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Mean and Variance of Discrete Random Variables
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
