1 ECON 240A Power 5. 2 Last Tuesday & Lab Two b Probability b Discrete Binomial Probability...

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ECON 240AECON 240A

Power 5Power 5

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Last Tuesday & Lab TwoLast Tuesday & Lab Two

ProbabilityProbability Discrete Binomial Probability Discrete Binomial Probability

DistributionDistribution

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The Normal DistributionThe Normal Distribution

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OutlineOutline

The cumulative distribution function, The cumulative distribution function, or sigmoid (S-shaped curve)or sigmoid (S-shaped curve)

The normal distribution as an The normal distribution as an approximation to the binomialapproximation to the binomial

The standardized normal variable, zThe standardized normal variable, z sample meanssample means The distribution of the sample mean: The distribution of the sample mean:

the normal distributionthe normal distribution

Binomial Distribution

Five Flips of a Fair Coindensity cumulative

k n p0 5 0.03125 0.031251 5 0.15625 0.18752 5 0.3125 0.53 5 0.3125 0.81254 5 0.15625 0.968755 5 0.03125 1

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Probability Density Probability Density FunctionFunction

Five Flips of a Fair Coin

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0 1 2 3 4 5

Number of Heads

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Cumulative Distribution Cumulative Distribution FunctionFunction


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The probability of getting two or less The probability of getting two or less heads in five flips is 0.5heads in five flips is 0.5• can use the cumulative distribution functioncan use the cumulative distribution function• can use the probability density function and can use the probability density function and

add the probabilities for 0, 1, and 2 headsadd the probabilities for 0, 1, and 2 heads the probability of getting two heads or the probability of getting two heads or

three heads isthree heads is• can add the probabilities for 2 heads and can add the probabilities for 2 heads and

three heads from the probability density three heads from the probability density function function

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k n p0 5 0.03125 0.031251 5 0.15625 0.18752 5 0.3125 0.53 5 0.3125 0.81254 5 0.15625 0.968755 5 0.03125 1

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the probability of getting two heads or the probability of getting two heads or three heads is:three heads is:• can add the probabilities for 2 heads and can add the probabilities for 2 heads and

three heads from the probability density three heads from the probability density function function

• can use the probability of getting up to 3 can use the probability of getting up to 3 heads, P(3 or less heads) from the heads, P(3 or less heads) from the cumulative distribution function (CDF) and cumulative distribution function (CDF) and subtract the probability of getting up to subtract the probability of getting up to one head P(1 or less heads]one head P(1 or less heads]

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Probability Density Probability Density FunctionFunction


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k n p0 5 0.03125 0.031251 5 0.15625 0.18752 5 0.3125 0.53 5 0.3125 0.81254 5 0.15625 0.968755 5 0.03125 1

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0

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]3[ hp

]1[ hp

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For the Binomial For the Binomial DistributionDistribution

Can use a computer as we did in Can use a computer as we did in Lab TwoLab Two

Can use Tables for the cumulative Can use Tables for the cumulative distribution function of the distribution function of the binomial such as Table 1 in the binomial such as Table 1 in the text in Appendix B, p. B-1text in Appendix B, p. B-1• need a table for each p and n.need a table for each p and n.

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Normal Approximation to Normal Approximation to the binomialthe binomial

Fortunately, for large samples, we Fortunately, for large samples, we can approximate the binomial with can approximate the binomial with the normal distribution, as we saw the normal distribution, as we saw in Lab Twoin Lab Two

Forty Tosses of a Fair Coin

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Binomial Probability Density Function

Forty Tosses of a Fair Coin

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Number of Heads

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Binomial Cumulative Distribution Function

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The Normal DistributionThe Normal Distribution

What would the normal density What would the normal density function look like if it had the same function look like if it had the same expected value and the same expected value and the same variance as this binomial variance as this binomial distributiondistribution• from Power 4, E(h) = n*p =40*1/2=20from Power 4, E(h) = n*p =40*1/2=20• from Power 4, VAR[h] = n*p*(1-p) = from Power 4, VAR[h] = n*p*(1-p) =

40*1/2*1/2 =1040*1/2*1/2 =10

Normal Density Function, Mean 20, Variance 10

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Comparing the Normal Density with the Binomial Probability Distribution

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binomial

normal

Comparing the Binomial and Normal Distribution Functions

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binomial

normal

Comparing the Binomial and Normal (Mean = 19.5) Cumulative Distribution Functions

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1.20 3 6 9 12 15 18 21 24 27 30 33 36 39

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binomial

normal

Normal Approximation to Normal Approximation to the Binomial: De Moivrethe Binomial: De Moivre

P(a k b) P[(a – ½ - np)/ )1(pnp z [(b + ½ - np)/ )1(pnp]

This is the probability that the number of heads will fall in theinterval 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 thevariance of the binomial distribution. The parameter 1/2 is a continuity correction since we are approximating a discrete functionwith a continuous one, and was the motivation of using mean 19.5instead of mean 20 in the previous slide. Visually, this seemed to be a better approximation than using a mean of 20.

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Guidelines for using the Guidelines for using the normal approximationnormal approximation

n*p>=5n*p>=5 n*(1-p)>=5n*(1-p)>=5

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The Standardized Normal The Standardized Normal VariateVariate

Z~N(0, 1.0]Z~N(0, 1.0] E[z} = 0E[z} = 0 VAR[Z] = 1VAR[Z] = 1

Density Function for the Standardized Normal Variate

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Standard Deviations

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2]1/)0[(2/1*]2/1[)( zezf

Cumulative Distribution Function for a Standardized Normal Variate

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Standrd Deviations

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Normal Variate xNormal Variate x

– E(z) = 0E(z) = 0– VAR(z) = 1VAR(z) = 1

),(~ 2Nx

)(xE

2][ xVAR

xorzxz *,/)(


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2]/)[(2/1*)2/1()( xexf

Normal Cumulative Distribution Function, Mean=20, Variance = 10

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b

dxxfbF )()(

b


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2]/)[(2/1*)2/1()( xexf

b

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For the Normal For the Normal DistributionDistribution

Can use a computer as we did in Can use a computer as we did in Lab TwoLab Two

Can use Tables for the cumulative Can use Tables for the cumulative distribution function of the normal distribution function of the normal such as Table 3 in the text in such as Table 3 in the text in Appendix B, p. B-8Appendix B, p. B-8• need only one table for the need only one table for the

standardized normal variate Z.standardized normal variate Z.

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Sample MeansSample Means

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Sample Mean ExampleSample Mean Example

Rate of return on UC Stock Index FundRate of return on UC Stock Index Fund return equals capital gains or losses return equals capital gains or losses

plus dividendsplus dividends monthly rate of return equals price monthly rate of return equals price

this month minus price last month, this month minus price last month, plus dividends, all divided by the plus dividends, all divided by the price last monthprice last month

r(t) ={ p(t) -p(t-1) + d(t)}/p(t-1)r(t) ={ p(t) -p(t-1) + d(t)}/p(t-1)

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Rate of Return UC Stock Rate of Return UC Stock Index Fund,Index Fund, http://atyourservice.ucop.http://atyourservice.ucop.edu/edu/

Date Rate Sept 2003 -0.29 Oct 2003 5.84 Nov 2003 1.40 Dec 2003 5.01 Jan 2004 1.95 Feb 2004 1.40 March 2004 -0.85

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Table Cont.Table Cont.

Date Rate

April 2004 -2.19

May 2004 1.36

June 2004 1.02

July 2004 -3.51

Aug 2004 0.45

Rate of Return UC Stock Index Fund

-4

-2

0

2

4

6

8

Sep-03 Oct-03 Nov-03

Dec-03 J an-04 Feb-04 Mar-04 Apr-04 May-04

J un-04 J ul-04 Aug-04

Date

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Rate of Return UC Stock Rate of Return UC Stock Index Fund,Index Fund, http://atyourservice.ucop.http://atyourservice.ucop.edu/edu/

Date RateSept 2002 -3.51Oct 2002 7.84Nov 2002 5.13Dec 2002 -5.24Jan 2003 -2.51Feb 2003 -1.66March 2003 0.61

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Table Cont.Table Cont.

Date Rate

April 2003 8.43

May 2003 5.55

June 2003 1.34

July 2003 2.67

Aug 2003 2.26


-6

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-2

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Oct-02 Dec-02

Jan-03 Mar-03

May-03

Jun-03 Aug-03

Oct-03

Date

Rat

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Data ConsiderationsData Considerations

Time series data for monthly rate of Time series data for monthly rate of returnreturn

since we are using the fractional change since we are using the fractional change in price (ignoring dividends) times 100 in price (ignoring dividends) times 100 to convert to %, the use of changes to convert to %, the use of changes approximately makes the observations approximately makes the observations independent of one anotherindependent of one another

in contrast, if we used price instead of in contrast, if we used price instead of price changes, the observations would price changes, the observations would be correlated, not independentbe correlated, not independent

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, other than normal: ri

= sample mean:

),( 2f

74.112/)(03'

02'

Aug

Septi

irr

Last 12 months, mean = 1.05


-6

-4

-2

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2

4

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8

10

Jul-02 Sep-02

Oct-02 Dec-02

Jan-03 Mar-03

May-03

Jun-03 Aug-03

Oct-03

Date

Rat

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1.74

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What are the properties of What are the properties of this sample mean?this sample mean?

n

i

n

i

ixEcixfixcixfixccxE11

)]([*)]([*)(*)]([*)(*][

Note: Expected value of a constant, c, times a random variable, x(i), where i indexes the observation

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

Last Time

Properties of

Expected value:

Variance

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1

12

1

12*)12/1()]([12/1}12/)]([{)(i i

irEirErE

12

1

12

1

2222 12/12*)12/1()]([)12/1(}12/)({)(i i

irVARirVARrVAR

),(~)(,, 2firwherer

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Central Limit TheoremCentral Limit Theorem

As the sample size grows, no As the sample size grows, no matter what the distribution, f, of matter what the distribution, f, of the rate of return, r, the the rate of return, r, the distribution of the sample mean distribution of the sample mean approaches normalityapproaches normality

An interval for the sample mean

]}/)[(]/){[()( nbrnapbrap

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The rate of return, rThe rate of return, ri i , could be distributed as uniform

f[r(i)]

r(i)

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And yet for a large sample, And yet for a large sample, the sample mean will be the sample mean will be distributed as normaldistributed as normal

Cumulative Distribution Function for a Standardized Normal Variate

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Standrd Deviations

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Bottom LineBottom Line

We can use the normal distribution We can use the normal distribution to calculate probability statements to calculate probability statements about sample meansabout sample means

An interval for the sample mean

]}/)[(]/){[()( nbrnapbrap

calculate

chooseinfer

?, 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

n

i

nrirs1

2 )1/(])([

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Text p.253Normal compared to t

t-distribution

t distributionas smple size grows

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Appendix BTable 4p. B-9

1 ECON 240A Power 5. 2 Last Tuesday & Lab Two b Probability b Discrete Binomial Probability...

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