Distribusi Binomial, Poisson, dan Hipergeometrik

  

Distribusi

Binomial,

Poisson, dan

Hipergeometrik

CHAPTER TOPICS

• The Probability of a Discrete Random Variable  Covariance and Its Applications in Finance  Binomial Distribution  Poisson Distribution  Hypergeometric Distribution

  STAT I ST I K & PROBABI LI TAS

RANDOM VARIABLE

• Random Variable

   Outcomes of an experiment expressed numerically  E.g., Toss a die twice; count the number of times the number 4 appears (0, 1 or 2 times)

   E.g., Toss a coin; assign $10 to head and -$30 to a tail = $10 = -$30 _{STAT I ST I K & PROBABI LI TAS} T

DISCRETE RANDOM

  VARIABLE

• Discrete Random Variable

   Obtained by counting (0, 1, 2, 3, etc.)  Usually a finite number of different values  E.g., Toss a coin 5 times; count the number of tails (0, 1, 2, 3, 4, or 5 times)

  STAT I ST I K & PROBABI LI TAS

  DISCRETE PROBABILITY DISTRIBUTION EXAMPLE Event: Toss 2 Coins Count # Tails Probability Distribution Values Probability 1/4 = .25 T 1 2/4 = .50 2 1/4 = .25 T This is using the A Priori Classical

  T T _{STAT I ST I K & PROBABI LI TAS} Probability approach.

DISCRETE PROBABILITY DISTRIBUTION

  , P(X ) ] Pairs

• List of All Possible [X _{j j}

  = Value of random variable

   X _j

  ) = Probability associated with value

   P(X _j

•  Mutually Exclusive (Nothing in Common)

• Collective Exhaustive (Nothing Left Out)

   P X  _j

  1 P X  _j

  1

   

   STAT I ST I K & PROBABI LI TAS

SUMMARY MEASURES

• Expected Value (The Mean)

  

 Weighted average of the probability distribution

 E X

  X P X  

  µ   j j

   



_j

   E.g., Toss 2 coins, count the number of tails, compute expected value:

  X P X 

  µ _{j j}

 

   _j     0 .25 1 .5 2 .25

  1          _{STAT I ST I K & PROBABI LI TAS}

  

SUMMARY MEASURES

_(continued)

• Variance

   Weighted average squared deviation about the mean _{2 2 2}

   

  

      E

  X X P

  X σ µ µ

    _{j j}    

    

   E.g., Toss 2 coins, count number of tails, compute ₂ variance: ₂

   ^{X  P X} σ µ _{j j}

       _{2 2 2}

    ^{1 .2 5  1  1 .5  2  1 .2 5}

             

   _.5

_{STAT I ST I K & PROBABI LI TAS}

  

COVARIANCE AND ITS

_N

APPLICATION

  X  E X   Y  E Y  P X Y

  σ _{XY i i i i}            _{i  1} X : discrete random variable _th

  X : i outcome of _i

  X Y : discrete random variable _th Y : i outcome of Y _{i th} P X Y : probability of occurrence of the i  i i  _th

  X i outcome of an d the outcome of Y

_{STAT I ST I K & PROBABI LI TAS}

COMPUTING THE MEAN FOR

  INVESTMENT RETURNS Return per $1,000 for two types of investments _Investment

P(X ) P(Y ) Economic Condition Dow Jones Fund X Growth Stock Y

_{.2 .2 Recession -$100 -$200 i i .3 .3 Expanding Economy + 250 + 350} .5 .5 Stable Economy + 100 + 50

  E X    100 .2  100 .5  250 .3  $105

  µ   ^{X         }

  E Y    200 .2  50 .5  350 .3  $90

  µ   Y          _{STAT I ST I K & PROBABI LI TAS}

  

COMPUTING THE VARIANCE

FOR INVESTMENT RETURNS

_{P(X ) P(Y ) Economic Condition Dow Jones Fund X Growth Stock Y i i} Investment _{.5 .5 Stable Economy + 100 + 50} .2 .2 Recession -$100 -$200 .3 .3 Expanding Economy + 250 + 350 _{2 2 2}

  2  .2  100 105   .5 100 105   .3 250 105 

  σ _X           14, 725

   121.35 _{2 2} σ _{X 2 2} .2 200 90 .5 50 90 .3 350 90       

  σ _{Y         } 37, 900

  194.68  

_{STAT I ST I K & PROBABI LI TAS}

σ _Y

P(X

  

STAT I ST I K & PROBABI LI TAS

COMPUTING THE COVARIANCE

FOR INVESTMENT RETURNS

  _{.2 Recession -$100 -$200 i Y i}

^{) Economic Condition Dow Jones Fund X Growth Stock Y}

.5 Stable Economy + 100 + 50 _{.3 Expanding Economy + 250 + 350} ^Investment

             

  100 105 200 90 .2 100 105 50 90 .5 250 105 350 90 .3 23,300 ^XY

  σ

             

  The covariance of 23,000 indicates that the two investments are positively related and will vary together in the same direction.

VARIATION FOR INVESTMENT RETURNS

• Investment X appears to have a lower risk (variation) per unit of average payoff (return) than investment
• Investment X appears to have a higher average payoff (return) per unit of variation (risk) than investment Y

  STAT I ST I K & PROBABI LI TAS COMPUTING THE COEFFICIENT OF

  Y

    121.35 1.16 116%

  105 ^X _X CV X σ µ

       

  194.68 2.16 216%

  _Y 90 ^Y CV Y σ µ

     

• The expected value of the sum is equal to the sum of the expected values
• The variance of the sum is equal to the sum of the variances plus twice the covariance
• The standard deviation is the square root of the variance

  STAT I ST I K & PROBABI LI TAS SUM OF TWO RANDOM

  VARIABLES

        E X Y E X E Y   

    ^{2 2 2}

  2 _{X Y X Y XY} Var X Y

  σ σ σ σ _     

  _{X Y} σ σ _{ }

2 X Y

  

• The portfolio expected return for a two-asset

  investment is equal to the weighted average

  of the two assets
• Portfolio risk

  

STAT I ST I K & PROBABI LI TAS

PORTFOLIO EXPECTED

RETURN AND RISK

         

  1 where portion of the portfolio value assigned to asset

  E P wE X w E Y w

  X

     

      ^{2 2 2 2}

  1

  2

  1 _{P X Y XY} w w w w

  σ σ σ σ     

P(X

  

STAT I ST I K & PROBABI LI TAS

COMPUTING THE EXPECTED RETURN AND

RISK OF THE PORTFOLIO INVESTMENT

  _{.2 Recession -$100 -$200 i Y i} ^{) Economic Condition Dow Jones Fund X Growth Stock Y} .5 Stable Economy + 100 + 50 _{.3 Expanding Economy + 250 + 350} ^Investment

  

Suppose a portfolio consists of an equal investment in each of

X and Y:

        0.5 105 0.5 90

  97.5 E P   

           

^{2 2}

  0.5 14725 0.5 37900 2 0.5 0.5 23300 157.5 _P σ

     

DOING IT IN PHSTAT

• PHStat | Decision Making | Covariance and Portfolio Analysis

   Fill in the “Number of Outcomes:”  Check the “Portfolio Management Analysis” box  Fill in the probabilities and outcomes for investment X and Y  Manually compute the CV using the formula in the previous slide

• Here is the Excel spreadsheet that contains the results of the previous investment example:

  Microsoft Excel _{Worksheet STAT I ST I K & PROBABI LI TAS}

  STAT I ST I K & PROBABI LI TAS

  IMPORTANT DISCRETE PROBABILITY DISTRIBUTIONS Discrete Probability Distributions Binomial Hypergeometric Poisson

• ‘n’ Identical Trials

•  2 Mutually Exclusive Outcomes on Each Trial

• Trials are Independent

  STAT I ST I K & PROBABI LI TAS BINOMIAL PROBABILITY DISTRIBUTION

   E.g., 15 tosses of a coin; 10 light bulbs taken from a warehouse

   E.g., Heads or tails in each toss of a coin; defective or not defective light bulb

   The outcome of one trial does not affect the outcome of the other

  

BINOMIAL PROBABILITY

DISTRIBUTION

_(continued)

• Constant Probability for Each Trial

   E.g., Probability of getting a tail is the same each time we toss the coin

• 2 Sampling Methods

   Infinite population without replacement  Finite population with replacement

  

STAT I ST I K & PROBABI LI TAS

BINOMIAL PROBABILITY DISTRIBUTION FUNCTION

  _ n n X

  ! _X P X  p 1  p  

    X n

  X !  !

    P X : probability of successes given and

  X n p

    X : number of "successes" in sample X  0,1, , n

     p

  : the probability of each "success" _{Tails in 2 Tosses of Coin} n : sample size _{X P(X) 1/4 = .25 2 1/4 = .25} 1 2/4 = .50

  STAT I ST I K & PROBABI LI TAS

  

BINOMIAL DISTRIBUTION

CHARACTERISTICS

• Mean

   E X  np µ

   

  

  np   5 .1  .5

  µ  

   E.g., _{P(X) .6} n = 5 p = 0.1

• Variance and _.4 Standard Deviation
_{2 .2}  np 1  p ^X

  σ  

   _{1 2 3 4 5}

   np 1  p σ

     np 1  p  5 .1 1 .1   .6708

  σ

   E.g.,     

  

STAT I ST I K & PROBABI LI TAS

BINOMIAL DISTRIBUTION IN PHSTAT

• PHStat | Probability & Prob. Distributions | Binomial  Example in Excel Spreadsheet

  Microsoft Excel _Worksheet STAT I ST I K & PROBABI LI TAS

EXAMPLE: BINOMIAL DISTRIBUTION

  A mid-term exam has 30 multiple choice questions, each with 5 possible answers. What is

the probability of randomly guessing the answer

for each question and passing the exam (i.e.,

having guessed at least 18 questions correctly)?

Are the assumptions for the binomial distribution met? Yes, the assumptions are met. _{Microsoft Excel} Using results from PHStat: _{n  3 0 p  0 .2  6} Worksheet P ^{X  1 8  1 .8 4 2 4 5 1 0} _{STAT I ST I K & PROBABI LI TAS}    

  STAT I ST I K & PROBABI LI TAS POISSON DISTRIBUTION Siméon Poisson

POISSON DISTRIBUTION

• Discrete events (“successes”) occurring in a given area of opportunity (“interval”)  “Interval” can be time, length, surface area, etc.
• The probability of a “success” in a given “interval” is the same for all the “intervals”
• The number of “successes” in one “interval” is independent of the number of “successes” in other “intervals”
• The probability of two or more “successes” occurring in an “interval” approaches zero as the “interval” becomes smaller

   E.g., # customers arriving in 15 minutes  E.g., # defects per case of light bulbs _{STAT I ST I K & PROBABI LI TAS}

  

STAT I ST I K & PROBABI LI TAS

POISSON PROBABILITY

DISTRIBUTION FUNCTION

      ! : p ro b ab ility o f "su ccesses" g iven

  : n u m b er o f "su ccesses" p er u n it : ex p ected (averag e) n u m b er o f "su ccesse s" : 2 .7 1 8 2 8 (b ase o f n atu ral lo g s) ^X e P X

  X P X

  X X e λ

  λ λ λ ^

   E.g., Find the probability of 4 customers arriving in 3 minutes when the mean is 3.6.

    ^{3.6 4}

  3.6 .1912 4! e P X ^

   

• Example in Excel Spreadsheet

  STAT I ST I K & PROBABI LI TAS POISSON DISTRIBUTION IN PHSTAT  PHStat | Probability & Prob.

  Distributions | Poisson

  Microsoft Excel _Worksheet P X x x ^x

  ( | !   

   e

  

POISSON DISTRIBUTION

CHARACTERISTICS

P( ^{X )}
• Mean

  = 0.5 _{.4 .6} λ

  

   E X  µ λ _N   _{.2 X}

  1 ^{2 3 4 5} 

  X P X _{i i}    _{i  1 P( X )}

  = 6 _.6 λ

• Standard Deviation _{.2 .4} and Variance
_{2 X}  

  σ λ σ λ

   _{2 4 6 8 10} STAT I ST I K & PROBABI LI TAS

HYPERGEOMETRIC DISTRIBUTION

• “n” Trials in a Sample Taken from a Finite Population of Size N  Sample Taken Without Replacement  Trials are Dependent  Concerned with Finding the Probability of “X” Successes in the Sample Where There are “A” Successes in the Population

  STAT I ST I K & PROBABI LI TAS

  

HYPERGEOMETRIC

DISTRIBUTION FUNCTION

A N  A

      E.g., 3 Light bulbs were selected     from 10. Of the 10, there were 4

   X n

  X    

  P X    defective. What is the probability

  N   that 2 of the 3 selected are

    n

    defective?

  P X : probability that X successes given , n N , and A   n

  : sam ple size _{4 6}    

  N : population size

      _{2 1}     A : num ber of "successes" in population _{P 2   .30}

    ₁₀

    X : num ber of "successes" in sam ple

    ₃   X 

  0,1, 2, , n   

_{STAT I ST I K & PROBABI LI TAS}

HYPERGEOMETRIC DISTRIBUTION CHARACTERISTICS

• Mean

  

A

E X n  

  

  µ  

  N

• Variance and Standard Deviation

  Finite nA N  A N

   n ₂  

  

  Population 

  σ ₂ N N 

  1 Correction Factor nA N  A

   N n

    

  σ _{2 STAT I ST I K & PROBABI LI TAS} N N 

  1

HYPERGEOMETRIC DISTRIBUTION

  IN PHStat

•  PHStat | Probability & Prob. Distributions |

  Hypergeometric …
• Example in Excel Spreadsheet

  Microsoft Excel _Worksheet STAT I ST I K & PROBABI LI TAS

  Distribusi Normal