B. Right answers each of her students’ questions
incorrectly with probability 1/4, independent of other
questions. In each lecture, she is asked 0, 1, or 2 questions with
equal probability 1/3. Let X and Y be respectively the number
of questions she is asked and the number of questions she
answers wrong in a given lecture. Find pXY (x, y). What is the
probability that she answers at least one question incorrectly?
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Discrete Random Variables
Example 2.15
Consider a transmitter that is sending messages over a
computer network. Let Y be the length of a message, and X be
the travel time of the message. Suppose the PMF of Y is
pY (102 ) =
pY (104 ) =
5/6
1/6
Furthermore, X depends on Y statistically by
X = 10−4 Y, with probability 1/2
X = 10−3 Y, with probability 1/3
X = 10−2 Y, with probability 1/6
We want to know the PMF of X.
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Discrete Random Variables
Conditional Expectation
An expectation based on conditional probability is called
conditional expectation.
The conditional expectation of X conditioned on an event A is
E[X|A] =
X
x pX|A (x)
x∈X
The conditional expectation of X conditioned on Y = y is
E[X|Y = y] =
X
x pX|Y (x|y)
x∈X
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Discrete Random Variables
Total Expectation Theorem
The expectation of a random variable is the expectation of a
conditional expectation, i.e.
E[X] =
X
pY (y)E[X|Y = y]
y
or more concisely
E[X] = E[E[X|Y ]]
Note that E[X|Y ] is a function of Y , so it is a random variable.
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Discrete Random Variables
Proof
E[X] =
X
xpX (x)
x
=
X X
x
x
=
X X
x
x
=
=
X
X
xpX|Y (x|y)pY (y)
x
pY (y)
y
=
pX|Y (x|y)pY (y)
y
XX
y
pXY (x, y)
y
X
xpX|Y (x|y)
x
pY (y)E[X|Y = y]
y
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Discrete Random Variables
General Form of Total Expectation Theorem
Let A1 , . . . , An be a partition of sample space Ω. Then
E[X] =
n
X
P(Ai )E[X|Ai ]
i=1
E[X] =
X
xpX (x)
x
=
X X
x
x
=
=
X X
x
pX|Ai (x)P(Ai )
x
Ai
X
P(Ai )
X
X
xpX|Ai (x)
x
Ai
=
P({X = x} ∩ Ai )
Ai
P(Ai )E[X|Ai ]
Ai
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Discrete Random Variables
Example 2.16
Message transmitted by a computer in Boston through a data
network is destined for New York with probability 0.5, for
Chicago with probability 0.3, and for San Francisco with
probability 0.2. The transmission time X is random. The mean
transmission time is 0.05 seconds for a message destined for
New York, 0.1 seconds for a message destined for Chicago, and
0.3 seconds for a message destined for San Francisco. What is
E[X]?
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Discrete Random Variables
Mean and Variance of a Geometric Random Variable
A = {first attempt successful}
⇒ Ac = {first attempt unsuccessful}
E[X] = P(A)E[X|A] + P(Ac )E[X|Ac ]
= p · 1 + (1 − p)(1 + E[X])
1
⇒ E[X] =
p
E[X 2 ] = P(A)E[X 2 |A] + P(Ac )E[X 2 |Ac ]
= p · 12 + (1 − p)E[(1 + X)2 ]
= p + (1 − p)(1 + 2E[X] + E[X 2 ])
2
1
⇒ E[X 2 ] = 2 −
p
p
1−p
⇒ var(X) = E[X 2 ] − E[X]2 =
p2
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Discrete Random Variables
Example 2.17 Programming
You write a program over and over, and each time there is a
probability p that it works correctly, independent of previous
attempts. What is the mean and variance of X, the number of
tries until the program works correctly?
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Discrete Random Variables
Independence of a Random Variable and an Event
Random variable X and event A are independent if
P({X = x} ∩ A) = P(X = x)P(A)
for every x ∈ X .
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Discrete Random Variables
Invariance of PMF
Suppose random variable X and event A are independent.
⇒ P({X = x} ∩ A) = P(X = x)P(A)
⇒
P({X = x} ∩ A)
= P(X = x)
P(A)
⇒ pX|A (x) = pX (x)
The PMF of X is invariant with the occurrence of A.
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Discrete Random Variables
Example 2.19
Consider 2 independent tosses of a fair coin. Let X be the
number of heads and A be the event that the number of heads
is even. Show that X and A are not independent.
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Discrete Random Variables
Independent Random Variables
Random variables X and Y are said to be independent if
pXY (x, y) = pX (x)pY (y)
for all x ∈ X and y ∈ Y. This is denoted by
X⊥
⊥Y
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Discrete Random Variables
Invariance of Probability
X⊥
⊥Y
⇒ pXY (x, y) = pX (x)pY (y)
⇒
pXY (x, y)
= pY (y)
pX (x)
⇒ pY |X (y|x) = pY (y)
⇒ pX|Y (x|y) = pX (x)
Prof. C. Chen
Discrete Random Variables
Properties
X⊥
⊥Y
⇒ E[XY ] =
X
pXY (x, y)xy
x,y
=
X
pX (x)pY (y)xy
x,y
=
X
pX (x)x
x
X
pY (y)y
y
= E[X]E[Y ]
var(X + Y ) = E[(X + Y )2 ] − E[X + Y ]2
= E[X 2 + 2XY + Y 2 ] − (E[X]2 + 2E[X]E[Y ] + E[Y ]2 )
= E[X 2 ] − E[X]2 + E[Y 2 ] − E[Y ]2
= var(X) + var(Y )
Prof. C. Chen
Discrete Random Variables
Example 2.20 Variance of a Binomial Random Variable
The variance of a binomial random variable
X ∼ binomial(n, p)
is
var(X) = np(1 − p)
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Discrete Random Variables
Proof
Consider n tosses of a coin (head showing up with probability
p). The number of heads
X = H1 + · · · + Hn
where Hi is the Bernoulli random variable for toss i, is a
binomial random variable
X ∼ binomial(n, p)
We have
var(X) = var(H1 ) + · · · + var(Hn )
=
n
X
p(1 − p)
i=1
= np(1 − p)
Prof. C. Chen
Discrete Random Variables
Example 2.21 Approval Rate
We wish to estimate the approval rate of Trump. We ask n
persons at random. Define
(
Ai =
1, if the ith person approves Trump
0, otherwise
The approval rate of the sampled n persons is
Rn =
A1 + · · · + An
n
Suppose A1 , . . . , An are independent Bernoulli random variables
with mean p and variance p(1 − p). Find the mean and variance
of Rn .
Prof. C. Chen
Discrete Random Variables
Example 2.22 Simulation
The probability of a well-defined event, say A, can be difficult to
compute. Yet, whether A occurs in a trial is easy to decide. In
this case, we can estimate the probability of A by simulation.
Prof. C. Chen
Discrete Random Variables
Quality
Carry out n trials. Associate trial i with a Bernoulli random
variable
(
1, if outcome i is in A
Ai =
0, otherwise
Consider the relative frequency
Fn =
⇒ E[Fn ] =
var(Fn ) =
A1 + · · · + An
n
n
1X
E[Ai ] = P(A)
n i=1
n
1
1 X
var(Ai ) = [P(A)(1 − P(A))]
2
n i=1
n
Prof. C. Chen
n→∞
−−−→
Discrete Random Variables
0