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- 1. Probability Distributions Istanbul Bilgi University FEC 512 Financial Econometrics-I Asst. Prof. Dr. Orhan Erdem
- 2. Some Common Probability Distributions Probability Distributions Discrete Continuous Probability Probability Distributions Distributions Normal Binomial Uniform Poisson Lognormal Lecture 3-2 FEC 512 Probability Distributions
- 3. Discrete Probability Distribution Example random variable X = total number of Experiment: toss 2 coins, tails in two tosses. Probability distribution X Probability T 0 0.25 1 0.50 T 2 0.25 T T Lecture 3-3 FEC 512 Probability Distributions
- 4. The Binomial Distribution Probability Distributions Discrete Probability Distributions Binomial Poisson Lecture 3-4 FEC 512 Probability Distributions
- 5. The Binomial Distribution Characteristics of the Binomial Distribution: A trial has only two possible outcomes – “success” or “failure” There is a fixed number, n, of identical trials The trials of the experiment are independent of each other The probability of a success, p, remains constant from trial to trial If p represents the probability of a success, then (1-p) = q is the probability of a failure Lecture 3-5 FEC 512 Probability Distributions
- 6. Binomial Distribution Settings A manufacturing plant labels items as either defective or acceptable A firm bidding for a contract will either get the contract or not A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer or reject it Lecture 3-6 FEC 512 Probability Distributions
- 7. Counting Rule for Combinations A combination is an outcome of an experiment where x objects are selected from a group of n objects n! C= n x x! (n − x )! where: n! =n(n - 1)(n - 2) . . . (2)(1) x! = x(x - 1)(x - 2) . . . (2)(1) 0! = 1 (by definition) Lecture 3-7 FEC 512 Probability Distributions
- 8. Binomial Distribution Formula n! x n−x P(x) = pq x ! (n − x )! P(x) = probability of x successes in n trials, with probability of success p on each trial Example: Flip a coin four times, let x = # heads: x = number of ‘successes’ in sample, n=4 (x = 0, 1, 2, ..., n) p = 0.5 p = probability of “success” per trial q = (1 - .5) = .5 q = probability of “failure” = (1 – p) n = number of trials (sample size) x = 0, 1, 2, 3, 4 Lecture 3-8 FEC 512 Probability Distributions
- 9. Binomial Distribution The shape of the binomial distribution depends on the values of p and n P(X) n = 5 p = 0.1 Mean .6 .4 .2 Here, n = 5 and p = .1 0 X 0 1 2 3 4 5 n = 5 p = 0.5 P(X) .6 .4 Here, n = 5 and p = .5 .2 X 0 0 1 2 3 4 5 Lecture 3-9 FEC 512 Probability Distributions
- 10. Binomial Distribution Characteristics Mean µ = E(x) = np Variance and Standard Deviation σ = npq 2 σ = npq Where n = sample size p = probability of success q = (1 – p) = probability of failure Lecture 3-10 FEC 512 Probability Distributions
- 11. Binomial Characteristics Examples Mean = (5)(.1) = 0.5 µ = np n = 5 p = 0.1 P(X) .6 .4 σ = npq = (5)(.1)(1 − .1) .2 = 0.6708 0 X 0 1 2 3 4 5 µ = np = (5)(.5) = 2.5 n = 5 p = 0.5 P(X) .6 .4 σ = npq = (5)(.5)(1 − .5) .2 = 1.118 X 0 0 1 2 3 4 5 Lecture 3-11 FEC 512 Probability Distributions
- 12. A binomial tree of asset prices “Example” We wish to know the value of an asset after two time periods. Each of the time periods the asset may rise (a success) with a probability of 0.5 or it may fall (a failure) with a probability of 0.5. Assume asset price movement in one time period is independent of that in the other time period. Su2 Su (60.50) (55) Sud=Sdu (49.5) S=50 Sd Sd2 (45) (40.50) T2 T0 T1 Lecture 3-12 FEC 512 Probability Distributions
- 13. A binomial tree of asset prices “Example” If the asset had previously risen by a factor u, it would either rise again by u to Su2 or would fall by d to Sud. If the asset had previoulsy fallen to Sd, it could rise by u to Sud or fall further to Sd2 Suppose that u=1.1, d=0.9, and S=50 Hence the expected value can be calculated as: µ = (60.50 * 0.25) + (49.50 * 0.50) + (40.50 * 0.25) = 50 The variance is σ2 = (60.5 – 50)2 * 0.25 + (49.50 – 50)2 * 0.50 + (40.5 – 50)2 * 0.25 = 50.25 Lecture 3-13 FEC 512 Probability Distributions
- 14. The Poisson Distribution Probability Distributions Discrete Probability Distributions Binomial Poisson Lecture 3-14 FEC 512 Probability Distributions
- 15. The Poisson Distribution To use binomial distribution, we must be able to count the # successes and failures. Although in many situations you may be able to count # successes, you often cannot count # failures. Example: An emergency call center could easily count the # calls its unit respond to in 1 hour, but how could it determine how many calls it didnt receive? Lecture 3-15 FEC 512 Probability Distributions
- 16. Characteristics of the Poisson Distribution The outcomes of interest are rare relative to the possible outcomes The average number of outcomes of interest per time or space interval is λ The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest The probability of that an outcome of interest occurs in a given segment is the same for all segments Lecture 3-16 FEC 512 Probability Distributions
- 17. Poisson Distribution Formula − λt ( λt ) e x P( x ) = x! where: t = size of the segment of interest x = number of successes in segment of interest λ = expected number of successes in a segment of unit size e = base of the natural logarithm system (2.71828...) Lecture 3-17 FEC 512 Probability Distributions
- 18. Poisson Distribution (continued) Ex. Page requests arrive at a 0,18 0,17 webserver at an average rate 0,16 of 5 every second. If the 0,15 number of requests in a 0,14 0,13 second has a Poisson 0,12 distribution, find the probability 0,11 0,10 that 15 requests will be made p(X=x) 0,09 0,08 in a given second. 0,07 e −5 515 0,06 P ( X = 15 ) = = 0.00016 0,05 0,04 15! 0,03 0,02 0,01 0,00 Here is what the distribution 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 function for the above example x=number of page requests in a second looks like Lecture 3-18 FEC 512 Probability Distributions
- 19. Poisson Distribution Characteristics Mean µ = λt Variance and Standard Deviation σ = λt 2 σ = λt λ = number of successes in a segment of unit size where t = the size of the segment of interest Lecture 3-19 FEC 512 Probability Distributions
- 20. Graph of Poisson Probabilities 0.70 Graphically: 0.60 λ = .05 and t = 100 0.50 λt = 0.40 P(x) X 0.50 0.30 0 0.6065 0.20 1 0.3033 0.10 2 0.0758 0.00 3 0.0126 0 1 2 3 4 5 6 7 4 0.0016 x 5 0.0002 P(x = 2) = .0758 6 0.0000 7 0.0000 Lecture 3-20 FEC 512 Probability Distributions
- 21. Poisson Distribution Shape The shape of the Poisson Distribution depends on the parameters λ and t: λt = 0.50 λt = 3.0 0.70 0.25 0.60 0.20 0.50 0.15 0.40 P(x) P(x) 0.30 0.10 0.20 0.05 0.10 0.00 0.00 1 2 3 4 5 6 7 8 9 10 11 12 0 1 2 3 4 5 6 7 x x Lecture 3-21 FEC 512 Probability Distributions
- 22. The Normal Distribution Probability Distributions Continuous Probability Distributions Normal Uniform Lognormal Lecture 3-22 FEC 512 Probability Distributions
- 23. The Normal Distribution The distribution whose pdf is given by ( x−µ )2 − f(x) 1 2σ 2 f ( x) = e 2π σ σ ‘Location is determined by x the mean, µ µ Spread is determined by the standard deviation, σ Lecture 3-23 FEC 512 Probability Distributions
- 24. The Normal Distribution f(x) ‘Bell Shaped’ Symmetrical σ The random variable has x an infinite theoretical µ range: + ∞ to − ∞ Lecture 3-24 FEC 512 Probability Distributions
- 25. Many Normal Distributions By varying the parameters µ and σ, we obtain different normal distributions Lecture 3-25 FEC 512 Probability Distributions
- 26. The Normal Distribution Shape f(x) Changing µ shifts the distribution left or right. Changing σ increases or decreases the σ spread. µ x Lecture 3-26 FEC 512 Probability Distributions
- 27. Finding Normal Probabilities Probability is the Probability is measured by the area area under the curve! under the curve f(x) P (a ≤ x ≤ b) a b x Lecture 3-27 FEC 512 Probability Distributions
- 28. Probability as Area Under the Curve The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below f(x) P( −∞ < x < µ) = 0.5 P(µ < x < ∞ ) = 0.5 0.5 0.5 x µ P(−∞ < x < ∞) = 1.0 Lecture 3-28 FEC 512 Probability Distributions
- 29. Empirical Rules What can we say about the distribution of values around the mean? There are some general rules: f(x) µ ± 1σ encloses about 68% of x’s σ σ x µ−1σ µ µ+1 − +1σ +1 68.26% Lecture 3-29 FEC 512 Probability Distributions
- 30. The Empirical Rule (continued) µ ± 2σ covers about 95% of x’s µ ± 3σ covers about 99.7% of x’s 3σ 3σ 2σ 2σ µ x µ x 95.44% 99.72% Lecture 3-30 FEC 512 Probability Distributions
- 31. The Standard Normal Distribution Also known as the “z” distribution Mean is defined to be 0 Standard Deviation is 1 f(z) 1 z 0 Values above the mean have positive z-values, values below the mean have negative z-values Lecture 3-31 FEC 512 Probability Distributions
- 32. Transformation to the Standard Normal Distribution Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normal distribution (z) x −µ z= σ Lecture 3-32 FEC 512 Probability Distributions
- 33. Example If x is distributed normally with mean of 100 and standard deviation of 50, the z value for x = 200 is x − µ 200 − 100 z= = = 2.0 σ 50 This says that x = 200 is two standard deviations (2 increments of 50 units) above the mean of 100. Lecture 3-33 FEC 512 Probability Distributions
- 34. Comparing x and z units µ = 100 σ = 50 100 200 x 0 2.0 z Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z) Lecture 3-34 FEC 512 Probability Distributions
- 35. The Standard Normal Table The Standard Normal table in the textbooks gives the probability from the mean (zero) up to a desired value for z .4772 Example: P(0 < z < 2.00) = .4772 z 0 2.00 Lecture 3-35 FEC 512 Probability Distributions
- 36. The Standard Normal Table (continued) The column gives the value of z to the second decimal point z 0.00 0.01 0.02 … 0.1 The row shows the 0.2 . . value of z to The value within the . the first table gives the .4772 2.0 decimal point probability from z = 0 up to the desired z 2.0 P(0 < z < 2.00) = .4772 value Lecture 3-36 FEC 512 Probability Distributions
- 37. Z Table example Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6) Calculate z-values: x −µ 8 −8 z= = =0 σ 5 8 8.6 x x − µ 8.6 − 8 0 0.12 Z z= = = 0.12 σ 5 P(8 < x < 8.6) = P(0 < z < 0.12) Lecture 3-37 FEC 512 Probability Distributions
- 38. Z Table example (continued) Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6) µ=8 µ=0 σ=5 σ=1 x z 8 8.6 0 0.12 P(8 < x < 8.6) P(0 < z < 0.12) Lecture 3-38 FEC 512 Probability Distributions
- 39. Solution: Finding P(0 < z < 0.12) Standard Normal Probability P(8 < x < 8.6) Table (Portion) = P(0 < z < 0.12) z .00 .01 .02 .0478 0.0 .0000 .0040 .0080 0.1 .0398 .0438 .0478 0.2 .0793 .0832 .0871 Z 0.00 0.3 .1179 .1217 .1255 0.12 Lecture 3-39 FEC 512 Probability Distributions
- 40. Lower Tail Probabilities Suppose x is normal with mean 8.0 and standard deviation 5.0. Now Find P(7.4 < x < 8) Z 8.0 7.4 Lecture 3-40 FEC 512 Probability Distributions
- 41. Lower Tail Probabilities (continued) Now Find P(7.4 < x < 8)… The Normal distribution is symmetric, so we use the .0478 same table even if z-values are negative: P(7.4 < x < 8) = P(-0.12 < z < 0) Z = .0478 8.0 7.4 Lecture 3-41 FEC 512 Probability Distributions
- 42. Distributions of Portfolio Returns Example. Assume that the stock index in a country has an annual return distribution that is normal with µ = 0.15 and σ = 0.30. What is probability that in a given year the stock index will exceed an annual return of 100%? What is the probability that the index will produce a negative return in a given year? We first need to transform the normal variable into a standard normal. X − 0.15 1.00 − 0.15 z= = = 2.83 0.30 0.30 Looking up 2.83 in the normal table, we find that F(z) is 0.9977. So 1 – F(z) = 0.0023. For finding the probability of a negative return, the transformation yields X − 0.15 0 − 0.15 z= = = −0.50 0.30 0.30 This time we are interested in F(z), which is 0.3085. Lecture 3-42 FEC 512 Probability Distributions
- 43. Linear Combinations of Two Normal Random Variables Let X~N(µX,σX2 ) and Y~N(µY,σY2 ) and σXY=cov(X,Y). If Z=aX+bY where a,b are constants, then Z~N(µZ,σZ2 ) where µZ=a µX +b µY σZ2=a2 σX2 +b2 σY2 +2abσXY=a2 σX2 +b2 σY2 +2abσX σYρ Lecture 3-43 FEC 512 Probability Distributions
- 44. Kurtosis and Skewness of Normal Distribution The skewness of a normal distribution is 0. Why? The kurtosis of a normal distribution is 3. Hence 3 is a benchmark value for tail thickness of a bell- shaped distribution. If kurt(X)>3, the dist. has thicker tails than norm. dist. If kurt(X)<3, the dist. has thinner tails than norm. dist. Lecture 3-44 FEC 512 Probability Distributions
- 45. The Uniform Distribution Probability Distributions Continuous Probability Distributions Normal Uniform Lognormal Lecture 3-45 FEC 512 Probability Distributions
- 46. The Uniform Distribution The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable Lecture 3-46 FEC 512 Probability Distributions
- 47. The Uniform Distribution (continued) The Continuous Uniform Distribution: 1 if a ≤ x ≤ b b−a f(x) = 0 otherwise where f(x) = value of the density function at any x value a = lower limit of the interval b = upper limit of the interval Lecture 3-47 FEC 512 Probability Distributions
- 48. Uniform Distribution Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6: 1 f(x) = 6 - 2 = .25 for 2 ≤ x ≤ 6 f(x) .25 x 2 6 Lecture 3-48 FEC 512 Probability Distributions
- 49. The Lognormal Distribution Probability Distributions Continuous Probability Distributions Normal Uniform Lognormal Lecture 3-49 FEC 512 Probability Distributions
- 50. The Lognormal Distribution Let Z ~N(µ,σ2), and X=eZ r.v. X is said to be log- normally distributed with parameters µ and σ2 lnX~N(µ,σ2) or In other words, X is lognormal if its “ln” is normally distributed. If X,Y is lognormally distributed, their linear combination(i.e. Portfolio of two stocks) may not be lognormal. Lecture 3-50 FEC 512 Probability Distributions
- 51. Lecture 3-51 FEC 512 Probability Distributions
- 52. The median of X is eµ, and the expected value of X σ2 µ+ is . The expectation is larger than the 2 e median because the lognormal distribution is right- skewed, and the skew. is more extreme with larger values of σ. Lecture 3-52 FEC 512 Probability Distributions
- 53. Example Let rt = ln( Pt / Pt −1 ) denote the log-return on an asset and assume that rt ~ N(µ,σ2). Let R = ( P P P ) − t −1 t t t −1 denote the simple monthly return, since we know e rt = 1 + Rt . Since rt is rt = ln(1 + Rt ) that and normally distributed e t = 1 + Rt is log-normally r ............. distributed. Lecture 3-53 FEC 512 Probability Distributions
- 54. Sample Moments Above we introduced the four statistical moments mean,variance, skewness, kurtosis. Given a pdf, we are able to calculate these stat. Moments according to the fomulae. In practical applications however, we are faced with the situation that we observe realizations of a pdf(e.g. The daily return of the IMKB-100 index over the last year), but we do not know the distribution that generates these returns. So taking expectation is impossible But having the observations x1,…,xn, we can try to estimate the “true moments” out of the sample. These estimates are called sample moments. Lecture 3-54 FEC 512 Probability Distributions
- 55. Mean (Arithmetic Average) The Mean is the arithmetic average of data values Sample mean n = Sample Size n ∑x x1 + x 2 + L + x n i x= = i =1 n n Lecture 3-55 FEC 512 Probability Distributions
- 56. Variation Measures of variation give information on the spread or variability of the data values. Same center, different variation Lecture 3-56 FEC 512 Probability Distributions
- 57. Variance Average of squared deviations of values from the mean Sample variance: n ∑ (x − x) 2 i s2 = i =1 n -1 Sample standard deviation: n ∑ (x − x) 2 i s= i=1 n -1 Lecture 3-57 FEC 512 Probability Distributions
- 58. Calculation Example: Sample Standard Deviation Sample Data (Xi) : 10 12 14 15 17 18 18 24 n=8 Mean = x = 16 (10 − x ) 2 + (12 − x ) 2 + (14 − x ) 2 + L + (24 − x ) 2 s= n −1 (10 − 16) + (12 − 16) + (14 − 16) + L + (24 − 16) 2 2 2 2 = 8 −1 126 = = 4.2426 7 Lecture 3-58 FEC 512 Probability Distributions
- 59. Comparing Standard Deviations Data A Mean = 15.5 s = 3.338 11 12 13 14 15 16 17 18 19 20 21 Data B Mean = 15.5 s = .9258 11 12 13 14 15 16 17 18 19 20 21 Data C Mean = 15.5 s = 4.57 11 12 13 14 15 16 17 18 19 20 21 Lecture 3-59 FEC 512 Probability Distributions
- 60. Coefficient of Variation Measures relative variation Always in percentage (%) Shows variation relative to mean Is used to compare two or more sets of data measured in different units Sample C.V. s CV = ⋅ 100% x Lecture 3-60 FEC 512 Probability Distributions
- 61. Comparing Coefficient of Variation Stock A: Average price last year = $50 Standard deviation = $5 s $5 CVA = ⋅ 100% = ⋅ 100% = 10% x $50 Both stocks have the same Stock B: standard deviation, but Average price last year = $100 stock B is less variable relative Standard deviation = $5 to its price s $5 CVB = ⋅ 100% = ⋅ 100% = 5% x $100 Lecture 3-61 FEC 512 Probability Distributions
- 62. Sharpe Ratio Let Dt=Rt-Rf where Rf is the riskfree rate of return. T T ∑ 1 D = ∑ Dt (Dt − D) 2 T t =1 σD = t =1 T −1 Sharpe Ratio is (D) S= σD Reading: “The Sharpe Ratio”, William F. Sharpe Lecture 3-62 FEC 512 Probability Distributions
- 63. Skewness The moment coefficient of skewness is derived by calculating the third moment about the mean and dividing by the cube of standard deviation : ( ) ∑ X −X 3 n −1 3 2 ∑( ) X −X n −1 Lecture 3-63 FEC 512 Probability Distributions
- 64. Shape of a Distribution Describes how data is distributed Symmetric or skewed Right-Skewed Symmetric Left-Skewed Mean = Median Median < Mean Mean < Median (Longer tail extends to left) (Longer tail extends to right) Lecture 3-64 FEC 512 Probability Distributions
- 65. Kurtosis Skewness indicates the degree of symmetry in the frequency distribution Kurtosis indicates the peakedness of that distribution Lecture 3-65 FEC 512 Probability Distributions
- 66. Kurtosis (continued) ∑(X − X ) 4 n −1 4 ∑(X − X ) 2 n −1 Lecture 3-66 FEC 512 Probability Distributions
- 67. About the Probability Distribution of Returns The assumption that period returns(e.g. Daily, monthly, annually) are normally distributed is inconsistent with the limited liability feature of most financial instruments, R≥-1. There are a number of empirical facts about return distributions While normal distribution is perfectly symmetric about its mean, 1. dailt stock returns are frequently skewed to the right. And few of them are skewed to the left. The sample daily return distributions for many individuals stocks 2. exhibit “excess kurtosis” or “fat tails”. i.e. There is more probability in the tails than would be justified by normal distribution. The extent of this excess kurtosis diminishes substantially, however , when monthly data is used. Lecture 3-67 FEC 512 Probability Distributions
- 68. Emprical Return Distributions: IMKB-100 Daily 500 Series: XU100RETURNS Sample 1/04/1993 12/29/2004 400 Observations 2968 Mean 0.002654 300 Median 0.001996 Maximum 0.194510 Minimum -0.181093 200 Std. Dev. 0.031281 Skewness 0.139399 Kurtosis 6.290761 100 Jarque-Bera 1348.812 Probability 0.000000 0 -0.1 0.0 0.1 0.2 Lecture 3-68 FEC 512 Probability Distributions
- 69. Emprical Return Distributions: IMKB-100 Monthly 24 Series: XU100RETURNS Sample 1993M01 2004M12 20 Observations 144 16 Mean 5.899510 Median 5.043156 Maximum 79.78386 12 Minimum -39.03413 Std. Dev. 17.21817 8 Skewness 0.903614 Kurtosis 5.585146 4 Jarque-Bera 59.69430 Probability 0.000000 0 -40 -20 0 20 40 60 80 Pay attention to Skewness and Kurtosis. Lecture 3-69 FEC 512 Probability Distributions
- 70. Value At Risk Lecture 3-70 FEC 512 Probability Distributions

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