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ECON 240A
Power 5
1
Last Week
 Probability
 Discrete Binomial Probability Distribution
2
The Normal Distribution
3
Outline
 The normal distribution as an
approximation to the binomial
 The standardized normal variable, z
 sample means
4
Binomial Distribution
Five Flips of a Fair Coin
density
cumulative
k
n
p
0
5 0.03125 0.03125
1
5 0.15625
0.1875
2
5
0.3125
0.5
3
5
0.3125
0.8125
4
5 0.15625 0.96875
5
5 0.03125
1
Probability Density Function
Five Flips of a Fair Coin
0.35
0.3
Probability
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
Number of Heads
4
5
6
Cumulative Distribution Function
Five Flips of a Fair Coin
1
0.9
0.8
Probability
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of Heads
4
5
7
Cumulative Distribution Function
 The probability of getting two or less heads
in five flips is 0.5
• can use the cumulative distribution function
• can use the probability density function and
add the probabilities for 0, 1, and 2 heads
 the probability of getting two heads or
three heads is
• can add the probabilities for 2 heads and three
heads from the probability density function 8
Cumulative Distribution Function
Five Flips of a Fair Coin
1
0.9
0.8
Probability
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of Heads
4
5
9
Binomial Distribution
Five Flips of a Fair Coin
density
cumulative
k
n
p
0
5 0.03125 0.03125
1
5 0.15625
0.1875
2
5
0.3125
0.5
3
5
0.3125
0.8125
4
5 0.15625 0.96875
5
5 0.03125
1
Cumulative Distribution Function
 the probability of getting two heads or
three heads is:
• can add the probabilities for 2 heads and three
heads from the probability density function
• can use the probability of getting up to 3
heads, P(3 or less heads) from the cumulative
distribution function (CDF) and subtract the
probability of getting up to one head P(1 or less
heads]
11
Probability Density Function
Five Flips of a Fair Coin
0.35
0.3
Probability
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
Number of Heads
4
5
12
Binomial Distribution
Five Flips of a Fair Coin
density
cumulative
k
n
p
0
5 0.03125 0.03125
1
5 0.15625
0.1875
2
5
0.3125
0.5
3
5
0.3125
0.8125
4
5 0.15625 0.96875
5
5 0.03125
1
Binomial Distribution
Five Flips of a Fair Coin
density
cumulative
k
n
p
0
5 0.03125 0.03125
1
5 0.15625
0.1875
2
5
0.3125
0.5
3
5
0.3125
0.8125
4
5 0.15625 0.96875
5
5 0.03125
1
Cumulative Distribution Function
Five Flips of a Fair Coin
1
p[h  3] 0.9
0.8
Probability
0.7
p[h  1]
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of Heads
4
5
15
For the Binomial Distribution
 Can use a computer as we did in Lab Two
 Can use Tables for the cumulative
distribution function of the binomial such
as Table 1 in the text in Appendix B, p. B-1
• need a table for each p and n.
16
17
Normal Approximation to the
binomial
 Fortunately, for large samples, we can
approximate the binomial with the normal
distribution, as we saw in Lab Two
18
Binomial Probability Density Function
Forty Tosses of a Fair Coin
0.14
0.12
0.08
0.06
0.04
0.02
Number of Heads
39
36
33
30
27
24
21
18
15
12
9
6
3
0
0
Probabilty
0.1
Binomial Cumulative Distribution Function
Forty Tosses of a Fair Coin
1
0.9
0.8
0.6
0.5
0.4
0.3
0.2
0.1
Number of Heads
39
36
33
30
27
24
21
18
15
12
9
6
3
0
0
Probability
0.7
The Normal Distribution
 What would the normal density function
look like if it had the same expected value
and the same variance as this binomial
distribution
• from Power 4, E(h) = n*p =40*1/2=20
• from Power 4, VAR[h] = n*p*(1-p) = 40*1/2*1/2
=10
21
Normal Density Function, Mean 20, Variance 10
0.14
0.12
Density
0.1
0.08
0.06
0.04
0.02
0
0
10
20
30
Number of Heads
40
50
Comparing the Normal Density with the Binomial
Probability Distribution
0.14
0.12
binomial
normal
0.08
0.06
0.04
0.02
Number of Heads
39
36
33
30
27
24
21
18
15
12
9
6
3
0
0
Density
0.1
Comparing the Binomial and Normal Distribution
Functions
1
0.9
0.8
binomial
normal
0.6
0.5
0.4
0.3
0.2
0.1
Number of Heads
39
36
33
30
27
24
21
18
15
12
9
6
3
0
0
Probability
0.7
Comparing the Binomial and Normal (Mean = 19.5)
Cumulative Distribution Functions
1.2
binomial
normal
0.8
0.6
0.4
0.2
Number of Heads
39
36
33
30
27
24
21
18
15
12
9
6
3
0
0
Probability
1
Normal Approximation to the
Binomial: De Moivre
P(a  k  b)  P[(a – ½ - np)/ np(1  p)  z  [(b + ½ - np)/ np(1  p) ]
This is the probability that the number of heads will fall in the
interval a through b, as determined by the normal cumulative
distribution function, using a mean of n*p, and a standard deviation
equal to the square root of n*p*(1-p), i.e. the square root of the
variance of the binomial distribution. The parameter 1/2 is a
continuity correction since we are approximating a discrete function
with a continuous one, and was the motivation of using mean 19.5
instead of mean 20 in the previous slide. Visually, this seemed to
be a better approximation than using a mean of 20.
Guidelines for using the normal
approximation
 n*p>=5
 n*(1-p)>=5
27
The Standardized Normal Variate
 Z~N(0, 10]
 E[z} = 0
 VAR[Z] = 1
28
f ( z)  [1 / 2 ] * e
1/ 2[( z 0) /1]2
Density Function for the Standardized Normal Variate
0.45
0.4
0.35
Density
0.3
0.25
0.2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
Standard Deviations
2
3
4
5
Cumulative Distribution Function for a Standardized Normal
Variate
1
0.9
0.8
Probabilty
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-5
-4
-3
-2
-1
0
1
Standrd Deviations
2
3
4
5
Normal Variate x
x ~ N ( ,  )
2
E (x )  
VAR[ x]  
2
z  ( x   ) /  , orz *    x
– E(z) = 0
– VAR(z) = 1
f ( x)  (1 /  2 ) * e
1/ 2[( x   ) /  ]2
Normal Density Function, Mean 20, Variance 10
0.14
0.12
Density
0.1
0.08
0.06
0.04
0.02
0
0
10
20
30
Number of Heads
40
50
b
F (b) 
 f ( x)dx

Normal Cumulative Distribution Function, Mean=20, Variance = 10
1
0.9
0.8
Probability
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
10
20
b
30
Number of Heads
40
50
f ( x)  (1 /  2 ) * e
1/ 2[( x   ) /  ]2
Normal Density Function, Mean 20, Variance 10
0.14
0.12
Density
0.1
0.08
0.06
0.04
0.02
0
0
10
20
b
30
Number of Heads
40
50
For the Normal Distribution
 Can use a computer as we did in Lab Two
 Can use Tables for the cumulative
distribution function of the normal such as
Table 3 in the text in Appendix B, p. B-8
• need only one table for the standardized
normal variate Z.
35
36
Sample Means
37
Sample Mean Example
 Rate of return on UC Stock Index Fund
 return equals capital gains or losses plus
dividends
 monthly rate of return equals price this
month minus price last month, plus
dividends, all divided by the price last
month
 r(t) ={ p(t) -p(t-1) + d(t)}/p(t-1)
38
Rate of Return UC Stock Index Fund,
http://atyourservice.ucop.edu/
Date
Sept 2002
Oct 2002
Nov 2002
Dec 2002
Jan 2003
Feb 2003
March 2003
Rate
-3.51
7.84
5.13
-5.24
-2.51
-1.66
0.61
39
Table Cont.
Date
Rate
April 2003
8.43
May 2003
5.55
June 2003
1.34
July 2003
2.67
Aug 2003
2.26
40
Rate of Return UC Stock Index Fund
10
8
6
Rate
4
2
0
Jul-02 Sep- Oct-02 Dec- Jan-03 Mar-2
02
02
03
-4
-6
Date
May- Jun-03 Aug- Oct-03
03
03
Data Considerations
 Time series data for monthly rate of return
 since we are using the fractional change in
price (ignoring dividends) times 100 to
convert to %, the use of changes
approximately makes the observations
independent of one another
 in contrast, if we used price instead of
price changes, the observations would be
correlated, not independent
42
Cont.
 assume a fixed target, i.e. the central
tendency of the rate is fixed, not time
varying
 Assume the rate has some distribution, f,
2
f
(

,

)
other than normal: ri =
Aug '03
 sample mean:
r   r (i ) / 12  1.74
i  Sept '02
Rate of Return UC Stock Index Fund
10
8
6
Rate
4
2
1.74
0
Jul-02 Sep- Oct-02 Dec- Jan-03 Mar-2
02
02
03
-4
-6
Date
May- Jun-03 Aug- Oct-03
03
03
What are the properties of this
sample mean?
45
Note: Expected value of a constant, c, times a random
variable, x(i), where i indexes the observation
n
n
i 1
i 1
E[cx]   c * x(i ) * f [ x(i )]  c *  x(i ) * f [ x(i )]  c * E[ x(i )]
Note: Variance of a constant times a random variable
VAR[c*x] = E{cx - E[c*x]}2 = E{c*[x-Ex]}2 =
E{c2[x -Ex]2 } = c2 *E[x-Ex]2 = c2 *VARx
Properties of r , where, r (i) ~ f ( , )
2
 Expected value:
12
12
i 1
i 1
E (r )  E{ [r (i )] / 12}  1 / 12 E[r (i )]  (1 / 12) *12  
 Variance
12
VAR(r )  VAR{ r (i ) / 12}  (1 / 12)
i 1
12
2
2
2
2
VAR
[
r
(
i
)]

(
1
/
12
)
*
12



/ 12

i 1
Central Limit Theorem
 As the sample size grows, no matter what
the distribution, f, of the rate of return, r, the
distribution of the sample mean
approaches normality
48
 An interval for the sample mean
p(a  r  b)  p{[( a   ) /  n]  r  [(b   ) /  n ]}
The rate of return, ri , could be
distributed as uniform
f[r(i)]
r(i)
50
And yet for a large sample, the sample
mean will be distributed as normal
Cumulative Distribution Function for a Standardized Normal
Variate
1
0.9
b
0.8
Probabilty
0.7
0.6
0.5
0.4
a
0.3
0.2
0.1
0
-5
-4
-3
-2
-1
0
1
Standrd Deviations
2
3
4
5
51
Bottom Line
 We can use the normal distribution to
calculate probability statements about
sample means
52
 An interval for the sample mean
p(a  r  b)  p{[( a   ) /  n]  r  [(b   ) /  n ]}
calculate
choose
infer
?, Assume we know,
or use sample
standard deviation, s
Sample Standard Deviation
 If we use the sample standard deviation, s,
then for small samples, approximately less
then 100 observations, we use Student’s t
distribution instead of the normal
s
n
2
[
r
(
i
)

r
]
/( n  1)

i 1
t-distribution
Text p.253
Normal
compared to t
t distribution
as smple size
grows
55
Appendix B
Table 4
p. B-9
56