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Variance and Standard Deviation
Christopher Croke
University of Pennsylvania
Math 115
Christopher Croke
Calculus 115
Expected value
The expected value E (X ) of a discrete random variable X with
probability function f (xi ) is:
E (X ) = Σi xi · f (xi ).
Christopher Croke
Calculus 115
Expected value
The expected value E (X ) of a discrete random variable X with
probability function f (xi ) is:
E (X ) = Σi xi · f (xi ).
This is a ”weighted average”.
Christopher Croke
Calculus 115
Expected value
The expected value E (X ) of a discrete random variable X with
probability function f (xi ) is:
E (X ) = Σi xi · f (xi ).
This is a ”weighted average”. For a uniform distribution, i.e.
f (xi ) = N1 where N = #{xi }, then it is just the usual average:
E (X ) =
Christopher Croke
ΣN
i=1 xi
.
N
Calculus 115
Expected value
The expected value E (X ) of a discrete random variable X with
probability function f (xi ) is:
E (X ) = Σi xi · f (xi ).
This is a ”weighted average”. For a uniform distribution, i.e.
f (xi ) = N1 where N = #{xi }, then it is just the usual average:
E (X ) =
ΣN
i=1 xi
.
N
Problem: What is the expected number of heads in four tosses?
Christopher Croke
Calculus 115
Expected value of a continuous random variable
For a continuous random variable with probability density function
f (x) on [A, B] we have:
Z
B
E (X ) =
xf (x)dx.
A
Christopher Croke
Calculus 115
Expected value of a continuous random variable
For a continuous random variable with probability density function
f (x) on [A, B] we have:
Z
B
E (X ) =
xf (x)dx.
A
Problem: Consider our random variable X which is the sum of the
coordinates of a point chosen randomly from [0, 1] × [0, 1]? What
is E (X )?
Christopher Croke
Calculus 115
Variance
The first first important number describing a probability
distribution is the mean or expected value E (X ).
Christopher Croke
Calculus 115
Variance
The first first important number describing a probability
distribution is the mean or expected value E (X ).
The next one is the variance Var (X ) = σ 2 (X ). The square root of
the variance σ is called the Standard Deviation.
Christopher Croke
Calculus 115
Variance
The first first important number describing a probability
distribution is the mean or expected value E (X ).
The next one is the variance Var (X ) = σ 2 (X ). The square root of
the variance σ is called the Standard Deviation.
For continuous random variable X with probability density function
f (x) defined on [A, B] we saw:
Z
B
E (X ) =
xf (x)dx.
A
Christopher Croke
Calculus 115
Variance
The first first important number describing a probability
distribution is the mean or expected value E (X ).
The next one is the variance Var (X ) = σ 2 (X ). The square root of
the variance σ is called the Standard Deviation.
For continuous random variable X with probability density function
f (x) defined on [A, B] we saw:
Z
B
E (X ) =
xf (x)dx.
A
(note that this does not always exist if B = ∞.)
Christopher Croke
Calculus 115
Variance
The first first important number describing a probability
distribution is the mean or expected value E (X ).
The next one is the variance Var (X ) = σ 2 (X ). The square root of
the variance σ is called the Standard Deviation.
For continuous random variable X with probability density function
f (x) defined on [A, B] we saw:
Z
B
E (X ) =
xf (x)dx.
A
(note that this does not always exist if B = ∞.)
The variance will be:
Z B
σ 2 (X ) = Var (X ) = E ((X − E (X ))2 ) =
(x − E (X ))2 f (x)dx.
A
Christopher Croke
Calculus 115
Variance
The first first important number describing a probability
distribution is the mean or expected value E (X ).
The next one is the variance Var (X ) = σ 2 (X ). The square root of
the variance σ is called the Standard Deviation.
For continuous random variable X with probability density function
f (x) defined on [A, B] we saw:
Z
B
E (X ) =
xf (x)dx.
A
(note that this does not always exist if B = ∞.)
The variance will be:
Z B
σ 2 (X ) = Var (X ) = E ((X − E (X ))2 ) =
(x − E (X ))2 f (x)dx.
A
Christopher Croke
Calculus 115
Problem:(uniform probability on an interval) Let X be the random
variable you get when you randomly choose a point in [0, B].
a) find the probability density function f .
b) find the cumulative distribution function F (x).
c) find E (X ) and Var (X ) = σ 2 .
Christopher Croke
Calculus 115
Problem:(uniform probability on an interval) Let X be the random
variable you get when you randomly choose a point in [0, B].
a) find the probability density function f .
b) find the cumulative distribution function F (x).
c) find E (X ) and Var (X ) = σ 2 .
Alternative formula for variance: σ 2 = E (X 2 ) − µ2 .
Christopher Croke
Calculus 115
Problem:(uniform probability on an interval) Let X be the random
variable you get when you randomly choose a point in [0, B].
a) find the probability density function f .
b) find the cumulative distribution function F (x).
c) find E (X ) and Var (X ) = σ 2 .
Alternative formula for variance: σ 2 = E (X 2 ) − µ2 . Why?
Christopher Croke
Calculus 115
Problem:(uniform probability on an interval) Let X be the random
variable you get when you randomly choose a point in [0, B].
a) find the probability density function f .
b) find the cumulative distribution function F (x).
c) find E (X ) and Var (X ) = σ 2 .
Alternative formula for variance: σ 2 = E (X 2 ) − µ2 . Why?
E ((X − µ)2 ) = E (X 2 − 2µX + µ2 )
Christopher Croke
Calculus 115
Problem:(uniform probability on an interval) Let X be the random
variable you get when you randomly choose a point in [0, B].
a) find the probability density function f .
b) find the cumulative distribution function F (x).
c) find E (X ) and Var (X ) = σ 2 .
Alternative formula for variance: σ 2 = E (X 2 ) − µ2 . Why?
E ((X − µ)2 ) = E (X 2 − 2µX + µ2 ) = E (X 2 ) + E (−2µX ) + E (µ2 )
Christopher Croke
Calculus 115
Problem:(uniform probability on an interval) Let X be the random
variable you get when you randomly choose a point in [0, B].
a) find the probability density function f .
b) find the cumulative distribution function F (x).
c) find E (X ) and Var (X ) = σ 2 .
Alternative formula for variance: σ 2 = E (X 2 ) − µ2 . Why?
E ((X − µ)2 ) = E (X 2 − 2µX + µ2 ) = E (X 2 ) + E (−2µX ) + E (µ2 )
= E (X 2 ) − 2µE (X ) + µ2 = E (X 2 ) − 2µ2 + µ2 = E (X 2 ) − µ2 .
Christopher Croke
Calculus 115
Variance - Continuous case
Problem Consider the problem of the age of a cell in a culture.
We had the probability density function was f (x) = 2ke −kx on
[0, T ] where k = ln(2)
T .
a) Find E (X ).
Christopher Croke
Calculus 115
Variance - Continuous case
Problem Consider the problem of the age of a cell in a culture.
We had the probability density function was f (x) = 2ke −kx on
[0, T ] where k = ln(2)
T .
a) Find E (X ).
b) Find σ 2 .
It is easier in this case to use the alternative definition of σ 2 :
Z B
2
2
2
σ = E (X ) − E (X ) =
x 2 f (x)dx − E (X )2 .
A
Christopher Croke
Calculus 115
Variance - Continuous case
Problem Consider the problem of the age of a cell in a culture.
We had the probability density function was f (x) = 2ke −kx on
[0, T ] where k = ln(2)
T .
a) Find E (X ).
b) Find σ 2 .
It is easier in this case to use the alternative definition of σ 2 :
Z B
2
2
2
σ = E (X ) − E (X ) =
x 2 f (x)dx − E (X )2 .
A
Problem: Consider our random variable X which is the sum of the
coordinates of a point chosen randomly from [0; 1] × [0; 1]? What
is Var (X )?
Christopher Croke
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Christopher Croke
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Why?
Christopher Croke
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Why?
E (X ) =
Σnk=1
Christopher Croke
n k n−k
k
p q
k
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Why?
E (X ) =
Σnk=1
n k n−k
k
p q
k
Now for any p and q this is
n k n−k ∂
=p
Σ
p q
k
∂p
Christopher Croke
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Why?
E (X ) =
Σnk=1
n k n−k
k
p q
k
Now for any p and q this is
n k n−k ∂
∂
=p
Σ
p q
= p (p + q)n
k
∂p
∂p
Christopher Croke
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Why?
E (X ) =
Σnk=1
n k n−k
k
p q
k
Now for any p and q this is
n k n−k ∂
∂
=p
Σ
p q
= p (p + q)n = pn(p + q)n−1 .
k
∂p
∂p
Christopher Croke
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Why?
E (X ) =
Σnk=1
n k n−k
k
p q
k
Now for any p and q this is
n k n−k ∂
∂
=p
Σ
p q
= p (p + q)n = pn(p + q)n−1 .
k
∂p
∂p
For our choice of p and q this is just np.
Christopher Croke
Calculus 115
Expected value for a binomial random variable
For X a binomial random variable with parameters n and p
E (X ) = np
Why?
E (X ) =
Σnk=1
n k n−k
k
p q
k
Now for any p and q this is
n k n−k ∂
∂
=p
Σ
p q
= p (p + q)n = pn(p + q)n−1 .
k
∂p
∂p
For our choice of p and q this is just np.
Problem: For our ballplayer hitting .300 coming to bat 4 times a
game what is the expected number of hits?
Christopher Croke
Calculus 115
Expected value
Problem: Two players play a game by drawing balls (without
replacement) in turn from an urn that has 2 red balls and 4 white
balls in it. A player wins when he draws a red ball. The player
going first will get $2 if he wins while the player going second will
get $3. Would you rather go first or second or does it matter at
all?
Christopher Croke
Calculus 115
Expected value
Problem: Two players play a game by drawing balls (without
replacement) in turn from an urn that has 2 red balls and 4 white
balls in it. A player wins when he draws a red ball. The player
going first will get $2 if he wins while the player going second will
get $3. Would you rather go first or second or does it matter at
all? How does this change if no one gets paid after 4 balls turn up
white?
Christopher Croke
Calculus 115
Expected value
Problem: Two players play a game by drawing balls (without
replacement) in turn from an urn that has 2 red balls and 4 white
balls in it. A player wins when he draws a red ball. The player
going first will get $2 if he wins while the player going second will
get $3. Would you rather go first or second or does it matter at
all? How does this change if no one gets paid after 4 balls turn up
white?
Christopher Croke
Calculus 115
variance in the discrete case
If f (xi ) is the probability distribution function for a random
variable with range {x1 , x2 , x3 , ...} and mean µ = E (X ) then:
Var (X ) = σ 2 = (x1 −µ)2 f (x1 )+(x2 −µ)2 f (x2 )+(x3 −µ)2 f (x3 )+...
It is a description of how the distribution ”spreads”.
Christopher Croke
Calculus 115
variance in the discrete case
If f (xi ) is the probability distribution function for a random
variable with range {x1 , x2 , x3 , ...} and mean µ = E (X ) then:
Var (X ) = σ 2 = (x1 −µ)2 f (x1 )+(x2 −µ)2 f (x2 )+(x3 −µ)2 f (x3 )+...
It is a description of how the distribution ”spreads”.
Note Var (X ) = E ((X − µ)2 ).
Christopher Croke
Calculus 115
variance in the discrete case
If f (xi ) is the probability distribution function for a random
variable with range {x1 , x2 , x3 , ...} and mean µ = E (X ) then:
Var (X ) = σ 2 = (x1 −µ)2 f (x1 )+(x2 −µ)2 f (x2 )+(x3 −µ)2 f (x3 )+...
It is a description of how the distribution ”spreads”.
Note Var (X ) = E ((X − µ)2 ).
The alternative description works for the same reason as in the
continuous case. i.e.
Var (X ) = E (X 2 ) − E (X )2
Christopher Croke
Calculus 115
variance in the discrete case
If f (xi ) is the probability distribution function for a random
variable with range {x1 , x2 , x3 , ...} and mean µ = E (X ) then:
Var (X ) = σ 2 = (x1 −µ)2 f (x1 )+(x2 −µ)2 f (x2 )+(x3 −µ)2 f (x3 )+...
It is a description of how the distribution ”spreads”.
Note Var (X ) = E ((X − µ)2 ).
The alternative description works for the same reason as in the
continuous case. i.e.
Var (X ) = E (X 2 ) − E (X )2
The standard deviation has the same units as X . (I.e. if X is
measured in feet then so is σ.)
Christopher Croke
Calculus 115
Problem: Remember the game where players pick balls from an
urn with 4 white and 2 red balls. The first player is paid $2 if he
wins but the second player gets $3 if she wins. No one gets payed
if 4 white balls are chosen.
Christopher Croke
Calculus 115
Problem: Remember the game where players pick balls from an
urn with 4 white and 2 red balls. The first player is paid $2 if he
wins but the second player gets $3 if she wins. No one gets payed
if 4 white balls are chosen.
We have seen that the payout and probabilities for the first player
are:
Payout Probability
8
2
15
1
0
15
6
−3
15
Christopher Croke
Calculus 115
Problem: Remember the game where players pick balls from an
urn with 4 white and 2 red balls. The first player is paid $2 if he
wins but the second player gets $3 if she wins. No one gets payed
if 4 white balls are chosen.
We have seen that the payout and probabilities for the first player
are:
Payout Probability
8
2
15
1
0
15
6
−3
15
2
.
The expected value was µ = − 15
Christopher Croke
Calculus 115
Problem: Remember the game where players pick balls from an
urn with 4 white and 2 red balls. The first player is paid $2 if he
wins but the second player gets $3 if she wins. No one gets payed
if 4 white balls are chosen.
We have seen that the payout and probabilities for the first player
are:
Payout Probability
8
2
15
1
0
15
6
−3
15
2
. What is the variance?
The expected value was µ = − 15
Christopher Croke
Calculus 115
Problem: Remember the game where players pick balls from an
urn with 4 white and 2 red balls. The first player is paid $2 if he
wins but the second player gets $3 if she wins. No one gets payed
if 4 white balls are chosen.
We have seen that the payout and probabilities for the first player
are:
Payout Probability
8
2
15
1
0
15
6
−3
15
2
. What is the variance?
The expected value was µ = − 15
Christopher Croke
Calculus 115
This often makes it easier to compute since we can compute µ and
E (X 2 ) at the same time.
Christopher Croke
Calculus 115
This often makes it easier to compute since we can compute µ and
E (X 2 ) at the same time.
Problem: Compute E (X ) and Var (X ) where X is a random
variable with probability given by the chart below:
X Pr (X = x)
1
0.1
2
0.2
3
0.3
4
0.3
5
0.1
Christopher Croke
Calculus 115
This often makes it easier to compute since we can compute µ and
E (X 2 ) at the same time.
Problem: Compute E (X ) and Var (X ) where X is a random
variable with probability given by the chart below:
X Pr (X = x)
1
0.1
2
0.2
3
0.3
4
0.3
5
0.1
We will see later that for X a binomial random variable σ 2 = npq.
Christopher Croke
Calculus 115
This often makes it easier to compute since we can compute µ and
E (X 2 ) at the same time.
Problem: Compute E (X ) and Var (X ) where X is a random
variable with probability given by the chart below:
X Pr (X = x)
1
0.1
2
0.2
3
0.3
4
0.3
5
0.1
We will see later that for X a binomial random variable σ 2 = npq.
Problem: What is the variance of the number of hits for our
batter that bats .300 and comes to the plate 4 times?
Christopher Croke
Calculus 115