Download Population Standard Deviation of a Discrete Random Variable

CHAPTER 16
Standard Deviation of a
Discrete Random Variable
First center (expected
value)
Now - spread
Standard Deviation of a
Discrete Random Variable
Measures how “spread out”
the random variable is
Summarizing data and
probability
Data
Histogram
measure of the
center: sample mean
x
measure of spread:
sample standard
deviation s
Random variable
Probability Histogram
measure of the
center: population
mean m
measure of spread:
population standard
deviation s
Example
x
0
100
p(x)
1/2 1/2
E(x) = 0(1/2) + 100(1/2) = 50
y
49 51
p(y)
1/2 1/2
E(y) = 49(1/2) + 51(1/2) = 50
Variance
Variation
n
s2 =
 (X
i
 X) 2
i=1
n-1
=
1805.703
= 53.1089
34
The deviations of the outcomes from the
mean of the probability distribution
xi - µ
Xi - X
s2 (sigma squared) is the variance of the
probability distribution
Variance
Variation
s2
k
=
n
s2 =
 (X
i
 X) 2
i=1
n-1
2
(
x

m
)
 P( X = x i )
 i
i =1
=
1805.703
= 53.1089
34
Economic
Scenario
Variation
s2
Profit X
($ Millions)
Probability P
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k
=
2
(
x

m
)
 P( X = x i )
 i
i =1
Example 3.65
3.65
3.65
s2 = (x1-µ)2 · P(X=x1) + (x2-µ)2 · P(X=x2) +
3.65
(x3-µ)2 · P(X=x3) + (x4-µ)2 · P(X=x4)
= (10-3.65)2 · 0.20 + (5-3.65)2 · 0.40 +
(1-3.65)2 · 0.25 + (-4-3.65)2 · 0.15 =
19.3275
P. 207, Handout 4.1, P. 4
Standard Deviation: of
More Interest then the
Variance
The population standard deviation is the square root
of the population variance
s  s

Standard Deviation
Standard
Deviation
Standard Deviation (s) =
Positive Square Root of the Variance
s =
s2
s2 = 19.3275
s, or SD, is the standard deviation of the
probability distribution
s (or SD) = s
2
s (or SD) = 19.3275  4.40 ($ mil.)
Probability Histogram
Probability
.40
.35
.30
.25
.20
.15
Lousy
s = 4.40
Good
OK
Great
.10
.05
-4
-2
0
2
4
Profit
µ=3.65
6
8
10
12
Finance and Investment
Interpretation
X = return on an investment (stock,
portfolio, etc.)
E(x) = m  expected return on this
investment
s is a measure of the risk of the
investment
Example
x
s2
0
1
1
3
p( x)
8
8
Compute the variance
k
=
2
(
x

m
)
 P( X = x i )
 i
i =1
2
3
8
:
3
1
8
m 
s 2  (0  1.5) 2  18   (1  1.5) 2  83   (2  1.5) 2  83   (3  1.5) 2  18 
 2.25 18   .25 83   .25 83   2.25 18 
 .75.
s  s

 .75  .866
Example (cont.)
m, s866
Specify the interval (m  s, m  s)
(1.5 - 1(.866), 1.5 + 1(.866))
(1.5 - .866, 1.5 + .866)
(.634, 2.366)
P(.634  x  2.366) =
p(1)+p(2)=3/8 + 3/8 = 3/4
68-95-99.7 Rule for
Random Variables
For random variables x whose probability
histograms are approximately moundshaped:
Pm  s  x  m  s  68
Pm  s  x  m  s  9
P(m 3s  x  m  3s  997
Rules for E(X), Var(X) and SD(X):
adding a constant a
If X is a rv and a is Example: a = -1
a constant:
 E(X+a) = E(X)+a
 E(X+a)=E(X-1)=E(X)-1
Rules for E(X), Var(X) and SD(X):
adding constant a (cont.)
Var(X+a) = Var(X)
SD(X+a) = SD(X)
Example: a = -1
 Var(X+a)=Var(X-1)=Var(X)
 SD(X+a)=SD(X-1)=SD(X)
Economic Profit X
Scenario ($ Millions)
Probability P
Economic Profit X+2
Scenario ($ Millions)
Great
x1 10
P(X=x1) 0.20
Great
Good
x2 5
P(X=x2) 0.40
OK
x3 1
Lousy
x4 -4
Probability P
P(X=x1) 0.20
Good
x1+ 10+2
2
x2+2 5+2
P(X=x3) 0.25
OK
x3+2 1+2
P(X=x3) 0.25
P(X=x4) 0.15
Lousy
x4+2 -4+2
P(X=x4) 0.15
P(X=x2) 0.40
E(x + a) = E(x) + a; Var(x + a)=Var (x); let a = 2
s = 4.40
Probability
-4
-2
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
0
2
4
Profit
m3.65
6
8
10
12
14
s = 4.40
Probability
-4
-2
0
2
4
6
8
Profit
m5.65
10
12
14
New Expected Value
Long (UNC-CH) way:
E(x+2)=12(.20)+7(.40)+3(.25)+(-2)(.15)
= 5.65
Smart (NCSU) way:
a=2; E(x+2) =E(x) + 2 = 3.65 + 2 = 5.65
New Variance and SD
Long (UNC-CH) way: (compute from
“scratch”)
Var(X+2)=(12-5.65)2(0.20)+…
+(-2+5.65)2(0.15) = 19.3275
SD(X+2) = √19.3275 = 4.40
Smart (NCSU) way:
Var(X+2) = Var(X) = 19.3275
SD(X+2) = SD(X) = 4.40
Rules for E(X), Var(X) and SD(X):
multiplying by constant b
E(bX)=b E(X)
 Var(b X) = b2Var(X)
 SD(bX)= |b|SD(X)
 Example: b =-1
 E(bX)=E(-X)=-E(X)
 Var(bX)=Var(-1X)=
=(-1)2Var(X)=Var(X)
 SD(bX)=SD(-1X)=
=|-1|SD(X)=SD(X)
Expected Value and SD of Linear
Transformation a + bx
Let X=number of repairs a new computer needs each year.
Suppose E(X)= 0.20 and SD(X)=0.55
The service contract for the computer offers unlimited repairs
for $100 per year plus a $25 service charge for each repair.
What are the mean and standard deviation of the yearly cost of
the service contract?
Cost = $100 + $25X
E(cost) = E($100+$25X)=$100+$25E(X)=$100+$25*0.20=
= $100+$5=$105
SD(cost)=SD($100+$25X)=SD($25X)=$25*SD(X)=$25*0.55=
=$13.75
Addition and Subtraction Rules
for Random Variables
 E(X+Y) = E(X) + E(Y);
 E(X-Y) = E(X) - E(Y)
 Var(X−Y)=Var(X)+Var(Y)
 SD(X −Y)= Var ( X )  Var (Y )
 When X and Y are
independent random
variables:
 Var(X+Y)=Var(X)+Var(Y)
 SD(X+Y)= Var ( X )  Var (Y )
SD’s do not subtract:
SD(X−Y)≠ SD(X)−SD(Y)
SD(X−Y)≠ SD(X)+SD(Y)
SD’s do not add:
SD(X+Y)≠ SD(X)+SD(Y)
In general, if X and Y are RVs,
Var(X+Y) ≠ Var(X)+Var(Y)
SD(X+Y) ≠ SD(X)+SD(Y)
Special case (previous slide): If RVs are
Independent:
Variances add: Var(X+Y)=Var(X)+Var(Y)
but…
Standard Deviations still DO NOT add:
SD(X+Y)≠SD(X)+SD(Y)
Pythagorean Theorem of Statistics: variances add;
standard deviations do not add
a2 + b2 = c 2
Var(X)+Var(Y)=Var(X+Y)
c2
Var(X)
a2
Var(X+Y)
a
c
SD(X+Y)
SD(X)
b
SD(Y)
b2
Var(Y)
a+b≠c
SD(X)+SD(Y) ≠SD(X+Y)
Pythagorean Theorem of Statistics: variances add;
standard deviations do not add
32 + 42 = 52
Var(X)+Var(Y)=Var(X+Y)
25
Var(X)
9
Var(X+Y)
3
5
SD(X+Y)
SD(X)
4
SD(Y)
16
Var(Y)
3+4≠5
SD(X)+SD(Y) ≠SD(X+Y)
Motivation for
Var(X-Y)=Var(X)+Var(Y)
 Let X=amount automatic dispensing machine
puts into your 16 oz drink (say at McD’s)
 A thirsty, broke friend shows up.
Let Y=amount you pour into friend’s 8 oz cup
 Let Z = amount left in your cup; Z = ?
 Z = X-Y
Has 2 +
components
Var(Y)
 Var(Z) = Var(X-Y) = Var(X)
For random variables, X+X≠2X
 Let X be the annual payout on a life insurance policy.
From mortality tables E(X)=$200 and SD(X)=$3,867. If
the payout amounts are doubled, what are the new
expected value and standard deviation?
The risk to the
 Double payout is 2X. E(2X)=2E(X)=2*$200=$400
insurance co. when
 SD(2X)=2SD(X)=2*$3,867=$7,734 doubling the payout
 Suppose insurance policies are sold to 2 people.
The
(2X) is not
the same
annual payouts are X1 and X2. Assume theas2the
people
risk when
behave independently. What are the expected
selling value
policies to 2
and standard deviation of the total payout?
people.
 E(X1 + X2)=E(X1) + E(X2) = $200 + $200 = $400
SD(X1 + X2 )= Var ( X1  X 2 )  Var ( X1 )  Var ( X 2 )
 (3867)2  (3867)2  14,953,689  14,953,689
 29,907,378  $5,468.76