Transition Density Function and Partial Di erential Equations

Transcription

1 Transition Density Function and Partial Di erential Equations In this lecture Generalised Functions - Dirac delta and heaviside Transition Density Function - Forward and Backward Kolmogorov Equation Similarity Reduction Method 1

2 The Dirac delta function The delta function denoted (x) ; is a very useful object in applied maths and more recently in quant nance. It is the mathematical representation of a point source e.g. force, payment. Although labelled a function, it is more of a distribution or generalised function. Consider the following de nition for a piecewise function ( 1 ; x 2 f (x) = 2 ; 2 0; otherwise 2

3 Now put the delta function equal to the above for the following limiting value (x) = lim f (x)!0 What is happening here? As decreases we note the hat narrows whilst becoming taller eventually becoming a spike. Due to the de nition, the area under the curve (i.e. rectangle) is xed at 1, i.e. 1 ; which is independent of the value of : So mathematically we can write in integral terms Z 1 1 f (x) dx = Z 2 = 1 1 f (x) dx + Z 2 2 = 1 for all : f (x) dx + Z 1 2 f (x) dx 3

4 Looking at what happens in the limit! 0; the spike like (singular) behaviour at the origin gives the following de nition with the property (x) = Z 1 ( 1 x = 0 0 x 6= 0 1 (x) dx = 1: 4

5 There are many ways to de ne (x) : Consider the Gaussian/Normal distribution with pdf G (x) = 1 p 2 exp 1 2 x 2! The function takes its highest value at x = 0; as jxj! 1 there is exponential decay away from the origin. If we stay at the origin, then as decreases, G (x) exhibits the earlier spike (as it shoots up to in nity), so The normalising constant will always be unity. lim!0 G (x) = (x) : 1 p 2 2 : ensures that the area under the curve 5

6 The graph below shows G (x) for values = 2:0 (royal blue); 1:0 (red); 0:5 (green); 0:25 (purple); 0:125 (turquoise); the Gaussian curve becomes slimmer and more peaked as decreases. 6

7 G (x) is plotted for = 0:01 Now generalise this de nition by centering the function f (x) at any point x 0 : So x x 0 = lim!0 f Z 1 x x x 0 dx = 1: 1 x 0 7

8 The gure will be as before, except that now centered at x 0 and not at the origin as before. So we see two de nitions of (x) : Another is the Cauchy distribution L (x) = 1 x

9 So here (x) = lim!0 1 x Now suppose we have a smooth function g (x) and consider the following integral problem Z 1 1 g (x) (x x 0) dx = g (x 0 ) This sifting property of the delta function is a very important one. 9

10 Heaviside Function The Heaviside function, denoted by H (), is a discontinuous function whose value is zero for negative parameters and one for positive arguments H (x) = ( 1 x > 0 0 x < 0 Some de nitions have and H (x) = 8 >< >: 1 x > x = 0 0 x < 0 H (x) = ( 1 x > 0 0 x 0 It is an example of the general class of step functions. 10

11 Probability Distributions At the heart of modern nance theory lies the uncertain movement of nancial quantities. For modelling purposes we are concerned with the evolution of random events through time. A di usion process is one that is continuous in space, while a random walk is a process that is discrete. The random path followed by the process is called a realization. Hence when referring to the path traced out by a nancial variable will be termed as an asset price realization. The mathematics can be achieved by the concept of a transition density function and is the connection between probability theory and di erential equations. 11

12 Trinomial Random Walk A trinomial random walk models the dynamics of a random variable, with value y at time t: is a probability. y is the size of the move in y. 12

13 The Transition Probability Density Function The transition pdf is denoted by p y; t; y 0 ; t 0 We can gain information such as the centre of the distribution, where the random variable might be in the long run, etc. by studying its probabilistic properties. So the density of particles di using from (y; t) to y 0 ; t 0 : Think of (y; t) as current (or backward) variables and y 0 ; t 0 as future ones. The more basic assistance it gives is with P a < y 0 < b at t 0 Z b y at t = p y; t; y 0 ; t 0 dy 0 a i.e. the probability that the random variable y 0 lies in the interval a and b;at a future time t 0 ; given it started out at time t with value y: 13

14 p y; t; y 0 ; t 0 satis es two equations: Forward equation involving derivatives with respect to the future state y 0 ; t 0 : Here (y; t) is a starting point and is xed. Backward equation involving derivatives with respect to the current state (y; t) : Here y 0 ; t 0 is a future point and is xed. The backward equation tells us the probability that we were at (y; t) given that we are now at y 0 ; t 0 ; which is xed. The mathematics: Start out at a point (y; t) : We want to answer the question, what is the probability density function of the position y 0 of the di usion at a later time t 0? This is known as the transition density function written p y; t; y 0 ; t 0 and represents the density of particles di using from (y; t) to y 0 ; t 0 : How can we nd p? 14

15 Forward Equation Starting with a trinomial random walk which is discrete we can obtain a continuous time process to obtain a partial di erential equation for the transition probability density function (i.e. a time dependent PDF). So the random variable can either rise or fall with equal probability < 1 2 and remain at the same location with probability 1 2: Suppose we are at y 0 ; t 0 ; how did we get there? 15

16 At the previous step time step we must have been at one of y 0 + y; t 0 or y 0 y; t 0 t or y 0 ; t 0 t : t So p y; t; y 0 ; t 0 = p y; t; y 0 + y; t 0 t + (1 2) p y; t; y 0 ; t 0 t +p y; t; y 0 y; t 0 t 16

17 Taylor series expansion gives (omit the dependence on (y; t) in your working as they will not change) p y 0 + y; t 0 t = p y 0 ; @t0t + p y 0 ; t 0 t = p y 0 ; + ::: p y 0 y; t 0 t = p y 0 ; 0t Substituting into the 0y + 2 p 02y2 0y y2 + ::: p y 0 ; t 0 = p y 0 ; @t0t 0y + 2 p 02y2 + (1 2) p y 0 ; 0t+ + p y 0 ; @t 0y + 2! p 02y2! 17

18 0t @t 0 = y Now take limits. This only makes sense if y2 t is O (1) ; i.e. y 2 O (t) and letting y; t! 0 gives 0 = 02; where c 2 = y2 t : This is called the forward Kolmogorov equation. Also called Fokker Planck equation. It shows how the probability density of future states evolves, starting from (y; t) : 18

19 Backward Equation The backward equation tells us the probability that we are at y 0 ; t 0 given that we were at an initial state (y; t) : y 0 ; t 0 are now xed and (y; t) are variables. 19

20 So the probability of being at y 0 ; t 0 given we are at y at t in the past is linked to the probabilities of being at y + y; t + t; y 0 ; t 0 ; y; t + t; y 0 ; t 0 and y y; t + t; y 0 ; t 0 : The backward equation is particularly important in the context of - nance, but also a source of much confusion. Illustrate with the real life example that Wilmott uses. Wilmott uses a Trinomial Random Walk: So a concrete example! 20

21 At 7pm you are at the o ce - this is the point (y; t) At 8pm you will be at one of three places: x The Pub - the point (y + y; t + t) ; x Still at the o ce - the point (y; t + t) ; x Madame Jojo s - the point (y y; t + t) We are interested in the probability of being tucked up in bed at midnight y 0 ; t 0 ; given that we were at the o ce at 7pm: 21

22 22

23 Remember p y; t; y 0 ; t 0 represents the probability of being at the future point y 0 ; t 0 ; i.e. bed at midnight, given that you started at (y; t) ; the o ce at 7pm. Looking at the earlier gure, we can only get to bed at midnight via either the pub the o ce Madame Jojo s at 8pm. 23

24 What happens after 8pm doesn t matter - we don t care, you may not even remember! We are only concerned with being in bed at midnight. The earlier gure shows many di erent paths, only the ones ending up in our bed are of interest to us. 24

25 In words: The probability of going from the o ce at 7pm to bed at midnight is the probability of going to the pub from the o ce and then to bed at midnight plus the probability of staying in the o ce and then going to bed at midnight plus the probability of going to Madame Jojo s from the o ce and then to bed at midnight 25

26 p y; t; y 0 ; t 0 = p y + y; t + t; y 0 ; t 0 + (1 2) p y; t + t; y 0 ; t 0 +p y y; t + t; y 0 ; t 0 : Now since y 0 ; t 0 do not change, drop these for the time being and use a TSE on the right hand side p (y; t) = which simpli es to p (y; y + 2 p 2y2 + ::: (1 2) p (y; t) t + p (y; @y y + 2! p 2y2 + ::: y 2 p 2:! + 26

27 Putting y2 t = O (1) and taking limit gives the backward equation or commonly = 1 p 2 2: c2@2 2 = 0: 27

28 The above can be expressed mathematically as p y; t; y 0 ; t 0 = p y + y; t + t; y 0 ; t 0 + (1 2) p y; t + t; y 0 ; t 0 + p y y; t + t; y 0 ; t 0 : Performing a Taylor expansion gives which becomes 0 + y2@2 2 + ::: 2 p y2 2 + ::: and letting y2 t = c 2 where c is non-zero and nite as t; y! 0; + p 2 = 0 28

29 Di usion Equation The model 0 = 02 for the unknown function p = p y 0 ; t 0 : The idea is to obtain a solution in terms of Gaussian curves. Let s drop the primed notation. We assume a solution of the following form exists: p (y; t) = t f y t where ; are constants to be determined. So put = y t which allows us to obtain the = 1 = yt 1 29

30 we can now say therefore p (y; t) = t @ = t f 0 () : 1 t = 2 @! t f 0 () = t f 0 () = t f 0 () f 0 () 2 f t f () + 1 f () we can use the chain rule @ f @t = yt 1 f 0 () 30

31 so = t 1 f () yt 1 f 0 () and then substituting these expressions in to the pde gives We know from that t 1 f () yt 1 f 0 () = c 2 t 2 f 00 : hence the equation above becomes y = t t 1 f () t 1 f 0 () = c 2 t 2 f 00 : For the similarity solution to exist we require the equation to be independent of t; i.e. 1 = 2 =) = 1=2; therefore f 1 2 f 0 = c 2 f 00 31

32 thus we have so far p = t f yp t which gives us a whole family of solutions dependent upon the choice of : We know that p represents a pdf, hence ZR p (y; t) dy = 1 = Z! du = dy= p t so the integral be- change of variables u = y= p t comes t +1=2Z 1 1 R t f yp dy t f (u) du = 1 which we need to normalize independent of time t: This is only possible if = 1=2: 32

33 So the D.E becomes 1 2 f + f 0 = c 2 f 00 : We have an exact derivative on the lhs, i.e. and we can integrate once to get 1 d 2d (f) = c2 f (f) = c2 f 0 + K: We set K = 0 in order to get the correct solution, i.e. d d (f) = f + f 0, hence 1 2 (f) = c2 f 0 which can be solved as a simple rst order variable separable equation: f () = A exp 1 4c 22 33

34 A is a normalizing constant, so write Z A exp 1 R 4c 22 d = 1: Now substitute x = =2c; so 2cdx = d 2cA Z exp x 2 dx = 1; R {z } = p which gives A = 1=2c p : Returning to becomes p y 0 ; t 0 = p (y; t) = t 1=2 f () 0 1 2c p t 0 y02 4t 0 c 2 This is a pdf for a variable y that is normally distributed with mean zero and standard deviation c p 2t; which we ascertained by the following 1 A : 34

35 comparison: 1 y 02 22t 0 c 2 : 1(x ) i.e. 0 and 2 2t 0 c 2 : If the random variable y 0 has value y at time t then we can generalize to p y; t; y 0 ; t 0 1 = 2c q (t 0 t) exp y 0 y 2 4c 2 (t 0 t) 1 A 35

36 At t 0 = t this is now a Dirac delta function y 0 y : This particle is known to start from q(y; t) and di uses out to y 0 ; t 0 with mean y and standard deviation c 2 (t 0 t): 36