# Definition A r.v. X has a normal distribution with mean µ and variance σ 2, where µ R, and σ > 0, if its density is f(x) = 1. 2σ 2.

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1 Chapter 6 Brownian Motion 6. Normal Distribution Definition 6... A r.v. X has a normal distribution with mean µ and variance σ, where µ R, and σ > 0, if its density is fx = πσ e x µ σ. The previous definition makes sense because f is a nonnegative function and πσ e x µ σ dx =. Note that by the changes of variables x µ σ = t πσ e x µ σ dx = π e t By a change to polar coordinates, e x x dx = e y dx dy = 0 = π e r r dr = π e r 0 = π. 0 dt. π e r 0 r dθ dr Theorem 6.. If X has a normal distribution with parameters µ and σ, then E[X] = µ and VarX = σ. Proof. By the change of variables x µ σ = t, E[X] = x πσ e x µ σ dx = µ + tσ πσ e t = µ because the function t πσ e t πσ e t By the change of variables x µ σ E[X µ ] = x µ = σ t d π e t = σ dt + σ t πσ e t dt = µ, is odd. = t and integration by parts, πσ e x µ σ π e t dt = σ. 6 dt dx = σ t π e t dt Q.E.D.

2 6 CHAPTER 6. BROWNIAN MOTION The following formulas could be useful: x e dx = π, xe x dx = 0, x e x dx = π A normal r.v. with mean zero and variance one is called a standard normal r.v. Probabilities for a standard normal r.v. can be found in table at the back of the book. Given a standard normal r.v. Z, to find PZ a, where a > 0, we do To find P Z a, where a > 0, we do P[Z a] = P [Z a] = P[Z a]. P[ Z a] = P[ a Z a] = P [Z a] P[Z a] = P [Z a] P [Z a] = P [Z a]. We usually will denote the density of a standard normal r.v. by φx = π e x. We denote the cumulative distribution function of a standard normal r.v. by Φx = x π e t dt. Theorem 6.. The moment generating function of a r.v. X with a normal distribution with parameters µ and σ, is M X t = e tµ+ t σ. Proof. By the change of variables x µ σ = y, M X t = = e tµ+ t σ etx πσ e x µ σ π e y σt dy = e tµ+ t σ dx = etµ+tσy π e y A standard normal r.v. Z has moment generating function: M Z t = e t.. dy Q.E.D. Theorem 6.3. If X has a normal distribution with parameters µ and σ, then Z = X µ σ a standard normal r.v. Proof. The moment generating function of Z = X µ σ is has M X t = E[e t X µ σ ] = e tµ σ E[e tx σ ] = e t, which is the moment generating function of a standard normal r.v. Q.E.D. Theorem 6.4. Let X,..., X n be independent r.v. s with a normal distribution. Then, n i= X i has a normal distribution. Proof. Let µ i = E[e tx i ] and let σi moment generating function of n i= X i is M n = exp = VarX i. Then, E[e tx i ] = exp tµ i + t σi. So, the i= X t = n i i= M X i t = n i= exp tµ i + t σi t n i= µ i + t n i= σ i.

3 6.. NORMAL DISTRIBUTION 63 Now, exp t n i= µ i + t n i= σ i with mean n i= µ i and variance n is the moment generating function of a normal distribution i= σ i. Since the moment generating function determines the distribution, we conclude that n i= X i has a normal distribution with mean n i= µ i and variance n i= σ i. Q.E.D. Example 6.. If X is a normal random variable with parameters µ = and σ = 4, compute P[5 X > 3]. Solution: We standardize the variable: Z = X 4 3 P[5 X > 3] = P[3 < X 5] = P 4 has a standard normal distribution. So, Z 5 4 = P[ < Z ] = P[Z ] P[Z ] = = Example 6.. A manufacturer produces an item which has a label weight of 0.4 grams. Let X denote the weight of a single mint selected at random from the production line. X has a normal distribution with mean.37 and variance 0.6. Find P[X < 0.4]. Solution: Z = X µ σ has a standard normal distribution. So, [ ] Z = P[Z.45] = P[X < 0.4] = P Example 6.3. Let Z be a standard normal r.v. Find c such that P [ Z c] =.95. Solution: Using the previous formula, 0.95 = P [ Z c] = P [Z c]. So, P[Z c] = = From the normal table, c =.96. Exercise 6.. The height, X, that a college high jumper will clear each time she jumps is a normal random variable with mean 6 feet and variance inches. What is the probability the jumper will clear 6 feet 4 inches in on a single jump? Exercise 6.. The length of the time required to complete a college achievement test is found to be normally distributed with mean 70 minutes and standard deviation minutes. When should the test be terminated if we wish to allow sufficient time of 90 % of the students to complete the test? Exercise 6.3. Suppose that X is a standard normal random variable with parameters µ = and σ = 4. Find P [X > ]. Example 6.4. If X and Y are independent identically distributed standard normal random variables, compute Pr{ X + Y 3}. Solution: X + Y has normal distribution with mean zero and variance. So, P[ X + Y 3] = P[ N0, 3 ] = P[ 0.70 N0,.] = = 0.73.

4 64 CHAPTER 6. BROWNIAN MOTION Example 6.5. If X and Y are independent identically distributed standard normal random variables, compute the density of U = + X + Y. Solution: By the previous theorem, U has a normal distribution. E[U] = and VarU = VarX + VarY =. So, the density of U is u f U u = e π u = e 4 π. 6. Central Limit Theorem If Y is a normal r.v. with mean µ and variance σ, then Y µ has a standard normal distribution. The central limit theorem says that we can do something like that for sums of σ i.i.d.r.v. s. Let X,..., X n be i.i.d.r.v. s with mean µ and variance σ. Then, [ n ] n E X i = E[X i ] = nµ and i= n Var X i = i= i= n VarX i = nσ. The central limit theorem says that for n large enough n i= X i E[ n i= X n i] i= Var n i= X = X i µ i σ n has approximately a normal distribution. Precisely, the central limit theorem says that: i= Theorem 6.5. Let {X i } i= be a sequence of i.i.d.r.v. s with finite second moment. Then, for each a R, [ n lim P i= X ] i E[X i ] n σ a = Φa, n where Φa = P[N0, a]. and The central limit theorem is also true for open intervals and bounded intervals: [ n lim P i= X ] i E[X i ] < a = P [Z < a] n nσ [ lim P a n n i= X i E[X i ] nσ ] b = P[a Z b], where Z denotes a r.v. with a standard normal distribution. Instead of n i= X i, we can use x = n n i= X i, and get that ] lim P n [ x µ σ/ n a = P[N0, a].

5 6.. CENTRAL LIMIT THEOREM 65 For most distributions, the central limit theorem gives a good approximation to the probability if n 30. For some distributions n = 0 is enough. A common application of the central limit theorem is the estimation of probabilities of a binomial distribution with large n. In this case, Theorem 6.6. Let S n be the a r.v. with a binomial distribution with parameters n and p, then [ ] lim P S n np a = Φa. n np p When handling discrete distributions, a continuity correction can be made. If a is an integer, then for each 0 t < [ ] S P [S n a] = P [S n a + t] = P n np a+t np Φ a+t np. np p np p np p Since each 0 t < gives a different approximation, the average of these t s is taken. So, a+0.5 np P [S n a] Φ np p a 0.5 np P [S n < a] Φ np p b+0.5 np P [a S n b] Φ a 0.5 np Φ np p np p b 0.5 np P [a < S n < b] Φ a+0.5 np Φ np p np p Example 6.6. Let S be the number of heads in 000 independent tosses of a fair coin. Find the probability that 480 S 55. Solution: S has a binomial distribution with parameters n = 000 and p =. Then, E[S] = np = 000 = 500 and VarS = np p = 000 = 50. So, P[480 S 55] P[ Z ] = Φ Φ.965 = = Exercise 6.4. Let X be the number of heads in 0000 independent tosses of a fair coin. Find the probability that 4900 X 550. Exercise 6.5. One thousand independent throws of a fair coin will be made. Let X be the number of heads in these thousand independent throws. Find the smallest value of a so that P [500 a X a] Exercise 6.6. Let X be the number of heads in 0,000 independent tosses of a fair coin. Find c so that the probability that 5000 c X c is 95 %. Exercise 6.7. One thousand independent rolls of a fair die are made. Let X be the number of times that the number 6 appears. Find the probability that 50 X < 00.

6 66 CHAPTER 6. BROWNIAN MOTION Exercise 6.8. One thousand independent rolls of a fair die will be made. number of sixes which will appear. Find the probability that 60 X 90. Let X be the Exercise 6.9. One thousand independent rolls of a fair die will be made. Let X be the sum of these thousand independent rolls. Find the value of a so that P [350 a X a] = 0.9. Exercise 6.0. It is assumed that the probability that a student in college has a GPA in high school of A- or above is 5%. Suppose that 500 students from that college are chosen at random, what is the probability that 0 students or less of these 500 students have a GPA in high school of A- or above. Exercise 6.. Let Y,..., Y 00 independent identically distributed random variables with exponential distribution of parameter λ =. Find approximately Pr[Y + + Y 00 0]. Exercise 6.. One has 00 bulbs whose light times are independent exponentials with mean 5 hours. If the bulbs are used one at time, with a failed bulb being immediately replaced by a new one, what is the probability that there is still a working bulb after 55 hours. Exercise 6.3. From past experience a professor knows that the test score of a student taking her final examination is a random variable with mean 70 and variance 5. Assuming that the distribution of grades is approximately bell shaped, what can be said about the probability that student s score is between 60 and 80? Exercise 6.4. A manufacturer of booklets packages them in boxes of 000. It is known that, on the average, the booklets weigh ounce, with a standard deviation of 0.05 oz. The manufacturer is interested in calculating P [00 booklets weigh more 00.4 oz], a number hat would help detect whether too many booklets are being put in a box. Estimate last probability. Exercise 6.5. The scores of a reference population in the Wechsler Intelligence Scale top 5% of the population? Exercise 6.6. Suppose that the weight in ounces of major league baseball is a random variable with mean µ = 5 and standard deviation σ = /5. Find the probability that a carton of 44 baseballs has a total weight less than 75 ounces. Exercise 6.7. A manufacturer packages some item in boxes of hundred. The weight of each item has mean oz. and standard deviation 0. oz. Approximate the probability that a box weights more than 0 oz.

7 6.3. BROWNIAN MOTION Brownian Motion Definition A stochastic process {Bt : t 0} is said to be a Brownian motion process with variance parameter σ > 0 if: i B0 = 0. ii independent increments For each 0 t < t < < t m, Bt, Bt Bt,..., Bt m Bt m are independent r.v. s. iii stationary increments For each 0 s < t, Bt Bs has a normal distribution with mean zero and variance σ t s. If σ =, we said that {Bt : t 0} is a standard Brownian motion. If {Bt : t 0} is a Brownian motion process with variance parameter σ > 0, then {σ Bt : t 0} is a standard Brownian motion. So, the study of Brownian motion reduces to the case of a standard Brownian motion. Unless, it is said otherwise, {Bt : t 0} will denote a standard Brownian motion. Theorem 6.7. Let {Bt : t 0} a standard Brownian motion. Then, the probability density function of Bt is f Bt x = e x t. πt Example 6.7. Let {Bt : t 0} a standard Brownian motion. Let 0 < s < t. Let a, b R. Show that: i CovBs, Bt = s. ii VarBt Bs = t s. iii VaraBs + bbt = a + b s + b t s. iv The distribution of abs + bbt is normal with mean zero. Solution: i Since Bs and Bt Bs are independent r.v. s, CovBs, Bt = CovBs, Bs + Bt Bs = CovBs, Bs + CovBs, Bt Bs = VarBs = s. ii Since Bt Bs has a normal distribution with mean zero and variance t s, VarBt Bs = t s. iii VaraBs + bbt = VaraBs + bbs + Bt Bs = Vara + bbs + bbt Bs = Vara + bbs + VarbBt Bs = a + b s + b t s iv Since Bs and Bt Bs are independent r.v. s and they have a normal distribution, a + bbs + bbt Bs = abs + bbt has a normal distribution.

8 68 CHAPTER 6. BROWNIAN MOTION Example 6.8. Suppose that {Xt : t 0} is a Brownian motion process with variance parameter σ = 9. Find: i P{X 5} ii Var3X X5. iii P{X X3 4} Solution: We have that {Bt : t 0} is a standard Brownian motion, where Bt = Xt 3. i We have that ii P{X } = P{B 4} = P{N0, 4 } = P{Z.884} = Var3X X5 = Var9B 6B5 = Var3B 6B5 B = = 7 iii X X3 has a normal distribution with mean zero and variance VarX X3 = Var3B 6B3 = Var 3B 6B3 B = = 54 So, P{X X3 4} = P{Z 4 54 } = P{Z } = Example 6.9. Let {Bt : t 0} a standard Brownian motion. Let 0 < s < t < u. Show that E[BsBtBu] = 0. Solution: Let X = Bs, let Y = Bt Bs and let let Z = Bu Bt. Then, Z and Y and Z are independent r.v. s with mean zero and E[X ] = s, E[Y ] = t s and let let E[Z ] = u t. So, E[BsBtBu] = E[XX + Y X + Y + Z] = E[X 3 + X Y Z + XY + X Z + XY Z] = E[X 3 ] + E[X ]E[Y ]E[Z] + E[X]E[Y ] + E[X ]E[Z] + E[X]E[Y ]E[Z] = 0 Exercise 6.8. Let {Bt : t 0} a standard Brownian motion. Let 0 < s < t < u. Let a, b, c R. Show that: i VaraBs + bbt + cbu = a + b + c s + b + c t s + c u t. ii The distribution of abs + bbt + cbu is normal with mean zero. Exercise 6.9. Find the density function of X = Bs + Bt, where 0 s < t? Exercise 6.0. Find the density function of X = B 3B3 + 4B5. Exercise 6.. Suppose that {Xt : t 0} is a Brownian motion process with variance parameter σ = 8. Find: i P{ X4 X > 0}. ii Var3 + X4 X + X3. iii Cov3 + X4 X, 5 X3.

9 6.3. BROWNIAN MOTION 69 Recall that if X,..., X m is a r.v. with joint density function f X,...,X m x,..., x m, and h is a smooth transformation on the domain of X,..., X m, then the joint density of Y,..., Y m = hx,..., X m, is f Y,...,Y m y,..., y m = f X,...,X m h y,..., y m Jh y,..., y m, where Jh y,..., y m is the determinant of the partial derivatives of the function h the Jacobian of the h. Theorem 6.8. Let {Bt : t 0} a standard Brownian motion. Let 0 < t <, t m. Then, the joint probability density function of Bt,..., Bt m is exp x t + x x t t + + xm x m t m t m f Bt,...,Bt mx,..., x m = π m. t t t t m t m Proof. Let X = Bt, X = Bt Bt,..., X m = Bt m Bt m. X, X,..., X m are independent r.v. s. X i has normal distribution with mean 0 and variance t i t i. So, the density of X i is f Xi x i = is πti t i e x i f X,...,X m x,..., x m = m i= f X i x i = m i= t i t i and the joint density function of X, X,..., X m x i πti t i e t i t i = π m t t t t m t m exp x t x t t x m t m t m Consider the transformation y, y,..., y m = hx, x,..., x m = x, x + x,..., x + + x m. We have that h y, y,..., y m = y, y y,..., y m y m and Jh y,..., y m =. Then, Y, Y,..., Y m = hx, X,..., X m = Bt, Bt,..., Bt m. The joint density of Y, Y,..., Y m is f Y,...,Y m y,..., y m = f X,...,X mh y,..., y m Jh y,..., y m = f X,...,X my, y y,..., y m y m = π exp y m t t t t m t m t y y ym y m t t t m t m Q.E.D. Markov property of the Brownian motion: Given 0 < s < < s k < t < < t m, we have that the distribution of Bt,..., Bt m given Bs,..., Bs k agrees with the distribution of Bt,..., Bt m given Bs k. Theorem 6.9. Let {Bt : t 0} a standard Brownian motion. Given 0 < s < < s k < t < < t m, we have that the conditional density of Bt,..., Bt m given Bs,..., Bs k agrees with the conditional density of Bt,..., Bt m given Bs k. Proof. Let X = Bs,..., X k = Bs k, Y = Bt,..., Y m = Bt m.

10 70 CHAPTER 6. BROWNIAN MOTION The joint density of X,..., X k, Y,..., Y m is f X,...,X k,y,...,y m x,..., x k, y,..., y m = π m+k s s s s k s k t s k t t t m t m exp x s x x x k x k y x k y y ym y m s s s k s k t s k t s t m t m. The joint density of X,..., X k is f X,...,X k x,..., x k = π exp x k s s s s k s k s x x x k x k s s s k s k So, the conditional density of Y,..., Y m given X,..., X k is f Y,...,Y m X,...,X k y,..., y m x,..., x k = π exp y x k y y ym y m m t s k t t t m t m t s k t t t m t m. The joint density of X k, Y,..., Y m is f Xm,Y,...,Y m x k, y,..., y m = The joint density of X k is π +m sk t s k s s t m t m exp x k s k y x k y y ym y m t s k t t t m t m. f Xk x k = πsk exp x k s k So, the conditional density of Y,..., Y m given X k is f Y,...,Y m X k y,..., y m x k = π exp y x k y y ym y m m t s k t t t m t m t s k t t t m t m. Therefore, the two conditional densities f Y,...,Y m X,...,X k y,..., y m x,..., x k and f Y,...,Y m X k y,..., y m x k agree. Q.E.D. Theorem 6.0. For 0 s < t, the distribution of Bt given Bs is normal with mean Bs and variance t s. Bt Bs d NBs, t s. Proof. By the computations in the previous theorem, the conditional density of Y = Bt given X = Bs is f Y X y x = exp y x. πt s t s This is the density of a normal r.v. with mean x and variance t s. Q.E.D. Example 6.0. Suppose that {Xt : t 0} is a Brownian motion process with variance parameter σ = 4. Find: i P{X3 X = /} ii P{X4 X > X = /}.

11 6.3. BROWNIAN MOTION 7 Solution: We have that {Bt : t 0} is a standard Brownian motion, where Bt = Xt. i So, P{X3 X = } = P{B3 B = } = P{B3 B B = } = P{B3 B B = } = P{B3 B } = P{Z 0.5} = ii Since X and X4 X are independent r.v. s P{X4 X > X = /} = P{X4 X > } = P{B4 B > } = P{Z > } = P{Z > } = Theorem 6.. For 0 s < t, the distribution of Bs given Bt is normal with mean s st s Bt and variance. Bs Bt d N s st s Bt,. t t t t Proof. Let X = Bs and let Y = Bt. By the computations in two theorems above, the joint density of X, Y f X,Y x, y = exp x y x. π st s s t s The marginal density of Y is f Y y = πt exp y t. The conditional density of X given Y = y is f X Y x y = = π st s t = π st s t exp x π st s s y x t s = exp y πt t π st s t y x y + xy x t s t s t s t s exp exp = exp π st s t tx sy + xy st s tt s t s x sy t st s t, exp y x y x t s t s which is the density of a normal r.v. with mean sy t and variance st s t. Q.E.D. Example 6.. Suppose that {Xt : t 0} is a Brownian motion process with variance parameter σ = 3. Find: P{X X = 5 4 } Solution: We have that { P X X = 5 } { = P B B = 5 } { = P B 3 3 B = 5 } 3. We have that [ E B B = 5 3] =

12 7 CHAPTER 6. BROWNIAN MOTION and So, { P B 3 3 B = 5 } 3 = P{Z Var B B = 5 3 = } = P{Z } = Hitting Time Let {Bt : t 0} be a standard Brownian motion. Let a > 0. Let T a denote the first time the Brownian motion process hits a. Then for t > 0, So, P{Bt a} = Pr{T a t} Pr{Bt a T a t} + Pr{T a > t} Pr{Xt a T a > t} = Pr{T a t} + Pr{T a > t}0 = Pr{T a t}. Pr{T a t} = P{Bt a} = P{N0, a t } = Φ a t = a π t e y dy. The density of T a is We also have that f Ta t = e a t π a t 3/ = ae a t. πt 3/ Pr{T a t} = P{Bt a} = P{ Bt a}. Example 6.. Let T a be the time until a standard Brownian motion process hits a. Show that E[T a ] =. Solution: because lim t E[T a ] = ae a t πt / a πt / 0 tf Ta t dt = = and 0 t ae a t dt = ae a t dt =, πt 3/ πt / a πt / dt =. Exercise 6.. Suppose that you own one share of a stock whose price changes according to a standard Brownian motion process. Suppose that you purchased the stock at a price a and the present time price is b, where b < a. You decide to sell the stock when it reaches the price a. What is the average time it takes the stock to recover to the original purchase price? Example 6.3. Let T a be the time until a standard Brownian motion process hits a. Calculate the following: i P{T 8}. ii The median of T. 0

13 6.4. HITTING TIME 73 Solution: i P{T 8} = P{B8 } = P ii Let m be the median of T. Then, { Z 8 } = P{Z } = / = P{T m} = P{Bm }. So, 0.75 = P{Bm < } = P{Z < m }, m = and m = Exercise 6.3. Let T a be the time until a standard Brownian motion process hits a. Calculate the following probabilities and percentiles: i P{T 3 8}. ii The 99-th percentile of T 3. Problem 6.. # 5, November 000. You own one share of a stock. The price is 8. The stock price changes according to a standard Brownian motion process, with time measured in months. Calculate the probability that the stock reaches a price of at some time within the next 4 months. A B C D E Solution: Let T 3 be the hitting time of 3. We need to find Theorem 6.. Given a > 0 > b, then Pr{T 3 4} = Φ3/ = P{Bt hits a before b} = b a + b. Example 6.4. You own one share of a stock. The price is 8. The stock price changes according to a standard Brownian motion process, with time measured in months. Calculate the probability that the stock reaches a price of before it reaches. Solution: 6 9. Let {Bt : t 0} be a standard Brownian motion. The maximum process over [0, t] is M t = max 0 s t Bs. For a > 0, and So, the density of M t is M t has the density of Bt. Pr{M t a} = Pr{T a t} = P{ Bt a} Pr{M t a} = P{ Bt a} = a 0 f Mt a = { πt e a t if a 0 0 if a < 0 πt e x t dx.

14 74 CHAPTER 6. BROWNIAN MOTION Example 6.5. E[M t ] = and Solution: E[M t ] = t π 0 and VarM t = tπ π a e a t πt t da = e a t π E[M t ] = E[Bt ] = t. a= 0 = t π Example 6.6. Suppose that {Xt 0 t < } is a Brownian motion process with variance parameter 9. Calculate the following: i P{max t [0,] Xt 4}. ii The median value of max t [0,] Xt. iii The mean value of max t [0,] Xt. Solution: i ii P{max t [0,] Xt 4} = P{max t [0,] Bt 4} = P{ B 4} 3 3 = P{ 3 4 Z 3 4 } = E[max Xt] = 3E[max Bt] = 3 t [0,] t [0,] π = 6. π iii Let m be mean value of max t [0,] Xt. Then, So, P{Z / = P{max t [0,] Xt m} = P{max t [0,] Bt m/3} = P{ B m/3} = P{B m/3} = P{Z m 3 }. 3 m } = 3/4, m 3 = and m =.8668 Continuity of the Brownian motion: In some sense, a function of t, the Brownian motion is a continuous function on t. A Brownian motion can be used to give probabilities to continuous function. The price of a stock is a continuous function. A Brownian motion can be used to estimate probabilities related with the price of a stock over time. 6.5 Modeling Stock Prices Definition X is said to have a lognormal distribution with parameters µ and σ, if ln X has a normal distribution with mean µ and variance σ. If X is said to have a lognormal distribution with parameters µ and σ, then X has the distribution of e µ+σz, where Z is a standard normal r.v. Example 6.7. Let X be a r.v. σ = 5. Find Pr{X 3}. with lognormal distribution with parameters µ = and

15 6.5. MODELING STOCK PRICES 75 Solution: Pr{X 3} = Pr{ln X ln 3} = Pr{Z ln 3 5 } = Pr{Z } = Example 6.8. Let X be a r.v. with lognormal distribution with parameters µ and σ. Show that: σ µ+ i E[X] = e. ii E[X ] = e µ+σ. iii VarX = e µ+σ e σ. iv For t > 0, Pr{X t} = Φ ln t µ σ, where Φt = Pr{Z t}. and Solution: We have that X has the distribution of e µ+σz. So, E[X] = E[e µ+σz σ µ+ ] = e E[X ] = E[e µ+σz ] = e µ+σ. Example 6.9. Let X be a lognormal r.v. with E[X] = e 5 and E[X ] = e 4. Find: i PrX e 7. ii The third quartile of X. So, Solution: We have that e 5 σ µ+ = E[X] = e, e 4 = E[X ] = e µ+σ. 5 = µ + σ, 4 = µ + σ Hence, µ = 3 and σ = 4. Thus, PrX e 7 = Prln X m = PZ = PZ = Let q be the third quartile of X. Then, and = 0.75 = PrX m = Prln X 7 = PZ ln m 3 4 ln m 3 and m = Definition We say that a stochastic process {Xt : t 0} is a Brownian motion with drift coefficient µ and variance parameter σ if Xt = σbt + µt, where {Bt : t 0} is a standard Brownian motion. If {Xt : t 0} is a Brownian motion with drift coefficient µ and variance parameter σ, then: i X0 = 0. ii {Xt : t 0} has stationary and independent increments iii Xt is normally distributed with mean µt and variance σ t.

16 76 CHAPTER 6. BROWNIAN MOTION Definition We say that a stochastic process {Xt : t 0} is a geometric Brownian motion with drift coefficient µ and variance parameter σ if Xt = e σbt+µt, where {Bt : t 0} is a standard Brownian motion. Suppose that a share of a certain stock is currently i.e. t = 0 selling for P 0. Set P t = P 0 Zt = P 0 e σbt+µt, where Zt is a geometric Brownian motion with drift coefficient µ and variance parameter σ and Bt is a standard Brownian motion. {P t : t 0} is called a price process model. The geometric Brownian motion is used to model prices of stock, because it is assumed that the rate of interest earned over disjoint interval of times are independent r.v. s. The difference P t P s is the interest I[s, t] earned over the period [s, t]. The effective P t P s rate of interest earned in the period [s, t] is, i.e. this is the interest earned per unit of investment in that period of time. I[s, t] = For the price process and 0 < t < < t m, P s P t P s. P s I[t, t ] = P t P t P t = e σbt Bt +µt t,..., I[t m, t m ] = P tm P t m P t m = e σbtm Bt m +µt m t m. So, I[t, t ],..., I[t m, t m ] are independent r.v. s. The average accumulation of P t is At = E[P t] = P 0 e µt+ σ t. Given an accumulation function At, the force of interest at time t is The force of interest of At = E[P t] is δ t = A t At = d ln At. dt δ t = d dt ln At = µ + σ. Exercise 6.4. To model the price of a share of stock with current price of 0, we use the model P t = 0Xt, where Xt is a geometric Brownian motion model with µ = 0 and σ = i Calculate the probability that the price hits 0e 0.5 by time. ii Calculate the median time that it is required for the price to hits 0e 0.5. iii Calculate the mean time that it is required for the price to hits 0e 0.5. iv Calculate the probability that the maximum price of the stock over the time interval [0, ] exceeds 0e v Calculate the median value of the maximum price of the stock over the time interval [0, ]. vi Calculate the median value of the maximum price of the stock over the time interval [0, ]. vii Calculate the expected value and the variance of the price of the stock at the time. viii Calculate the probability that the stock increases by more than 0% in a two year period. Exercise 6.5. To model the price of a share of stock with current price of 0, we use the model P t = Xt, were Xtis a geometric Brownian motion model with µ = and

17 6.5. MODELING STOCK PRICES 77 σ = i Calculate the probability that the price increases by more than 0% before it decreases by more than 0%. ii Calculate the probability that the price increases by more than 5% before it decreases by more than 0%. Exercise 6.6. Suppose that stock currently sells for 00 and that the price process {P t : t 0}is modeled by a multiple of a geometric Brownian motion, where the drift parameter µ = 0.0 and the variance parameter is σ = 0.0. At a force of interest of δ = 0.0 the present value of the price at time t is e δt P t. Calculate the following: i E[P ], VarP and Pr{P 00e 0.03 }. ii The probability the maximum present value of the price over [0, ] exceeds 00e iii The probability the present value of the price hits 00e 0.4 before time 5. iv The probability the present value of the price hits 00e 0.0 before it hits 00e 0.. Problem 6.. # 7, Sample Test. You are given: The logarithm of the price of a stock can be modeled by Brownian motion with drift coefficient µ = 0 and variance parameter σ = The price of the stock at time t = 0 is 0. Calculate the probability that the price of the stock will be or greater at some time between t = 0 and t =. A 0.03 B 0.8 C 0.73 D 0.36 E Solution: Let P t be the price of the stock at time t. P t = 0e 0.04Bt, where Bt is a standard Brownian motion. We need to find Pr{sup 0 t P t } = Pr{sup 0 t 0e 0.04Bt } = Pr{sup 0 t Bt 5 ln/0} = Pr{ N0, 5 ln/0} = Problem 6.3. # 8, November 00. The value of currency in country M is currently the same as in country N i.e., unit in country M can be exchanged for unit in country N. Let Ct denote the difference between the currency values in country M and N at any point in time i.e., unit in country M will exchange for + Ct at time t. Ct is modeled as a Brownian motion process with drift 0 and variance parameter 0.0. An investor in country M currently invests in a risk free investment in country N that matures at.5 units in the currency of country N in 5 years. After the first year, unit in country M is worth.05 in country N. Calculate the conditional probability after the first year that when the investment matures and the funds are exchanged back to country M, the investor will receive at least.5 in the currency of country M. A 0.3 B 0.4 C 0.5 D 0.6 E 0.7 Solution: We have that Ct = 0.0Bt = 0.Bt, where Bt0 is a standard Brownian motion. We have to find

18 78 CHAPTER 6. BROWNIAN MOTION P{.5 C5 >.5 C =.05} = P{ > + 0.B5 + 0.B =.05} = P{0 > B5 + 0.B =.05} = P{0 > B5 B =.5} = P{B5 B < 0.5 B =.5} = P{B5 B < 0.5} = P{N0, 0.5} = 0.4. Problems. Let {Bt : t 0} be a standard Brownian motion. i Find the mean and the variance of B + B5 3B3. ii Find Pr B + B5 3B3 3.. Let {Bt : t 0} be a standard Brownian motion. Show that for 0 < s < t < u < v, the variance of abs + bbt + cbu + dbv is a + b + c + d s + b + c + d t s + c + d u t + d v u. 3. Let {Xt : t 0} be a Brownian motion process with variance parameter σ = 3. Find the following: i Pr X 5. ii Pr X3 X4 5. iii Pr X X + 3X Let {Bt : t 0} be a standard Brownian motion process. Find the following: i Pr B 3 B =. ii Pr B3 B4 B = 3. iii Pr0 B3 4 B5 = Let {Xt : t 0} be a Brownian motion process with variance parameter σ = 9. Find the following: i Pr3 X 9 X = 6. ii Pr X X5 X3 = 3 6. Let {Bt : t 0} be a standard Brownian motion. Let t > 0 and let M t = sup 0 s t Bs be the maximum process over t. Find: i Pr M 9 5. ii The median of M Let {Bt : t 0} be a standard Brownian motion. Let t > 0 and let M t = sup 0 s t Bs be the maximum process over t. Find: i PrM 4 5. ii The density of M 4. iii The median of M 4. iv The first and the third quartile of M 4. v The mean and the variance of M 4.

19 6.5. MODELING STOCK PRICES Let {Bt : t 0} be a standard Brownian motion. Let T a be the first time that the Brownian motion hits a. Find the following: i PrT 4. ii The density of T. iii The median of T. iv Pr T 3 5. v The first and the third quartile the 5 % and 75 % percentile of T Use a Geometric Brownian Motion model with µ = 0 and σ = 0.09 to model the ratio P t/0, where P t is the price at time t for a share of a stock currently selling at 0. What is the probability that the stock increases by more than 0% in a two-year period? 0. You own one share of a stock. The price is. The stock price P t changes according to the model P t = e 0.5Bt, where Bt, t 0, is a standard Brownian motion and the time is measured in months. i Calculate the probability that the stock reaches a price of 30 at some time within the next 4 months. ii Calculate the median of the time that it takes the price to reach 30.. You own one share of a stock. The current price of the stock is 0. Let P t be the P t stock price at time t. changes according to a Geometric Brownian Motion model P 0 with µ = 0. and σ = 0.4, with time measured in months. Find: i The mean and the variance of the price of the stock one year from now. ii The probability that the price of the stock 6 months from now is more than 30.. You own share of stock currently worth 0. Assume that the change in value of this share over time follows a standard Brownian motion process where time is measured in months. What is the probability that the price three months from now is greater than 05? 3. You own one share of a stock. The price is 50. The stock price P t changes according to a geometric Brownian Motion model with µ = 0 and σ = 0.36, with time measured in months. Let S be the time, P t hits 70. Show that the distribution of S is that of T a, where T a is the time that a standard Brownian motion hits a > 0, for some a. Find: i a. ii Pr S 4. iii The median of S. iv The first and the third quartile the 5 % and 75 % percentile of S. 4. Using a Geometric Brownian Motion model with µ = 0, σ = 0.09 to model the ratio P t/0, where P t is the price at time for a share of a stock currently selling at 0, calculate the probability that the price hits 5 by time. 5. You own one share of a stock. The price is 8. Let P t be the stock price at time t. changes according to a Geometric Brownian Motion model with µ = 0. and σ = 0.49, P t P 0

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