Homework Assignments

Transcription

1 Homework Assignments Week 1 (p. 57) #4.1, 4., 4.3 Week (pp 58 6) #4.5, 4.6, 4.8(a), 4.13, 4.0, 4.6(b), 4.8, 4.31, 4.34 Week 3 (pp 15 19) #1.9, 1.1, 1.13, 1.15, 1.18 (pp 9 31) #.,.6,.9 Week 4 (pp 36 37) #3.1, 3.4, 3.5, Supp. #1,

2 Chapter 1 Probability Probabilities and Events Random Variables and Expected Values Conditional Probability Covariance and Correlation

3 Chapter Normal Random Variables Continuous Random Variables Normal Random Variables Properties p of Normal Random Variables.4 The Central Limit Theorem

4 Normal Distributions ib ti (x ) - 1 f(x) e, (- <x< )

5 The Normal Distribution E(X) = E( Z + ) = E(Z) + = Var(X) = Var( Z + m) = E(Z) =

6 The Empirical i Rule ( Rule) l) 68% of the data 95% of the data 99.7% of the data

7 Rule in Math terms 1 1 x ( ) e 1 1 x ( ) e x ( ) 3 e dx dx dx

8 Converting to a Standard d Normal Distribution z = x

9 Example.3a 3a IQ examination scores for sixth graders are normally distributed with mean value 100 and standard deviation 14.. What is the probability that a randomly chosen sixth grader has an IQ score greater than 130? 9

10 Example.3a 3a Solution. Let X be the score of a randomly chosen sixth grader. Then X PX { 130} P PZ {.113} 1 (.113)

11 Example.3d 3d Starting at some fixed time, let S(n) denote the price of a certain security at the end of n additional weeks, n 1. A popular model for the evolution of these prices assumes that the price ratios S(n)/S(n 1) for n 1 are independent and identically distributed ib t d (i.i.d.) (iid) lognormal l random variables. Assuming this model, with lognormal parameters μ =.0165 and σ =.0730, what is the probability that 11

12 Example.3d 3d (a) the price of the security increases over each of the next two weeks; (b) the price at the end of two weeks is higher than it is today? Solution. Let Z be a standard normal random variable. It is easy to see that 1

13 (a) P Example.3d 3d S(1) S(1) 1 Pln 0 S(0) S(0) P Z P Z P Z

14 Example.3d 3d Therefore, the probability that the price is up after one week is Since the successive price ratios are independent, by the multiplication rule, the probability that the price increases over each of the next two weeks is.5894 =

15 Example.3d 3d (b) P S () () (1) S S 1 P 1 S(0) S(1) S(0) S (1) S (1) P ln ln 0 S(0) S(0) 0 (.0165) PZ

16 Example.3d 3d (b) P S() PZ S (0).0730 P Z P Z

17 Section.4 The Central Limit Theorem 17

18 Markov s Inequality Suppose X is a non-negative negative RV with finite mean. Then for any a>0 P X a E X a

19 Markov s Inequality Example: The average amount of waiting time at a post office is 5 minutes. Estimate the probability that you have to wait more than 0 minutes. [ ] 5 PX ( 0) EX.5 0 0

20 Chebyshev s Inequality Suppose that t X is a RV with finite it mean and finite variance. The for any k > 0, P X k 1 k P X k 1 1 k

21 Chebyshev s Inequality Chebyshev s Inequality Proof: ( ) X P X k P k X E E k X Var k 1 k k k

22 Variance of the Sample Mean Let X 1, X,, X n be a sample of size n drawn from this population. Then each X k has the same distribution as the entire population and E(X ) = μ & Var(X ) = σ k k Let X = n (X + X X ) 1 n n be the sample average.

23 Variance of the Sample Mean By linearity of expectation E(X ) = μ When X 1,X,, X n are drawn with replacement, they are independent and each X k has variance σ. Then nσ σ σ Var(X ) = = SD(X ) = n n n n n n

24 Proof: E( X ) n n n X i E ( X i ) i1 i1 n E n n n Var( X ) n n n Xi Var( Xi) i1 i1 n Var n n n n

25 Sample Variance The RV S defined dfi d by n (X X) i n i=1 S = n 1 is called the sample variance. Theorem: E[S ] = σ

26 Proof: Sample Variance n _ n X - X = X - X X + (X ) i n i i n n i=1 i=1 n _ n _ i n i n i=1 i=1 = X - X X + n (X ) n _ i n n n = X - X n X + n (X ) i=1 n i=1 _ i n = X - n (X )

27 n 1 n i i=1 1 E S = E X - X n-1 n 1 = E X ne Xn n 1 i i=1 1 n 1 1 = ne X -ne X n-1

28 Sample Variance E S = n σ σ - μ -n n -μ n-1 nσ - nμ - σ +nμ = n-1 nσ - σ = n-1 = σ

29 P( X-μ <ε) 1asn Weak klaw of large numbers Theorem: Let X 1,X, be a sequence of independent random variables with the same distribution. Let denote the common expected value = E(X i ). And let X n = Then for every > 0: X + X X 1 n n nn

30 Weak Law of large numbers Proof: Let μ = E(X i ) and σ = SD(X i ) <.. Then E(X n) = μ and SD(X n) =. Apply Chebyshev s inequality to n P X n P X n n n For a fixed ε right hand side tends to 0 as n tends to infinity. σ X n n

31 The Central Limit Theorem Let S n = X X n be the sum of identical, independent d random variables with = E(X i ) and = SD(X i ). Then for large n, the distribution of S n is approximately normal with mean E(S n ) = n and SD(S n ) = n 1/. Furthermore, Sn n X n lim P x lim P x ( x ) n n n n

32 Given: The Central Limit Theorem 1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation.. Simple random samples all of size n are selected from the population. (The samples are selected so that t all possible samples of the same size n have the same chance of being selected.)

33 The Central Limit Theorem - cont Conclusions: 1. The distribution of sample mean x n will, as the sample size increases, approach a normal distribution. ib ti. The mean of the sample means is the population mean µ. 3. The standard deviation of all sample means is n

34 Practical lrules Commonly Used 1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger.. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n (not just the values of n larger than 30).

35 Example.3a* 3a* IQ examination scores for sixth graders are normally distributed with mean value 100 and standard deviation 14.. What is the probability that (a) A randomly chosen sixth grader has an IQ score greater than 130? (b) The average IQ score of four randomly chosen sixth graders is greater than 130? 35

36 Example.3a* 3a* Solution. (a) Let X be the score of a randomly chosen sixth grader. Then X PX { 130} P PZ {.113} 1 (.113)

37 Example.3a* 3a* (b) X PX { 130} P PZ { 4.3} 1 (4.3)

38 Approximation of a Binomial Distribution with a Normal Distribution If np 5 and n(1 p) 5 then µ = np and np(1 p) and the random variable has a (normal) distribution.

39 Binomial Probability Distribution 1. The procedure must have fixed number of trials.. The trials must be independent. 3. Each trial must have all outcomes classified into two categories. 4. The probability of success remains the same in all trials.

40 Example.4a 4a A fair coin is tossed 100 times. What is the probability that heads appears fewer than 40 times? Solution. If X denotes the number of heads, then X is a binomial random variable with parameters n = 100 and p = 1/. Since np = 50 5 we have np(1 p) = 5 5, and so 40

41 Example.4a 4a A fair coin is tossed 100 times. What is the probability that heads appears fewer than 40 times? X PX ( 40) P 5 5 P { Z } ( ).08 41

42 Example.4a 4a A graphic calculator computing binomial probabilities gives the exact solution.0176, and so the preceding is not quite as accurate as we might like. However, we could improve the approximation by using continuity correction: Write desired probability as P{X < 39.5}. This gives 4

43 Example.4a 4a X PX ( 39.5) P 5 5 PZ {.10} (.10)

44 Definition When we use the normal distribution (which is a continuous probability bbili distribution) ib i as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a discrete whole number x in the binomial distribution by representing the single value x by the interval from x 0.5 to x (that is, adding and subtracting 0.5)

45 x = at least 1 (includes 1 and above) x = more than 1 (doesn t include 1) x = at most 1 (includes 1 and below) x = fewer than 1 (doesn t include 1) x = exactly 1

46 Notation for Normal Distribution Definition: If RV X has a normal distribution with mean μ and standard deviation σ, we say X has a normal distribution with parameters μ and σ and denote X ~ N(μ, σ )

47 Lognormal Distribution Definition: Let X ~ N(μ, μ σ ) and let Y = e X. Then Y is said to have a lognormal distribution with parameters μ and σ. Theorem: If Y has a lognormal distribution with parameters μ and σ, then Y has pdf (ln x μ) σ 1 f(x)= e, x>0 Y xσ π

48 Proof: Lognormal Distribution Y F(x)=Pe <x Y = P Y < ln x ln x 1 (t μ) σ = e dt σ π f(x)=f(x)= Y Y 1 xσ π e (ln x μ) μ σ

49 Lognormal Distribution Theorem: If Y has a lognormal distribution with parameters μ and σ, then σ E Y =e μ+ μ+σ σ Var(Y) = e (e 1)

50 Lognormal Distribution Proof: Left as exercises. Hint: Consider the Laplace transform zx f(z) = e f(x)dx Y

51 Example: Suppose that the price S of a particular stock at closing has a lognormal distribution with E[S] = $0 and Var(S) = 4. What is probability that the price exceed $? Solution: We need to find μ and σ first. E S =e =0 σ μ+ σ μ + =ln 0 μ+σ σ Var(S) = e (e 1) = 4 ln 0 σ e (e 1) = 4

52 σ 400(e 1) = 4 σ =ln σ μ + =ln 0 ln 101 ln 1.01 μ =ln

53 P(S > ) = 1 P(S ) =1 P(ln S ln ) l ln 1.01 ln ln 0 =1 Φ ln 1.01

54 P(S > ) 1 (1.01) = =.156

55 Calculations We have seen that the terminal py payoff of a call option is in the form (X K) + = max{x K, 0}. Let s calculate expectations and variance of an RV in the form of (X K) + where X is a RV and K is a constant

56 Expectation involving Normal RV Theorem: Suppose X ~ N(μ, σ ). Then ( K ) / K E ( X K ) e ( K )

57 Expectation involving Normal RV Proof: 1 ( x) / E ( X K ) ( xk ) e dx 1 K ( ) ( x) / x Ke dx

58 Expectation involving Normal RV Let t = (x μ)/σ or x = μ + σ t 1 ( x) / K E ( X K) ( xk) e dx 1 / t ( K te ) dt K t / t / e dt te dt K K K

59 Expectation involving Normal RV K / t E ( X K) ( K) lim te dt M M K / ( ) lim M K t K e M K K ( ) ( K ) / K e

60 Expectation involving Lognormal RV Theorem: Suppose Y has a lognormal distribution with parameters μ and σ, and K > 0. Then / ln K ln K E ( Y K) e K

61 Expectation involving Lognormal RV Proof: 1 1 (ln x ) / E ( Y K ) ( x K ) e dx x 1 1 ( ) x K (ln x) / x K e dx

62 Expectation involving Lognormal RV Let t = (ln x μ)/σ or x = e μ+σ t 1 t t / ln K E ( Y K) ( e K) e dt t t / K t / 1 e e e dt e e dt ln K ln K / ( t) / 1 ln ee e dtk ln K K

63 Expectation involving Lognormal RV Let z = t σ or t = z + σ / z / 1 lnk E ( Y K) e e dzk ln K / ln K ln K e K / ln K ln K e K

64 Variance involving a Lognormal RV Exercise: Suppose Y has a lognormal distribution with parameters μ and σ, and K > 0. Let ω = (μ ln(k))/ σ. Prove that ( ) / a) E ( YK) e w Ke w K ( w) ( ) / b) Var ( Y K) e w Ke w K ( w) / e w K w

65 Chapter 3 Geometric Brownian Motion 3.1 Geometric Brownian Motion (GBM) 3. GBM as a Limit of Simpler Models 3.3 Brownian Motion 3.4 Exercises

66 Section Geometric Brownian Motion (GBM) 66

67 Notations Suppose that we are interested in the price of some security, say a stock, as it evolves over time. Let the present time be time 0. For 0 y <, let S(y) denote the price of the security a time y from the present. S(y) is known as a stochastic process.

68 Geometric Brownian Motion (GBM) Definition: We say that price process S(y) follows a geometric Brownian motion with drift parameter μ and volatility parameter σ if For t 0, the RV S(t + y) / S(y) is independent of all prices up to time y 0 ln{s(t + y) / S(y)} ~ N(t μ, t σ )

69 Geometric Brownian Motion (GBM) The series of security prices follows a GBM if the ratio of the price a time t in the future to the present price is Independent of the past history of prices, A lognormal probability distribution with parameters μt and σ t

70 Implications of the GBM Model The assumption that S(y) follows a GBM implies Markov property: For fixed μ and σ, it is only the current price not the history of past prices that affects probabilities of future prices. Probabilities concerning the ratio of the price at time t in the future to the current price is independent of the current price. Thus, the probability a given security doubles in price in the next month is the same no matter whether the current price is 10 or 5.

71 Markov property The Markov property implies that the probability distribution ib i of the price at any particular future time is not dependent on the particular path followed by the price in the past. The Markov property of stock prices is consistent with the weak form of market efficiency. This states that the present price of a stock impounds all the information contained in a record of past prices.

72 Implications of the GBM Model Under GBM, the expected price grows at the rate of μ + σ / E S() t S(0) e t /

73 Section 3. 3 GBM as a Limit of Simpler Models 73

74 Modified Bernoulli Distribution Let X be a modified Bernoulli random variable with parameter p. X is equal to 1 with probability p and to 1 with probability (1 p). Then, EX1 p( 1) (1 p) p1 1 ( 1) (1 ) 1 E X p p Var( X ) E[ X ] E[ X ] 1 (p 1) 74

75 A Simpler Model For a fixed time t > 0, positive integer n, let Δ = t/n denote a small increment of time and suppose that, t every Δ time units, the price of a security either goes up by the factor u with probability bilit p or goes down by the factor d with probability 1 p. 75

76 Stock Price in One Step time=0 p time=δ us Stock S 1-p ds

77 Stock price in n (=4) steps us u S u 3 S u ds u 4 S u 3 ds CRR Model S uds ud S u d S ds d S ud 3 S d 3 S d 4 S

78 By the CLT and with carefully chosen p, d, and u, the simpler model is going to approach a GBM as we let Δ become smaller and smaller. In order to accomplish this, we need to determine three parameters p, u, and d by matching the mean and variance (from the market). We could take d = 1/u. 78

79 Let S i = S(iΔ) be the stock price at time iδ for i= 0,1,,,n. Suppose X 1, X,, X n be i.i.d. of modified Bernoulli RVs with parameter p. Let Y n = X 1 + X + + X n. Write u = exp(α), then d = exp( α). Then Xn Sn S 1 S n n1 Sn 1 S S u S e S e X n X n 79

80 S S e n n1 0 X n X S e n1 n e S e n n1 n Se Se 0 ( X X ) X ( ) Y n X1 X X n n 80

81 We need Note that lim Yn Y N( t, t ) n E[ Y ] E[ Y ] ne[ X ] n(p1) n n n Var Y Var Y nvar X n p ( n ) ( n ) ( n ) [1 ( 1) ] For large n, we would like to have n (p1) t n [1 ( p1) ] t 81

82 Divide id both sides by n, ( p1) [1 ( p 1) ] By the 1 st eqn., the nd eqn. is equivalent to When n,, Δ 0. Hence 8

83 To summarize, we must have ( p 1) Solve for α and p, p

84 Take u e e d e e p

85 Section 3.3 Brownian Motion (BM) 85

86 Brownian Motion (BM) Definition: We say that price process S(y) follows a Brownian motion (BM) with drift parameter μ and volatility parameter σ if For t 0, the RV S(t + y) S(y) is independent of all prices up to time y 0 S(t + y) S(y) ~ N(t μ, t σ )

87 Brownian Motion (BM) BM shares with GBM the Markov property that future price depends on the present and all past prices only through the present price In BM, the difference of prices has a normal, p distribution, not the logarithm of their ratios.

88 History of Brownian Motion (BM) BM is named after the Scottish botanist Robert Brown who first described d the random motions of pollen grains suspended in water (187). The mathematics ti of this process were formalized by Bachelier in an option pricing context (1900), and by Einstein (1905). A mathematically concise definition and a rigorous theory of BM, was developed by Norbert Wiener in a series of papers originating in BM is also known as Wiener process.

89 Important Properties of BM Finiteness Continuity Markov Martingale Quadratic variation Normality

90 Major Flaws Negative stock price: the price of a stock is a normal random variable which can theoretically become negative. The assumption that a price difference over an interval of fixed length has the same normal distribution no matter what the price at the beginning of the interval does not seem totally reasonable.

91 Is GBM a Better Model for Stocks? In GBM model, the logarithm of the stock's price that is a normal RV, the model does not allow for negative stock prices. In GBM model, it is the percentage change in price, not the absolute change, whose probabilities do not depend on the present price.