Ordinary and Stochastic Di erential Equations Lecture notes in Mathematics PhD in Economics and Business. B. Venturi G. Casula, A.

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1 Ordinary and Stochastic Di erential Equations Lecture notes in Mathematics PhD in Economics and Business B. Venturi G. Casula, A. Pili

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3 Contents Part. Di erential Equations v Chapter. Ordinary di erential equations vii. De nition vii 2. First order di erential equations viii 3. Solution of rst order ODE ix 4. The separation of variables method ix 5. Examples of non-linear functions xii 6. Particular solution xiv 7. Singular integral xiv Chapter 2. Linear di erential equations xvii. ODE with non-constant coe cients xvii 2. Second order di erential equations xix 3. How to trasform a second order ODE xxiii Chapter 3. Systems of di erential equations xxv. First order system xxv 2. Stable and namic system xxvi 3. Market model with time expectation xxvi 4. Stability of the equilibrium point: knot and focus xxviii 5. Matrix method xxxi Part 2. Stochastic Di erential Equations xxxiii Chapter 4. Probability Space xxxv. De nition xxxv 2. Random Variable xxxvi 3. Stochastic process xxxvi 4. Examples of Stochastic Equations xxxviii 5. The Random Walk Process xl iii

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5 Part Di erential Equations

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7 CHAPTER Ordinary di erential equations. De nition A ordinary di erential equation (O.D.E.) is a mathematical expression within which we nd an unknown function y(x) and its rst derivative y0(x). Definition. An equation containing the derivatives of one or more dependent variables, with respect to one or more independent variables, is said to be a di erential equation. The order of the di erential equation is given by the maximum derivative that appears in the unknown function: if the highest-order derivative is the rst, we have a rst order di erential equation (and so on, if the highest-order derivative is the second, we have a second order di erential equation). For example: d 2 3 y dx + y = 0 2 dx is a second-order ordinary di erential equation, because we have d 2 y. The function y is de ned in an open interval from a to b in R: dx 2 y : (a; b) 2 R A di erential equation can be: () homogeneous, if the second term is zero: dx + y(x) = 0 (2) non-homogeneous, if we have a variable in the second term, like e x, which depends on x, which itself depends on the unknown function: + y(x) = ex dx vii

8 viii. ORDINARY DIFFERENTIAL EQUATIONS 2. First order di erential equations Let s consider: F (x; y(x); y 0 (x)) = 0 (2.) where: y(x) = unknown function y 0 (x) = rst derivative x = independent variable The 2. it s a rst order di erential equation (in its implicit form). In the explicit form we can nd at the rst side the rst derivative (y 0 ): = f(x; y(x)) (2.2) dx here f is the know function. Some examples of rst order di erential equation can be: = f(x); = y; = I(t) dx dx dx A di erential equation is said to be linear if the unknown function is a rst order function: dx + x2 y(x) = e x in this case, although x 2 isn t linear, this di erential equation is however linear: it s the degree of the unknow function y(x) which determines the linearity. A nonlinear ordinary di erential equation is simply one that isn t linear. Nonlinear functions of the dependent variable or its derivatives, such as sin y, cannot appear in a linear equation. Therefore: ( y)y 0 + 2y = e x (coe cient depends on y) d 2 y dx 2 + sin y = 0 (nonlinear function of y) d 4 y dx 4 + y 2 = 0 (power not ) are examples of nonlinear rst-, second-, and fourth-order ordinary di erential equations, respectively. We can observe the di erences between numerical and di erential equations: in both numerical (such as 3x + = 0) we have to nd an unknown quantity, x, which is a number; instead in the di erential equations we are looking for an unknown function, not a number.

9 4. THE SEPARATION OF VARIABLES METHOD ix 3. Solution of rst order ODE Solve a di erential equation means to nd the function g(x) (called solution or integral) that makes the expression identically satis ed. A function g(x) is a solution when it is substituted into 2.2 it reduces to an identity in a certain open interval (a; b) in R. Generally we can nd the solution by integration. What is the di erence between identity and equation? in the identity we can replace any number (2x + 3x = 5x) and will always remain true, regardless of the values assigned to variables; in the equation we need to nd an unknown factor, and there is a unique solution (3x + = 0). The equation becomes an identity only when we insert that particular solution (=3). 4. The separation of variables method Let y 0 = f(x)g(y) be an ordinary di erential equation as a product of two functions: the function f depends on the variable x and the function g depends on the variable y. The method separates the two variables y and x placing them in di erent sides of the equation; each sides is then integrated: Z y 0 = f(x)g(y) dx = f(x)g(y) = f(x)dx g(y) Z = f(x)dx + C g(y) Definition 2. A rst-order di erential equation of the form dx = f(x)g(y) is said to be separable or to have separable variables. For example, let s consider one of the most simple di erential equations: y 0 = y The function that is equal to its derivative is the exponential. Generally we can assume that if the di erential equation is lineare, its solution is an exponential. We can use the separation of variables to nd the solution: we can write the explicit form like a product at the second side (y )

10 x. ORDINARY DIFFERENTIAL EQUATIONS y 0 = y () y depends on x, we have a derivative at rst side: dx = y (2) we separate the di erentials: y = dx (3) and proceed with integration by parts. Integration is the inverse operation of the derivative. If we are looking for the unknown function, we have to integrate: Z Z y = dx At the rst side we have: Z y 0 Z y = dx When we have a function in the denominator and its derivative at numerator, we nd a growth rate. This is a derivative of a logarithm. The primitive is the function ln(y). ln(y) = x + C Ça va sans dire: when we integrate, we must not forget to add a constant of integration C. We aren t interested to ln(y), but we re looking for y, then we use the exponential (the inverse function of logarithm) to both sides: and we get: e ln(y) = e (x+c) y = e (x+c) for the properties of powers: y = e x e C Generally we can write e C in this elegant form: y = Ce x

11 4. THE SEPARATION OF VARIABLES METHOD xi This is the general solution of the rst order di erential equation y 0 = y.

12 xii. ORDINARY DIFFERENTIAL EQUATIONS 5. Examples of non-linear functions y0 = xy 2 y0 = p y In this case the link between y and its rst derivative y0 isn t linear and certainly we don t have an exponential solution. Example : y0 = xy 2 We apply the separation of variables method: Z dx = xy2 = x dx y Z 2 Z = x dx y 2 Z y 2 = x dx y = x+ + y = x2 2 y least common multiple at second side: = x2 2 + C This is the solution. y = y(x) = x 2 + 2C 2 2 x 2 + 2C Only when the link is linear we can nd exponential type solutions.

13 Example 2: 5. EXAMPLES OF NON-LINEAR FUNCTIONS xiii y0 = p y = y 2 Z dx = y 2 = dx y 2 Z Z = dx y 2 Z y 2 = dx y = x + C 2 2y 2 = x + C y 2 = x + C 2 x + C y(x) = 2 2

14 xiv. ORDINARY DIFFERENTIAL EQUATIONS 6. Particular solution We have a particular solution when assign a particular value to the solution. Let s de ne such as particular solution or integral of the di erential equation F (x; y; y0) = 0 each function y = F (x) obtained by assigning particular values to the arbitrary constant y = (x). For example: y 0 x 2 = 0 given the initial conditions P (4; 3 ): Z y 0 = x 2 dx = x2 = x 2 dx Z = x 2 dx y = x3 3 + C by sostituting of the initial conditions: we get C: 3 = C So the particular solution is: C = 2 C = axis. y = x it s exactly the point where the function intersects the x 7. Singular integral It s the solution that we can t obtain by assigning a value to the costant C. For example:

15 7. SINGULAR INTEGRAL xv y0 = 2 p y dx = 2p y 2 p y = dx 2 (y) 2 = dx Z Z (y) 2 = dx 2 y = x + C y 2 = x + C y = (x + C) 2 observe that y = 0 is a solution but this solution cannot be obtained by assigning a value to C from the general solution.

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17 CHAPTER 2 Linear di erential equations. ODE with non-constant coe cients Let s consider this di erential equation: y 0 + a (x) y = a 0 (x) (.) this is a linear, homogeneous, rst order di erential equation with a non-constant coe cient a (x) function of x. The general solution is: Z y = e R a (x)dx e R a (x)dx a 0 (x)dx + C To obtain this solution we multiply each sides for e R a (x)dx : e R a (x)dx dx + er a (x)dx a (x) y = e R a (x)dx a 0 (x) the rst term is the derivative of a product of functions: D e R a (x)dx y(x) = e R a (x)dx a 0 (x) We integrate each side and obtain: Z D e R Z a (x)dx y(x) dx = e R a (x)dx a 0 (x)dx Z e R a (x)dx y(x) = e R a (x)dx a 0 (x)dx + C Z y(x) = e R e R a (x)dx a a (x)dx 0 (x)dx + C Z y(x) = e R a (x)dx e R a (x)dx a 0 (x) + C This is the solution of.. Let s consider the next di erential equation with initial conditions assigned: dx = x y(x) con y(0) = xvii

18 xviii 2. LINEAR DIFFERENTIAL EQUATIONS In this case we have another linear, homogeneous, rst order di erential equation with a non-constant coe cient (x): Z y(x) = Z x dx ln y = x2 2 + C y(x) = Ce x2 2 When a particular value is substituted to constant C, the solution becomes a particular solution. In this case when y(0) =, C = : The solution is: y(x) = e x2 2. Ce 02 2 = C =.. The homogeneous case. Let s consider this linear, homogeneous, rst order di erential equation: dx + a (x)y(x) = 0 (.2) To nd the solution, we use the separation of variables method: dx = a (x)y(x) Z Z = a (x)dx y(x) Z ln y(x) = a (x)dx y(x) = Ce R a (x)dx We can write the general solution of.2 such as: y(t) = Ce R a (x)dx.2. The non homogeneous case. Let s consider this linear, non-homogeneous, rst order di erential equation: dx + a (x)y(x) = a 0 (x) (.3) To nd the solution, we always use the separation of variables method.

19 2. SECOND ORDER DIFFERENTIAL EQUATIONS xix 2. Second order di erential equations 2.. Homogeneous case with constant coe cients. d 2 x dt + a dx 2 2 dt + a x(t) = 0 (2.) To nd the solution, we assume that takes an exponential form : x(t) = Ce t First derivative: Second derivative: dx dt = Cet d 2 x dt = 2 C2 e t If we substitute the solution inside the di erential equation, obtain the characteristic equation (or characteristic polynomial): C 2 e t + a 2 Ce t + a Ce t = 0 Ce t ( 2 + a 2 + a ) = 0 The solution depends on the two eigenvalues and 2 : if they are di erent and real, we have two solutions. The general solution is a linear combination of this two solutions: x(t) = C e t + C 2 e 2t Speci cally, we have three cases: () if and 2 are di erent, the general solution is x(t) = C e t + C 2 e 2t (2) if and 2 are equal ( = 2 = ), the general solution is x(t) = C e t + C 2 te 2t (unlike before, the second term is multiplied by t) (3) if and 2 are complex conjugate: (a) = + i (b) 2 = i...so the general solution is: x(t) = C e t cos t + C 2 e t sin t 2 We are considering a linear di erential equation. 2 The sine and cosine functions allow us to transform the imaginary part of the solution in a real part using Euler s formula.

20 xx 2. LINEAR DIFFERENTIAL EQUATIONS EXAMPLE: d 2 x dt + 2dx 3x(t) = 0 2 dt Assume that the solution takes this form: x(t) = Ce t First derivative: Second derivative: dx dt = Cet d 2 x dt = 2 C2 e t substitute the solution inside the di erential equation: factoring Ce t : C 2 e t + 2Ce t 3Ce t = 0 Ce t = 0 We nd the solution by solving the characteristic equation: = 0 ;2 = b p b 2 4ac 2a The general solution is: = 2 p = = 2 = 3 x(t) = C e t + C 2 e 3t

21 2. SECOND ORDER DIFFERENTIAL EQUATIONS xxi 2.2. Non-homogeneous case with constant coe cients. d 2 x dt + a dx 2 2 dt + a x(t) = a 0 (t) (2.2) It s a non-homogeneous case because the second term is nonzero. Let s consider a 2 e a as costant coe cients: are real numbers. To nd the solution a good technique is to nd a linear combination of a 0 (t) with its rst and second derivatives. For example: if a 0 (t) = 6t 3 3t then we can nd values of a, b, c and d such that at 3 + bt 2 + ct + d is the solution. if a 0 (t) = 2 sin t + cos t, we look for values of A and B such that A sin t + B cos t is the solution. if a 0 (t) = 2e Bt, we look for values of A such that Ae Bt is the solution. EXAMPLE: d 2 x 5 dx + 4x(t) = dt 2 t2 2t dt First we nd the solution for the homogeneous part. The characteristic equation is: = 0 ;2 = b p b 2 4ac = = 2a 2 = 4 The general solution of homogeneous is: x(t) = C e t + C 2 e 4t Now we can nd the solution for the non-homogeneous part: the function on the right-hand side is a second-degree polynomial, so to nd the solution we have to try a general polynomial that is a function of the form: First derivative: Second derivative: x(t) = at 2 + bt + c x 0 (t) = 2at + b x 00 (t) = 2a

22 xxii 2. LINEAR DIFFERENTIAL EQUATIONS We substitute inside the di erential equation: 2a 5(2at + b) + 4(at 2 + bt + c) = t 2 2t 2a 0at 5b + 4at 2 + 4bt + 4c = t 2 2t 4at 2 t 2 + 4bt + 2t 0at 5b + 4c + 2a = t 2 2t (4a )t 2 + (4b 0a + 2) t + 2a 5b + 4c = 0 System: 8 < : 8 < : 4a = 4b 0a = 2 2a 5b + 4c = 0 4a = 0 4b 0a + 2 = 0 2a 5b + 4c = 0, Solution is: a = ; b = ; c = The particular solution of the non-homogeneous is: x(t) = 4 t2 + 8 t + 32 The general solution is a linear combination of two solutions: x(t) = C e t + C 2 e 4t + 4 t2 + 8 t + 32

23 3. HOW TO TRASFORM A SECOND ORDER ODE xxiii 3. How to trasform a second order ODE An n order di erential equation can be written as a system of n rst order equations. Let s consider this second order di erential equation: We can write as: dx dt dx 2 d 2 x dt + a dx 2 2 dt + a x(t) = a 0 (t) = x(t) dt = a (t)x (t) a 2 (t)x 2 (t) + a 0 (t) EXAMPLE: The next non linear, second order di erential equation associated to an economic tourism model 3 : d 2 P dp + ^ dt 2 dt + ^ sin(p ) = a + ^ cos(t) can be written as a system of two rst order di erential equations: dp = y dt = ^y ^ dt sin(p ) + a + ^ cos(t) 3 Dynamical Analysis of a tourism-based economy, Venturi, 2000

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25 CHAPTER 3 Systems of di erential equations. First order system.. General form. Let s consider this two rst order di erential equations in two variables: dx dt dt = ax(t) + by(t) = cx(t) + (t) the unknown functions are x(t) and y(t). We can solve this system by isolating y(t) in the rst equation: First derivative: y(t) = x0 (t) b ax(t) b y 0 (t) = x00 (t) ax 0 (t) b b This step is very useful because I can replace y(t) and y 0 (t) in the second equation by transforming the system into a second order di erential equation: y 0 (t) = cx(t) + (t) x 00 (t) ax 0 (t) x 0 (t) = cx(t) + d b b b x 00 (t) (a + d)x 0 (t) + (ad bc)x(t) = 0 ax(t) b EXAMPLE: x 0 (t) = 2x(t) + y(t) y 0 (t) = 4x(t) 3y(t) We isolate y(t) in the rst equation: First derivative: y(t) = x 0 (t) xxv 2x(t)

26 xxvi 3. SYSTEMS OF DIFFERENTIAL EQUATIONS y 0 (t) = x 00 (t) We replace it in the second equation: 2x 0 (t) x 00 (t) 2x 0 (t) = 4x(t) 3 (x 0 (t) 2x(t)) x 00 (t) 2x 0 (t) = 4x(t) 3x 0 (t) + 6x(t) x 00 (t) + x 0 (t) 2x(t) = 0 Characteristic equation: = 0 The general solution is: ;2 = b p b 2 4ac 2a = = 2 = 2 x(t) = C e t + C 2 e 2t 2. Stable and namic system We consider a di erential equation in vectorial form: where: dx dt = f(x) (2.) x = (x ; :::; x n ) f = (f ; :::; f n ) with f a function independent of time t. The 2. represents a namical system: a (constant) is an equilibrium point of the system if f(a) = 0. The equilibrium point of the system is solution of the equation f(x) = Market model with time expectation A typical economical example is the following: Qd = d2 p 2 dp 2p(t) + 40 dt 2 dt Q s = 3p(t) 5

27 3. MARKET MODEL WITH TIME EXPECTATION xxvii In equilibrium Q d = Q s d 2 p dt 2 2 dp 2p(t) + 40 = 3p(t) 5 dt d 2 p 2 dp 5p(t) = 45 dt 2 dt d 2 p dt + 2dp + 5p(t) = 45 2 dt On the left-hand side we have the homogeneous part, on the righthand side the non-homogeneous. First, we nd the solution of the homogeneous part. The characteristic equation is: = 0 The solution is: ;2 = b p b 2 4ac 2a = 2i = 2 = + 2i p(t) = C e t cos 2t + C 2 e t sin 2t Solution of the non-homogeneous part: d 2 p dt + 2dp + 5p(t) 2 dt = 45 5C = 45 C = 45 5 C = 9 The intertemporal equilibrium is: p(t) = C e t cos 2t + C 2 e t sin 2t + 9 If we assign the following initial conditions: p(0) = 2 p 0 (0) = C e 0 cos 0 + C 2 e 0 sin 0 = 3 C = 3

28 xxviii 3. SYSTEMS OF DIFFERENTIAL EQUATIONS e 0 (C cos 0 2C 2 cos 0 + 2C sin 0 + C 2 sin 0) = The solution is: When t goes to in nity: lim t! 3e t cos 2t = 0 lim t! 2e t sin 2t = 0 p(t) = 3e t cos 2t + 2e t sin 2t + 9 2C 2 C = 2C 2 3 = C 2 = 2...they goes to zero, then the solution goes to 9. This is a stable equilibrium point. 4. Stability of the equilibrium point: knot and focus In general, a linear system has a unique equilibrium point at the origin. An example of a linear system is the following: dx = 2x dt (t) + x 2 (t) _x (t) = 2x dx 2 or, alternatively: (t) + x 2 (t) = x dt (t) + 2x 2 (t) _x 2 (t) = x (t) + 2x 2 (t) we want to know if the equilibrium point is stable or not. From the rst equation I obtain x 2 (t) and its derivative: x 2 (t) = _x (t) 2x (t) _x 2 (t) = x (t) 2 _x (t) We replace it on the second equation: _x 2 (t) = x (t) + 2x 2 (t) x (t) 2 _x (t) = x (t) + 2 [ _x (t) 2x (t)] x (t) 2 _x (t) = x (t) + 2 _x (t) 4x (t) x (t) 4 _x (t) + 3x (t) = 0 Characteristic equation: = 0 = ;2 = 2 = 3

29 4. STABILITY OF THE EQUILIBRIUM POINT: KNOT AND FOCUS xxix This is the solution of the rst equation: x (t) = C e t + C 2 e 3t If we substitute the rst derivative ( _x (t) = C e t + 3C 2 e 3t ) inside x 2 (t) = _x (t) 2x (t) we obtain the solution of the second equation: x 2 (t) = C e t + 3C 2 e 3t 2 C e t + C 2 e 3t x 2 (t) = C e t + C 2 e 3t This is the solution of the system : x (t) = C e t + 3C 2 e 3t x 2 (t) = C 3 e t + C 4 e 3t In the phase diagram the solution is an unstable knot. For convenience, we can use c 3 and c 4 to distinguish them.

30 xxx 3. SYSTEMS OF DIFFERENTIAL EQUATIONS KNOT AND FOCUS

31 5. MATRIX METHOD xxxi 5. Matrix method Another resolutive method for these systems is the matrix method. Let s consider the following system: _x (t) = 2x (t) + x 2 (t) _x 2 (t) = x (t) 2x 2 (t) This is the matrix of the coe cient: 2 A = 2 Calculating the trace and the determinant of the matrix: T ra = 2 2 = 4 DetA = ( 2 2) ( ) = 4 = 3 We use the invariants of the matrix (with alternating signs) to write the characteristic equation:: = 0 = ;2 = 2 = 3 We have a stable knot.

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33 Part 2 Stochastic Di erential Equations

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35 CHAPTER 4 Probability Space. De nition A Probability Space is given by a tripla (; =; P ): = a space set of possible outcomes events of a random experiment; = = is a set of sets ( from a mathematical point of view is a -algebra); P = is a probability measure in =; P :! [0; ]! R =, -algebra, is a family set such that: () O 2 =; (2) If A 2 = then the complementary A c of A is in =; (3) The unions of a countable sets of A belongs to =. Example. We consider the roll of the dice = f; 2; 3; 4; 5; 6g Let = be the -algebra genereted by the sets for which we can evaluate a probability measure. E = even number = {2, 4, 6}, F = is a number bigger than 4 = {5, 6}, G = is the number 7. G belongs to =? Example 2. Example of a mesure of Probability. We de ne: P(E) = # of outcomes values of E/# all possible outcomes of P(E) =#{2, 4, 6}/#{, 2, 3, 4, 5, 6}=3/6=/2 P(F) = #{5, 6}/#{, 2, 3, 4, 5, 6} =2/6= /3 xxxv

36 xxxvi 4. PROBABILITY SPACE 2. Random Variable Let the probability space (; =; P ): we de ne a random variable X a mesure function from X :! R For all A 2 B(R), exist B 2 =: X (A) = B X is a variable which takes the values: a ; a 2 ; a 3 :::a n respectively with probability p ; p 2 ; p 3 :::p n P (X 2 A) = P! 2 :! 2 X (A) = P (B); A in R; B 2 = Let X be the random variable de ne as follows: X(!) = ;for! = or 2; X(!) = 0;for! = 3; 4; 5 and +X(!) = 6: The value of P (X 0) is given by: P (X 0) = P (X 2 f0; +g) = P (f! 2 : X(!) = 0gU[(f! 2 : X(!) = +g) = P (f! 2 : X (f0; +g)g) = P (f3; 4; 5; 6g) = 4=6 3. Stochastic process Stochastic process or sometimes random process (widely used) is a collection (a family) of random variables; this is often used to represent the evolution of some random value, or system, over time. This is the probabilistic counterpart to a deterministic process (or deterministic system). Instead of describing a process which can only evolve in one way (as in the case, for example, of solutions of an ordinary di erential equation), in a stochastic or random process there is some indeterminacy: even if the initial condition (or starting point) is known, there are several (often in nitely many) directions in which the process may evolve. 3.. Brownian Motion. It is called Brownian motion (or Wiener process) a process (B(t); t 0) which starts from zero, i.e. B(0) = 0, with the following properties: () a process with independent increments; in other words the increment B(t) B(s) is independent of B(u) B(v) for t > s u > v ;

37 3. STOCHASTIC PROCESS xxxvii (2) is a process with stationary increments, i.e., the distribution B(t) B(s) t > s 0 depends only on the distance t s and not by t and / or s separately. (3) is a Gaussian process increments: i.e. B(t) B(s) N(0; t s) 3.2. Wiener Process Normal distribution. W t W s N(0; t s) 0 s t f(x; ; ) = p 2 e (x ) = is the mean or expectation of the distribution (and also its median and mode) = is its standard deviation; its variance is therefore 2 n A random variable with a Gaussian distribution is said to be normally distributed and is called a normal deviate. If = 0 and =, the distribution is called the standard normal distribution or the unit normal distribution, and a random variable with that distribution is a standard normal deviate W t+dt W t = W t = N p dt Variance. In probability theory and statistics, variance measures how far a set of numbers is spread out. A variance of zero indicates that all the values are identical. Variance is always non-negative. A small variance indicates that the data points tend to be very close to the mean (expected value) and hence to each other, while a high variance indicates that the data points are very spread out from the mean and from each other. The square root of variance is called the standard deviation. Stochastic Equations Capital Markets Let (S(t); t 0) be a process which represents the value of an option: ds=s = dt + dx The market price deterministic contribution We consider a process (S(t), t 0) which represents the value of a prize at the time t. S + ds is the variation at the time dt ds = S(t + dt) S(t)) ds=s = dt trend o drift ( deriva )

38 xxxviii 4. PROBABILITY SPACE Deterministic part Stochastic contribution in a di erential equation: The market price dx= stochastic contribution of a process X is given by shock exogenous factors, unknown news. We consider gaussian shock: > 0, is called volatility dx N(0; c 2 ) Where c is the variance of the process dx. ds=s = dt + dx Where dx is a Gaussian variable with average zero and given variance. We assume what the variance depends on the interval dt dx N(0; dt) dx = Z p dt with Z N(0; ) EdX = 0 and V ar(dx) = dt, we have EdS = E(SdX + Sdt) = Sdt and V ar(ds) = EdS 2 (EdS) 2 = 2 S 2 dt We can known the behavior of S by estimation of and by the market dates. 4. Examples of Stochastic Equations Lemma di Ito X = f(t; Y ) dx = df(t; Y ) = f t (t; Y )dt + f Y (t; Y )dy + 2 f Y Y (t; Y )dy dy with: dtdt = 0 dw t dt = 0 dw t dw t = dt The Black-Scholes Di usion Equation dx=x = dt + dw t

39 4. EXAMPLES OF STOCHASTIC EQUATIONS xxxix The solution is given by: X(0) = X 0 X(t) = X 0 e 2 2 t+wt where: X = f(t; Y ) = X 0 e Y Y = t + W 2 2 t dy dy = 2 2 = 2 dt 2 = dw t = 2 dw t dx = df(t; Y ) = X 0 e Y dy + 2 ey dy dy dy = 2 dt + dw 2 t dtdy + dw tdy 2 dt + dw 2 t + dw t 2 dt + dw 2 t dx = df(t; Y ) = X 0 e Y 2 dt + dw 2 t = X 0 e Y dy + X 0 e Y dw t = XdY + XdW t + 2 ey 2 dw t

40 xl 4. PROBABILITY SPACE 5. The Random Walk Process Before analysing the stochastic calculation in its matematical formulation, we want to describe empirically the behaviour of a generic nancial security, which can be described, following the EMH hypothesis, as a random walk and then as a browian motion. The Random Walk is the easiest stochastic process which assumes that the variable under analysis changes over time in a total random and independent way into the surrounding space. In this section we will discuss the stu of that stochastic processes which comply to the EMH and on which the EMH itself is based. In this case our random walk is by de nition symmetric, because we have 50% of probability p that our variable (the price of the security) increases, and 50% (q = p) of probability which decreases. We can think to a series of evolution of the security price over time, which it can be indicated as follow: M K = kx R j k = ; 2; ::: con M 0 = 0 () j= This process M K = 0; ; 2::: is a symmetric random walk because for each increment the price can go up (+) or down ( ) by one unit and each option is equally likely. A very important property of the symmetric random walk is that its increments are indipendent: indeed, following the EMH, the returns of a generic security have the same property. Here we are referring to the increments (returns) of the process expressed in continuous by the formula: p ln (p p ) or ln p This is why we are always referring to a stochastic process in the continuous and not in the discreet case. Following the notation (), if we choose nonnegative integer numbers 0 = K 0 < K < :::K m the random variables R K = (R K R K2 ) ; (R K2 R K ) ; :::; R Km R Km are independent. Each of these increments (think to returns on shares) has expected value equal to zero and variance t i t i. It has expected value equal to zero for two main reasons:

41 5. THE RANDOM WALK PROCESS xli one intuitive, given the fact that we are working in the continuous, the increment dm! 0 is in nitesimal; the other one depends on the Martingale property, typical of the symmetric Random Walk: p + = E [p + ji ] = p p + + p = 0 The variance V ar (P j ) = EPj 2 = is equal to one because we have said that at each step (at each increment over time) the variable price (P ) can increase or decrease, so that each step is equal to one positive (+) or negative ( ) event. For this reason we can say that the variance of a symmetric Random Walk is a consequence of a standard Brownian motion which accumulate at rate per unit of time, so that the variance of the increment in any moment in time (from k to ` and for k < `) will be always equal to ` k. The concept of volatility in this case becomes variance or squared variance. The variance of the process is expressed as: Xt+ Xt+ V ar (R t R t ) = V ar (R j ) = = t i+ t i = (2) j=t And the squared variance: [R; R] k = j=r kx (R j R j ) 2 = t = (3) j= The squared variance is calculated step by step, then you to the power of two each step and nally you calculate the cumulative sum. Given that these increments (R j R j ) are equal to X j which can only be + or, the equation (3) can be only + (because of the power). If we substitute i = 0 in (2), this is equal to (3), but with di erent calculations: the variance is calculated making the average of all the increments and taking into account their probabilities; the squared variance is calculated along an individual increment of the process, without taking into account probabilities (+ or of the step).

42 xlii 4. PROBABILITY SPACE It is possible to calculate the variance of a Random Walk only theoretically because it requires to calculate the average for each step, realized or not. To have a clear idea of the di erence, you can think to the GARCH(p; q) process, composed by a term which is the conditioned variance (in our case is the variance as weighted average of the probabilities of all the increments) and the other term is the squared error which changes step by step, is independent and correspond to the concept of squared variance. In order to approximate a Brownian motion (which is a continuous process) we need to increase the temporal frequency and decrease the increment in scale, that is the step of a symmetric Random Walk. This means that t! but each temporal increment dt! 0, and this implies an higher temporal frequency. The increment of a variable (for instance, the price), which is its return, will be computed itself in the interval dt! 0. Also for a scaled symmetric Random Walk (that is the approximation of a Brownian motion) is is always valid the Martingale property for which: E W (n) (t + )jf (t) = W (n) (t) From a strictly statistical point of view, the Brownian motion cannot be the more appropriate stochastic process to being used in the EMH eld, which are based on the standard normal distribution as returns distribution. Also the Brownian motion for n! will have a distribution, evaluated at time t which converges to the standard normal distribution with average zero and variance equals to t. In order to demonstrate this central limit theorem, it is good to know which intuitively to generate random variables and create a process from a speci c distribution is to identify the moments generating function or characteristic function in the complex. The density function of the standard normal distribution is: f(x) = p e x2 2t 2t with average zero and variance equals to t. The moments generating function is:

43 5. THE RANDOM WALK PROCESS xliii g(u) = Given that: = Z + e ux f(x)dx Z + p exp ux 2t Z = e 2 u2 t + p exp 2t = e 2 u2 t x 2 dx 2t! (x ut) 2 dx 2t p 2t e (x ut)2 2t is the density function of a normal distribution with average ut and variance t, it integrates to. If t is such that nt is an integer, then the moments generating function for W (n) (t) is: g n (u) = Ee uw (n) (t) u = E exp pn M u t = E exp j=! u Xnt p X j n j= Ynt u = E exp pn X j Given that the random variables are independent, the equation above can be written as: Ynt j= u E exp pn X j = = Ynt 2 e p u n + 2 e p u n j= nt 2 e p u n + 2 e p u n (4) It is possible to prove that equation (4) converges to the moments generating function for n! : g(u) = e 2 u2 t

44 xliv 4. PROBABILITY SPACE If we consider the logaritm of g n (t): log g n (t) = nt log 2 e p u n + 2 e p u n it converges to: log g n (t) = 2 u2 t to simplify we can change the variable X = = p n so that: lim log g log 2 n(u) = t lim eux + e ux 2 n! X!0 X 2 If we set X = 0 in the expression on the right hand side, we obtain the indeterminate form 0=0 and we can use L Hôpital s rule to calculate rst the derivative of the numerator with respect to X, and then the derivative of the denominator with respect log 2 eux + 2 e ux = X2 = 2X u 2 eux u 2 e ux 2 eux + 2 e ux (numerator) lim log g 2 n(u) = t lim eux ue ux 2 n! X!0 2X 2 eux + e ux 2 = t u 2 lim 2 eux ue ux 2 X!0 X u (denominator) Given that lim X!0 2 eux + 2 e ux =. If we substitute X = 0, we obtain 0=0 and we apply the L Hôpital s rule once again. After all the calculations we obtain: lim log g n(u) = t u 2 n! 2 lim X!0 2 eux + u2 2 e ux = 2 u2 t If you take a closer look to these formulas, it is easy to see that the moments generating function is the exponential transformation of the generic variable u. It is strictly related with the cumulative distribution function because the integral expressed along all the R axis tells you

45 5. THE RANDOM WALK PROCESS xlv that the sum of probabilities which an event may occur with probability p(x) sum up to eventually. The Random Walk process/brownian motion follows an exponential law from the initial point t 0 = u 0. The variable X i exponentially grows over time. The value =2 makes the process independent itself (or rather: the collection of variables which compose it are independent). The idea is always the same: 50% of probability that the variable goes up, and 50% of probability that the variable goes down. Finally, this is implied in the result (=2) u 2 t where the last u 2 is raised to the power of two because re ects the fact that we have the usual two probabilities in the realisation of the next step: upward or downward, which can be considered as 2. From the central limit theorem seen before, it is possible to conclude that the limit of a binomial asset pricing model (think of its use in the computation of the options price in the discreet) conducts to an asset pricing in the continuous which follows a log-normal distribution. We make a model for the price of a share from time 0 to time t chosing an integer n and building a binomial model which give us n steps per unit of time. n t must give us an integer. We have: u n = + p n (up-factor) And: d n = p n (down-factor) where is theoretically always constant and positive and represents the volatility. The risk neutral probabilities are: ~p = + r d n u n d n = ~q = u n r u n d n = p n 2 p n = 2 p n 2 p n = 2 Once again it is ensured the independence of the process, the Martingale s idea and the increments of the process tipical of the White Noise. The share price at time t is given by the initial price S(0) plus the sum of the rst n t process realisations. The sum of increases H nt and decreases L nt is equal to n t:

46 xlvi 4. PROBABILITY SPACE nt = H nt + L nt (5) The Random Walk W nt is the number of increases minus the number of decreases in these n t realisations: W nt = H nt L nt (6) If we sum (5) and (6) and divide by 2: H nt = 2 (nt + W nt) If we subtract (5) and (6) and divide by 2: Then: L nt = 2 (nt + W nt) S n (t) = S(0)u Hnt n d Lnt n = S(0) + pn 2 (nt+wnt) p n (nt Wnt) 2 of: If n!, the distribution of S n (t) converges to the distribution S(t) = S(0) exp W (t) 2 2 t where W (t) is a N (0; t). The distribution of (7) is log-normal because, generally speaking, any variable of the form c e x, where c is a constant and x is normally distributed, it has a log-normal distribution. In (7) the variable X = W (t) 2 2 t has average 2 2 t and variance 2 t. This means that the variable has two equations: one represents the trend, its evolution and it is deterministic, the other one, which is stochastic, represents the (namic) noise around the trend. As mentioned before, the Brownian motion is the limit of a scaled Random Walk W (n) (t) for n!. Simplifying, the Brownian motion is the limit for n! of a symmetric Random Walk with ~p = ~q, which independent increments are a process known as White Noise. This is a proper concept of EHM, and we will see how it can be generalised. The only di erences between Random Walk and Brownian motion is that the former is a discreet and almost normal process, the latter is continuous and exactly normal. (7)

47 5. THE RANDOM WALK PROCESS xlvii Obviously, if you want to simulate a Brownian motion, we need to use a scaled Random Walk. The Brownian motion has increments which have average zero and variance equals to t t 0. The increments are independent and identically distributed as a Normal N (0; ). The random variables are W (t ); W (t 2 ); :::; W (t n ) and are jointly normally distributed. The moments generating function of the variables random vector can be calculated as in the case of the Random Walk.