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Mean and Variance of Discrete Random Variables
Expected Value Variance and Standard Deviation Practice Exercises
Expected Value of Discrete Random Variable
Suppose you and I play a betting game: we flip a coin and if it lands heads, Igive you a dollar, and if it lands tails, you give me a dollar.On average, how much am I expected to win or lose?
expected winnings = (−1)(
12
)︸ ︷︷ ︸
win -$1 half the time
+ (1)(
12
)︸ ︷︷ ︸
win $1 half the time
= 0
Arthur Berg Mean and Variance of Discrete Random Variables 2/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Expected Value of Discrete Random Variable
Suppose you and I play a betting game: we flip a coin and if it lands heads, Igive you a dollar, and if it lands tails, you give me a dollar.On average, how much am I expected to win or lose?
expected winnings = (−1)(
12
)︸ ︷︷ ︸
win -$1 half the time
+ (1)(
12
)︸ ︷︷ ︸
win $1 half the time
= 0
Arthur Berg Mean and Variance of Discrete Random Variables 2/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Expected Value of Discrete Random Variable
Suppose you and I play a betting game: we flip a coin and if it lands heads, Igive you a dollar, and if it lands tails, you give me a dollar.On average, how much am I expected to win or lose?
expected winnings = (−1)(
12
)︸ ︷︷ ︸
win -$1 half the time
+ (1)(
12
)︸ ︷︷ ︸
win $1 half the time
= 0
Arthur Berg Mean and Variance of Discrete Random Variables 2/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Expected Value of Discrete Random Variable
Suppose you and I play a betting game: we flip a coin and if it lands heads, Igive you a dollar, and if it lands tails, you give me a dollar.On average, how much am I expected to win or lose?
expected winnings = (−1)(
12
)︸ ︷︷ ︸
win -$1 half the time
+ (1)(
12
)︸ ︷︷ ︸
win $1 half the time
= 0
Arthur Berg Mean and Variance of Discrete Random Variables 2/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Expected Value of Discrete Random Variable
Suppose you and I play a betting game: we flip a coin and if it lands heads, Igive you a dollar, and if it lands tails, you give me a dollar.On average, how much am I expected to win or lose?
expected winnings = (−1)(
12
)︸ ︷︷ ︸
win -$1 half the time
+ (1)(
12
)︸ ︷︷ ︸
win $1 half the time
= 0
Arthur Berg Mean and Variance of Discrete Random Variables 2/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Definition of Expected Value of a Discrete Random Variable
DefinitionThe expected value of a discrete random variable X with probabilitydistribution p(x) is given by
E(X) , µ =∑
x
x pX(x) (?)
where the sum is over all values of x for which pX(x) > 0.
Note that in order for (?) to exist, the sum must converge absolutely; that is∑x
|x| pX(x) <∞ (??)
If (??) does not hold, we say the expected value of X does not exist.
Arthur Berg Mean and Variance of Discrete Random Variables 3/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Definition of Expected Value of a Discrete Random Variable
DefinitionThe expected value of a discrete random variable X with probabilitydistribution p(x) is given by
E(X) , µ =∑
x
x pX(x) (?)
where the sum is over all values of x for which pX(x) > 0.
Note that in order for (?) to exist, the sum must converge absolutely; that is∑x
|x| pX(x) <∞ (??)
If (??) does not hold, we say the expected value of X does not exist.
Arthur Berg Mean and Variance of Discrete Random Variables 3/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Definition of Expected Value of a Discrete Random Variable
DefinitionThe expected value of a discrete random variable X with probabilitydistribution p(x) is given by
E(X) , µ =∑
x
x pX(x) (?)
where the sum is over all values of x for which pX(x) > 0.
Note that in order for (?) to exist, the sum must converge absolutely; that is∑x
|x| pX(x) <∞ (??)
If (??) does not hold, we say the expected value of X does not exist.
Arthur Berg Mean and Variance of Discrete Random Variables 3/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Expected value of functions of random variables
The expected value of g(X) where g is any real-valued function is naturally
E(g(X)) =∑
x
g(x)p(x)
ExampleConsider the Bernoulli random variable
X =
{0, w.p. 1/21, w.p. 1/2
Compute E(X2 − 1
).
E(X2 − 1
)= (02 − 1)
(12
)+ ((1)2 − 1)
(12
)=−12
Arthur Berg Mean and Variance of Discrete Random Variables 4/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Expected value of functions of random variables
The expected value of g(X) where g is any real-valued function is naturally
E(g(X)) =∑
x
g(x)p(x)
ExampleConsider the Bernoulli random variable
X =
{0, w.p. 1/21, w.p. 1/2
Compute E(X2 − 1
).
E(X2 − 1
)= (02 − 1)
(12
)+ ((1)2 − 1)
(12
)=−12
Arthur Berg Mean and Variance of Discrete Random Variables 4/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Birthday Problem Revisited
65 people participated in the birthday game a few weeks back.I claimed that if no two birthdays matched, then I would pay everyone 30monopoly dollars, but otherwise each person would pay me one monopolydollar.Therefore either I would lose 65*30=1950 monopoly dollars, or I would win 65monopoly dollars.My expected earnings should be somewhere between -1950 and +65. Let’scompute it.My total earnings are represented by the following random variable
X =
{30, w.p. p−1950, w.p. 1− p
where
p ≈ 365(364)(363) · · · (301)36565 ≈ .9977
ThereforeE(X) = 30(.9977)− 1950(.0023) = 25.446
Arthur Berg Mean and Variance of Discrete Random Variables 5/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Birthday Problem Revisited
65 people participated in the birthday game a few weeks back.I claimed that if no two birthdays matched, then I would pay everyone 30monopoly dollars, but otherwise each person would pay me one monopolydollar.Therefore either I would lose 65*30=1950 monopoly dollars, or I would win 65monopoly dollars.My expected earnings should be somewhere between -1950 and +65. Let’scompute it.My total earnings are represented by the following random variable
X =
{30, w.p. p−1950, w.p. 1− p
where
p ≈ 365(364)(363) · · · (301)36565 ≈ .9977
ThereforeE(X) = 30(.9977)− 1950(.0023) = 25.446
Arthur Berg Mean and Variance of Discrete Random Variables 5/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Birthday Problem Revisited
65 people participated in the birthday game a few weeks back.I claimed that if no two birthdays matched, then I would pay everyone 30monopoly dollars, but otherwise each person would pay me one monopolydollar.Therefore either I would lose 65*30=1950 monopoly dollars, or I would win 65monopoly dollars.My expected earnings should be somewhere between -1950 and +65. Let’scompute it.My total earnings are represented by the following random variable
X =
{30, w.p. p−1950, w.p. 1− p
where
p ≈ 365(364)(363) · · · (301)36565 ≈ .9977
ThereforeE(X) = 30(.9977)− 1950(.0023) = 25.446
Arthur Berg Mean and Variance of Discrete Random Variables 5/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Birthday Problem Revisited
65 people participated in the birthday game a few weeks back.I claimed that if no two birthdays matched, then I would pay everyone 30monopoly dollars, but otherwise each person would pay me one monopolydollar.Therefore either I would lose 65*30=1950 monopoly dollars, or I would win 65monopoly dollars.My expected earnings should be somewhere between -1950 and +65. Let’scompute it.My total earnings are represented by the following random variable
X =
{30, w.p. p−1950, w.p. 1− p
where
p ≈ 365(364)(363) · · · (301)36565 ≈ .9977
ThereforeE(X) = 30(.9977)− 1950(.0023) = 25.446
Arthur Berg Mean and Variance of Discrete Random Variables 5/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Birthday Problem Revisited
65 people participated in the birthday game a few weeks back.I claimed that if no two birthdays matched, then I would pay everyone 30monopoly dollars, but otherwise each person would pay me one monopolydollar.Therefore either I would lose 65*30=1950 monopoly dollars, or I would win 65monopoly dollars.My expected earnings should be somewhere between -1950 and +65. Let’scompute it.My total earnings are represented by the following random variable
X =
{30, w.p. p−1950, w.p. 1− p
where
p ≈ 365(364)(363) · · · (301)36565 ≈ .9977
ThereforeE(X) = 30(.9977)− 1950(.0023) = 25.446
Arthur Berg Mean and Variance of Discrete Random Variables 5/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Birthday Problem Revisited
65 people participated in the birthday game a few weeks back.I claimed that if no two birthdays matched, then I would pay everyone 30monopoly dollars, but otherwise each person would pay me one monopolydollar.Therefore either I would lose 65*30=1950 monopoly dollars, or I would win 65monopoly dollars.My expected earnings should be somewhere between -1950 and +65. Let’scompute it.My total earnings are represented by the following random variable
X =
{30, w.p. p−1950, w.p. 1− p
where
p ≈ 365(364)(363) · · · (301)36565 ≈ .9977
ThereforeE(X) = 30(.9977)− 1950(.0023) = 25.446
Arthur Berg Mean and Variance of Discrete Random Variables 5/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Variance
Definition (variance)The variance of a random variable X with expected value µ is given by
var(X) , σ2 = E[(X−µ)2]
DefinitionThe standard deviation of a random variable X is, σ, the square root of thevariance, i.e.
sd(X) , σ =√
E [(X − µ)2] =√
var(X)
Arthur Berg Mean and Variance of Discrete Random Variables 6/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Variance
Definition (variance)The variance of a random variable X with expected value µ is given by
var(X) , σ2 = E[(X−µ)2]
DefinitionThe standard deviation of a random variable X is, σ, the square root of thevariance, i.e.
sd(X) , σ =√
E [(X − µ)2] =√
var(X)
Arthur Berg Mean and Variance of Discrete Random Variables 6/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Some Distributions
X1 =
{−1, w.p. 1/21, w.p. 1/2
X2 =
{−2, w.p. 1/22, w.p. 1/2
X3 =
{−3, w.p. 1/23, w.p. 1/2
Hence E(X1) = E(X2) = E(X3) = 0 and
var(X1) = (1− 0)2(
12
)+ (−1− 0)2
(12
)= 1
var(X2) = (2− 0)2(
12
)+ (−2− 0)2
(12
)= 4
var(X3) = (3− 0)2(
12
)+ (−3− 0)2
(12
)= 9
Therefore sd(X1) = 1, sd(X2) = 2, and sd(X3) = 3.Arthur Berg Mean and Variance of Discrete Random Variables 7/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Some Distributions
X1 =
{−1, w.p. 1/21, w.p. 1/2
X2 =
{−2, w.p. 1/22, w.p. 1/2
X3 =
{−3, w.p. 1/23, w.p. 1/2
Hence E(X1) = E(X2) = E(X3) = 0 and
var(X1) = (1− 0)2(
12
)+ (−1− 0)2
(12
)= 1
var(X2) = (2− 0)2(
12
)+ (−2− 0)2
(12
)= 4
var(X3) = (3− 0)2(
12
)+ (−3− 0)2
(12
)= 9
Therefore sd(X1) = 1, sd(X2) = 2, and sd(X3) = 3.Arthur Berg Mean and Variance of Discrete Random Variables 7/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Some Distributions
X1 =
{−1, w.p. 1/21, w.p. 1/2
X2 =
{−2, w.p. 1/22, w.p. 1/2
X3 =
{−3, w.p. 1/23, w.p. 1/2
Hence E(X1) = E(X2) = E(X3) = 0 and
var(X1) = (1− 0)2(
12
)+ (−1− 0)2
(12
)= 1
var(X2) = (2− 0)2(
12
)+ (−2− 0)2
(12
)= 4
var(X3) = (3− 0)2(
12
)+ (−3− 0)2
(12
)= 9
Therefore sd(X1) = 1, sd(X2) = 2, and sd(X3) = 3.Arthur Berg Mean and Variance of Discrete Random Variables 7/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Theorem 4.2
Theorem (mean and variance of aX + b)For any random variable X (discrete or not) and constants a and b,
1 E(aX + b) = a E(X) + b2 var(aX + b) = a2 var(X)
It follows that if X has mean µ and standard deviation σ, then
Y =x− µσ
has mean 0 and standard deviation 1. This is the standardized form of X.
TheoremIf X and Y are random variables, then E(X + Y) = E(X) + E(Y).
Arthur Berg Mean and Variance of Discrete Random Variables 8/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Theorem 4.2
Theorem (mean and variance of aX + b)For any random variable X (discrete or not) and constants a and b,
1 E(aX + b) = a E(X) + b2 var(aX + b) = a2 var(X)
It follows that if X has mean µ and standard deviation σ, then
Y =x− µσ
has mean 0 and standard deviation 1. This is the standardized form of X.
TheoremIf X and Y are random variables, then E(X + Y) = E(X) + E(Y).
Arthur Berg Mean and Variance of Discrete Random Variables 8/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Theorem 4.2
Theorem (mean and variance of aX + b)For any random variable X (discrete or not) and constants a and b,
1 E(aX + b) = a E(X) + b2 var(aX + b) = a2 var(X)
It follows that if X has mean µ and standard deviation σ, then
Y =x− µσ
has mean 0 and standard deviation 1. This is the standardized form of X.
TheoremIf X and Y are random variables, then E(X + Y) = E(X) + E(Y).
Arthur Berg Mean and Variance of Discrete Random Variables 8/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Theorem 4.3
Theorem (var(X) = E(X2)− µ2)
If X is a random variable with mean µ, then
var(X) = E(X2)− µ2
Proof.
var(X) = E[(X − µ)2]
= E(X2 − 2Xµ+ µ2)
= E(X2)− E(2Xµ) + E(µ2)
= E(X2)− 2µ2 + µ2
= E(X2)− µ2
Arthur Berg Mean and Variance of Discrete Random Variables 9/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Theorem 4.3
Theorem (var(X) = E(X2)− µ2)
If X is a random variable with mean µ, then
var(X) = E(X2)− µ2
Proof.
var(X) = E[(X − µ)2]
= E(X2 − 2Xµ+ µ2)
= E(X2)− E(2Xµ) + E(µ2)
= E(X2)− 2µ2 + µ2
= E(X2)− µ2
Arthur Berg Mean and Variance of Discrete Random Variables 9/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Chebyshev’s inequality
There are many random variables X with a given mean µ and a given varianceσ2, but they all must satisfy the following inequality.
TheoremChebyshev’s inequality Let X be a random variable with mean µ and varianceσ2. Then for any positive k,
P(|X − µ| < kσ) ≥ 1− 1k2
Letting k = 2 in Chebyshev’s inequality gives
P(µ− 2σ < X < µ+ 2σ) ≥ 1− 14
=34
That is, the interval from µ− 2σ to µ+ 2σ must contain at least 3/4 of theprobability mass.
Arthur Berg Mean and Variance of Discrete Random Variables 10/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Chebyshev’s inequality
There are many random variables X with a given mean µ and a given varianceσ2, but they all must satisfy the following inequality.
TheoremChebyshev’s inequality Let X be a random variable with mean µ and varianceσ2. Then for any positive k,
P(|X − µ| < kσ) ≥ 1− 1k2
Letting k = 2 in Chebyshev’s inequality gives
P(µ− 2σ < X < µ+ 2σ) ≥ 1− 14
=34
That is, the interval from µ− 2σ to µ+ 2σ must contain at least 3/4 of theprobability mass.
Arthur Berg Mean and Variance of Discrete Random Variables 10/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Chebyshev’s inequality
There are many random variables X with a given mean µ and a given varianceσ2, but they all must satisfy the following inequality.
TheoremChebyshev’s inequality Let X be a random variable with mean µ and varianceσ2. Then for any positive k,
P(|X − µ| < kσ) ≥ 1− 1k2
Letting k = 2 in Chebyshev’s inequality gives
P(µ− 2σ < X < µ+ 2σ) ≥ 1− 14
=34
That is, the interval from µ− 2σ to µ+ 2σ must contain at least 3/4 of theprobability mass.
Arthur Berg Mean and Variance of Discrete Random Variables 10/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Exercise 4.17
Arthur Berg Mean and Variance of Discrete Random Variables 11/ 12
Expected Value Variance and Standard Deviation Practice Exercises
Exercise 4.37
Arthur Berg Mean and Variance of Discrete Random Variables 12/ 12