STAT S315F Unit 4 (LN)(Sept 2016 Presentation) PDF

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THE OPEN UNIVERSITY OF HONG KONGSTAT S 31 5F Applied Probability Models for InvestmentUnit 4 Brownian Motion1. IntroductionDiffusion processes, which are random processes that vary continuously in continuous time.Brownian motion, and more general classes of diffusion, continue to be of great importa...

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Lecture Note

THE OPEN UNIVERSITY OF HONG KONG STAT S315F Applied Probability Models for Investment Unit 4 Brownian Motion 1. Introduction Diffusion processes, which are random processes that vary continuously in continuous time. Brownian motion, and more general classes of diffusion, continue to be of great importance in modern physics and chemistry. However, these models also have other applications. Continuous random motions arise in economics (if the price of a share is held to vary at random, but not so much at random as to allow the price to jump sizeably). In 1900, the research worker Louis Bachelier put forward the proposition that stock market prices could be modelled by a version of a random walk. He argued that these prices were subject continually to influences causing them to rise and fall, and that these influences were, to all intents and purposes, random. One very long record of stock market prices is the list of wheat prices in Chicago, measured in cents per bushel, and published at weekly intervals between January 1883 and September 1934 (with observations for 1915 to 1920 missing). A graph of the first two years of wheat prices in this series is shown in Figure 1.

Figure 1 The price of wheat in Chicago STAT S315F

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Although the wheat price was published only once a week, it could change at any time, and the price varies continuously (though rounded to the nearest cent). Therefore if X(t) is the price at time t, then {X(t); t ≥ 0} is a random process for which both time and X(t) are continuous, and it is an example of a diffusion process. In this section, a random walk will be used to model wheat prices. By considering properties of the random walk and examining what happens when the length of the interval between steps and the size of the steps are reduced, a continuous model for a random process in continuous time will be developed. Let X(t) denote the wheat price at time t relative to the value at time 0, so that X(0) = 0. Suppose that the price changes at times separated by a fixed interval of duration h, and that it is observed at each of the times 0, h, 2h, 3h, . . .. For the Chicago wheat prices, time is measured in weeks, a week being the interval between times when the current price was published, but h might be much smaller than 1. (The price might change by the day, or hourly, or every minute.) Next suppose that whenever a change occurs, the wheat price has probability 1/2 of increasing by an amount a, and probability 1/2 of decreasing by a, where a is some constant fixed over the whole period of observation. Finally, suppose that the changes are independent of one another. With these assumptions, X(t) can be modelled by a random walk: for n = 1, 2, X(nh) = X((n − 1)h) + Z n ,

where

. . .,

P( Z n = a) = P( Z n = −a) = 1/2, and Z 1 , Z 2 , . . .

are independent. It follows that: X(nh) = Z 1 + Z 2 + … + Z n . Since P( Z i = a) = P( Z i = −a) = 1/2, for i = 1, 2, . . ., E( Z i ) = 0, V( Z i ) = a 2 . Therefore E[X(nh)] = 0, and since Z 1 , Z 2 , …, Z n . are independent, V [X(nh)] = na 2 . Writing t = nh gives E[X(t)] = 0, V [X(t)] =

ta 2 . h

As a result, E[X(t)] = 0 and V [X(t)] is proportional to t. This random process should be expected to possess properties similar to those of the random walk of which it is, in a sense, the limit. For example, since the random walk is symmetric, the continuous random process should be symmetric: E[X(t)] = 0.

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Also, it should have the Markov property: if the value of X(u) is known, then further information about values before time u is not relevant to its development after time u. Moreover, X(u, u + t), the change in position between time u and time u + t, which is equal to X(u + t) − X(u), should have the same probability distribution as X(0, t) =X(t), and be independent of the history of the process before time u. The random process {X(t); t ≥ 0} is said to possess stationary independent increments. This idea is illustrated in Figure 2.

Figure 2 Stationary independent increments The requirements of a random process {X(t); t ≥ 0}, with X(0) = 0, that have been described, may for convenience be referred to as symmetry, stationarity and continuity. In the 1920s, Norbert Wiener was the first to show that it is possible to find a random process {X(t); t ≥ 0} satisfying the conditions of symmetry, stationarity and continuity simultaneously. Such a process is called a Brownian motion, or a Wiener process. The process is assumed to begin at time 0 with X(0) = 0. The Wiener process or Brownian motion is defined in the following box.

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Brownian motion The Wiener process or Brownian motion is a random process {X(t); t ≥ 0} with X(0) = 0 that has the following properties.  Every increment X(u, u + t) is normally distributed with mean 0 and variance

 2t , where  2 is a constant called the rate or diffusion coefficient of the process. 

For every pair of disjoint time intervals [ t1 , t2 ], [ t3 , t4 ], the increments X(t 1, t 2 ) and X( t3 , t4 ) are independent random variables.

 The random function t → X(t) is continuous. Remarks: Note that from the first property, the distribution of X(t) is X(t) ~N(0,  2 t ). From the second property, it follows that Brownian motion possesses the Markov property.

2. The Ordinary Brownian Motion By definition, for ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient

 2 , the distribution of the change in position in any time interval of duration t is normal with mean 0 and variance  2 t : for u ≥ 0, t > 0, X(u, u + t) = X(u+t)−X(u) ~ N(0,  2 t ). In particular, since X(0) = 0 by definition, X(t) = X(0, t) ~ N(0,  2 t ). The use of this result is illustrated in Example 1. Example 1 The vertical motion of a particle up and down a straight line may be modelled by ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient  2 = 0.3 cm 2 per second. The distribution of the particle’s position after 30 seconds is given by with t = 30: X(30) ~ N(0, 0.3 × 30) = N(0, 9).

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Therefore the probability that the particle will be more than 5 cm above its starting point after 30 seconds is 5 0  P( X (30)  5)  1     1  (1.67)  1  0.9525  0.048.  9 

Similarly, the probability that the particle will be within 2 cm of its starting point after 30 seconds is  20  20 P ( 2  X (30)  2)         (0.67)  ( 0.67)  0.497 . 9   9  

The probability that the particle will be more than 6 cm away from its starting point (in either direction) after 30 seconds is

  6  0  P ( X (30)  6)  2  1      0.046 .  9  

Example 2 (Using the Markov property) In Example 1, the vertical motion of a particle was modelled by ordinary Brownian motion with diffusion coefficient 

2

= 0.3 cm 2 per second.

By definition, X(0) = 0. Suppose that after 5 seconds the particle is observed to be 2 cm below its starting point, and that after 10 seconds it is observed to be 3 cm above its starting point: X(5) = −2 and X(10) = 3. Then the probability that the particle will be less than 5cm above its starting point after 30 seconds is given by the conditional probability: P(X(30) < 5 |X(10) = 3,X(5) = −2 and X(0) = 0). By the Markov property, this is equal to P(X(30) < 5 |X(10) = 3), and this probability is the same as P(X(30) − X(10) < 5 − 3) = P(X(10, 30) < 2). For t > 0, X(u, u + t) ~ N(0,  2 t ), so X(10, 30) ~ N(0, 0.3 × 20) = N(0, 6).  20  Hence the required probability is P( X (10, 30)  2)     (0.82)  0.7939 .  6 

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3. Conditional Distributions Now suppose that the values of X( t1 ) and X( t2 ) are observed. What can be said about the value of X(t) for t1 < t < t2 ? That is, what can be said about the history of the process between the times at which it is observed? In order to answer questions about X(t), the conditional distribution of X(t) given the values of X(t1 ) and X( t2 ) is required. This distribution is stated without proof in the following box. Conditional distributions If {X(t); t ≥ 0} is ordinary Brownian motion with diffusion coefficient  2 , then the conditional distribution of X(t) given X( t1 ) = x1 and X( t 2 ) = x2 , where t1< t < t 2 , is normal with mean c and variance  c2 given by

c 

x1 ( t2  t)  x2 (t  t1 ) t 2  t1

and

 c2 

(t2  t )(t  t1 ) 2  . t2  t1

Figure 3 Geometric interpretation of the conditional mean Remarks: In the special case where X( t 1) = X( t2 ) = 0, the conditioned random process conditional mean{(X(t) |X(t1) =X(t2) = 0); t1 ≤ t ≤ t2 } is known as a Brownian bridge. It describes a process on an interval that is constrained to have the same value at each end point.

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Example 3 In Example 1, the vertical motion of a particle was modelled by ordinary Brownian motion with diffusion coefficient 

2

= 0.3 cm 2 per second.

By definition, X(0) = 0. Suppose that after 30 seconds the particle is observed to be 9 cm above its starting point. What is the probability that after 10 seconds the particle was below its starting point? The probability required is P(X(10) < 0 |X(0) = 0 and X(30) = 9).

c 

0(30  10)  9(10  0) 3 30  0

and

 c2 

(30  10)(10  0) 0.3  2 . 30  0

So the conditional distribution of X(10) is N(3, 2), and hence  0 3  P(X(10) < 0 |X(0) = 0 and X(30) = 9) =    (2.12)  0.017 .  2 

Example 4 Suppose that the price of a share above its value at time 0 may be modelled as ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient  2 = 6 (pence) 2 per day. (a) Calculate the probability that after 25 days the share price will be more than 20 pence above the price at time 0. Solution The probability required is P(X(25) > 20). Using X(25) ~ N(0, 150), so  20  0  P (X (25)  20)  1    1  (1.63)  0.0516  150 

The share price is observed to be 20 pence below the starting price after 20 days, and 40 pence above the starting price after 50 days. (b) Given these observations, calculate the probability that the share price will be within 20 pence of the starting price after 100 days. Solution Since 100 days is later than all the given observations, the Markov property can be used. STAT S315F

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The probability required is

P(20  X (100)  20 | X (50)  40)  P( 60  X (50, 100)  20) As X(50, 100) ~ N(0, 300), so this probability is equal to   20  0    60  0        (1  0.8749)  (1  0.9997)  0.1248  300   300 

(c) Given these observations, calculate the probability that the share price was less than 10 pence above the starting price after 40 days. Solution Since 40 days lies between two of the times at which the price is known, we need to find the conditional distribution of X(40). The probability required is P( X (40)  10 | X (20)  20, X (50)  40) Consider:

c 

 20(50  40)  40(40  20)  20 50 20

and

 c2 

(50  40)(40  20)  6  40 . 50  20

hence the conditional distribution of X(40) is N(20, 40). Therefore the required  10  20  probability is equal to:    1  0.9429  0.0571  40 

3. Waiting Times Suppose that the vertical motion of a particle may be modelled as ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient  2 . In this section, the waiting time until a particular value a is attained is studied. This waiting time is denoted Wa . The probability that Wa , the waiting time until the value a is first attained, is greater than τ , is equal to the probability that X(w) is less than a for all w, 0 ≤ w ≤ τ : P( Wa > τ) = P(X(w) < a for all w, 0 ≤ w ≤ τ ). Thus the distribution of the waiting time Wa is related to the distribution of the random variable X(t) denoting the position of the particle at time t. STAT S315F

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There are three different types of realisation of the random process {X(t); t ≥ 0} that need to be considered. These correspond to the following three events. (1) [Wa > τ] (2) [Wa < τ and X(τ) < a] (3) [Wa < τ and X(τ) > a] These are the only three possibilities for the ordinary Brownian motion up to time τ. [ a = τ] and [X(τ) = a] Since Wa and X(t) are continuous random variables, the events W have probability 0 and can be discounted as virtually impossible. A realisation of type (1), (2) and (3) are illustrated in Figure 4 and 5 respectively.

Figure 4

Figure 5

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A realisation of type (1)

Two realisations of (a) type (2) and (b) type (3)

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Waiting Times For ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient  (for which X(0) = 0), the waiting time until the value a (a > 0) is first attained is denoted Wa . 2

  a    , where w > 0. The c.d.f. of Wa is FW (w )  P (Wa  w )  2 1    2  w   

The p.d.f. of Wa is fW (w) 

 a a2   , where w > 0. exp   2  w 2w  2 w 

Example 5 In Example 1, the vertical motion of a particle was modelled as ordinary Brownian motion with diffusion coefficient  2 = 0.3 cm 2 per second. The probability that the particle will not reach a point 5 cm above its starting point within the first minute can be found by finding the probability: P(W5 > 60).

 5    P(W5  60)  2 1      2(1  0.8810)  0.238  0.3 60    Therefore, P(W5 > 60) = 1 – 0.238 = 0.762

4. Brownian motion with drift For ordinary Brownian motion {X(t); t ≥ 0}, the distribution of X(t) is symmetric, so there is no preferred direction of motion. In this section, a model that allows for a preferred direction of motion is discussed briefly.

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Example 6 The variation in the price of a particular share may be modelled as Brownian motion with drift. The drift coefficient is 10 pence per month, and the diffusion coefficient is 400 (pence) 2 per month. Originally, the cost of the share to its purchaser was ￡2.50. So C(t), the price of the share t months after its purchase, is given by C(t) = 250 + 10t + X(t), where {X(t); t ≥ 0} is ordinary Brownian motion with diffusion coefficient 400 (pence) 2 per month. (a) Calculate the probability that one month after purchase, the price of the share will be lower than when it was bought. Solution The probability required is P(C(1) < 250). Since C(t) = 250 + 10t+X(t) and X(t) ~ N(0, 400t), it follows that C(1) = 250 + 10 +X(1) = 260 + X(1), where X(1) ~ N(0, 400). Therefore   10  0  P(C(1) < 250) = P(260 + X(1) < 250) = P(X(1) < −10) =   = 1 – 0.6915  400  = 0.3085

(b) It is known that four months after it was bought, the share was trading at ￡3. Calculate the probability that three months earlier (that is, one month after its purchase) its price was less than ￡2.50.

Solution The probability required is P(C(1) < 250 |C(0) = 250, C(4) = 300). Since C(t) = 250 + 10t + X(t), this probability is equal to P(X(1) < −10 |X(0) = 0, X(4) = 10). STAT S315F

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Since t = 1 lies between t = 0 and t = 4, the conditional distribution of X(1) is normal with mean c 

0(4  1)  10(1  0) (4  1)(1 0)  2.5 and variance c2   400  300 . 4 0 4 0

  10  2.5  Therefore P(X(1) < −10 |X(0) = 0, X(4) = 10) =   = 1 – 0.7642 = 0.236. 300  

5. Brownian motion with drift If the vertical movement of a particle is modelled as ordinary Brownian motion, then in any interval the particle is as likely to move some distance upwards as it is to move the same distance downwards. However, some quantities that fluctuate continuously up and down in a random manner do so in a different way. For instance, in a given interval, the value of a quantity may be as likely to double as it is to halve, and as likely to treble as it is to reduce to one third of its initial value, and so on. In this section, a model for quantities that fluctuate in this way is described. This model is called geometric Brownian motion. It is derived from ordinary Brownian motion with diffusion coefficient 

2

as described in the following box.

Geometric Brownian motion The random process {Z(t); t ≥ 0} obtained from ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient  2 through the relationship Z(t) = exp X(t), or equivalently ln Z(t) = X(t), is called geometric Brownian motion. By definition, X(0) = 0 for ordinary Brownian motion, so Z(0) = 1 for geometric Brownian motion. In addition, since Z(t) = exp X(t), it follows that Z(t) > 0 for all t. Therefore geometric Brownian motion is a model for a quantity that takes only positive values. Also, some properties of ordinary Brownian motion carry over to geometric Brownian motion; for example, geometric Brownian motion possesses the Markov property. Realisations of geometric Brownian motion are very spiky and impossible to represent exactly. A typical realisation of geometric Brownian motion {Z(t); t ≥ 0} is represented approximately in Figure 6.

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Figure 6

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A realisation of geometric Brownian motion

Notice that the runs up and down look very similar to periods of exponential growth and periods of exponential decay. The geometric Brownian motion model has been used to good effect in some financial contexts, for instance, in the US futures markets. By definition, for geometric Brownian motion {Z(t); t ≥ 0}, Z(t) = exp X(t), where {X(t); t ≥ 0} is ordinary Brownian motion with diffusion coefficient  2 . Thus the c.d.f. of Z(t) can be written in terms of the c.d.f. of X(t), as follows. For z > 0, P(Z(t) ≤ z) = P(exp X(t) ≤ z) = P(X(t) ≤ ln z). The distribution of Z(t) is called a lognormal distribution. Also, for t > 0, ln Z(t) = X(t) ~ N(0,  2 t) and the distribution of Z(t) is called a lognormal distribution. Therefore problems involving geometric Brownian motion {Z(t); t ≥ 0} can be converted into problems involving the underlying ordinary Brownian motion {X(t); t ≥ 0}.

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Example 7 The value of a price index at its base date (t = 0) is 100. Suppose that the ratio of the value of the index at time t to its base value is modelled as geometric Brownian motion {Z(t); t ≥ 0}, derived from ordinary Brownian motion {X(t); t ≥ 0} with diffusion coefficient 

2

= 0.25 per year.

The probability that the value of the index will be greater than 250 after 3 years is 250   P Z(3)    P(exp X (3)  2.5)  P ( X (3)  ln 2.5) 100    ln 2.5  0   1     1 0.8554  0.1446  0.75 

Therefore the probability that the value of the price index will be greater than 250 after 3 years is approximately 0.145.

Example 8 The value of the price index described in Example 7 is 100 at time 0. Suppose that afte...