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Introduction to Financial Mathematics Lecture Notes - MAP 5601 Department of Mathematics Florida State University Fall 2003 Table of Contents 1. Finite Probability Space 12. Elements of Continuous Probability Theor 123. Differential Equation 21 Lecture Notes - MAP 5601 map 5601 LecNotes. tex i 8/27/20031.

Finite Probability Spaces The toss of a coin or the roll of a die results in a finite number of possible outcomes. We represent these outcomes by a set of outcomes called a sample space. For a coin we might denote this sample space by {H, T} and for the die {1, 2, 3, 4, 5, 6}. More generally any convenient symbols may be used to represent outcomes.

Along with the sample space we also specify a probability function, or measure, of the likelihood of each outcome. If the coin is a fair coin, then heads and tails are equally likely. If we denote the probability measure by P, then we write P (H) = P (T) = 12. Defninition 1.1. A finite probability space is a pair (, P) where is the sample space set and P is a probability measure: If = {!

1, ! 2, ... , ! n}, then (i) 0 P (! i) 1 for all i = 1, ... , n (ii) n Pi = 1 P (! i) = 1. In general, given a set of A, we denote the power set of A by P (A). By definition this is the set of all subsets of A. For example, if A = {1, 2}, then P (A) = {; , {1}, {2}, {1, 2}}.

Here, as always, ; is the empty set. By additivity, a probability measure on extends toP if we set P (; ) = 0. Example. The division of {1, 2, ... , 6} into even and odd, {2, 4, 6} and {1, 3, 5}, is an example of partition. Defninition 1.2.

A partition of a set (of arbitrary cardinality) is a collection of nonempty disjoint subsets of whose union is. If the outcome of a die toss is even, then it is an element of {2, 4, 6}. In this way partitions may provide information about outcomes. Defninition 1.3. Let A be a partition of.

A partition B of is a refinement of A if every member of B is a subset of a member of A. For example B = {{1, 2}, {3}, {4, 5, 6}} is a refinement of {{1, 2, 3}, {4, 5, 6}}. Notice Lecture Notes - MAP 5601 map 5601 LecNotes. tex 1 8/27/2003 that a refinement contains at least as much information as the original partition. In the language of probability theory, a function on the sample space is called a random variable. This is because the value of such a function depends on the random occurrence of a point of. However, without this interpretation, a random variable is just function. Given a finite probability space (, P) and the real-valued random variable X: !

I Rwe define the expected value of X or expectation of X to be the weighted probability of its values. Definition 1.4. The expectation, E (X), of the random variable X: ! IR is by definition (X) = n Xi = 1 X (! i) P (! i) where = {! 1, ! 2, ... , ! n}.

We see in this definition an immediate utility of the property Pi = 1 P (! i) = 1. If X is identically constant, say X = C, then E (X) = C. When a partition of is given, giving more in general than just, we define a conditional expectation. Definition 1.5. Given a finite probability space (, P) and a partition of, A, we define the conditional expectation of the random variable X: ! IR with respect to the partition A at the point! 2 byE (X|A) (!) = P 2 A (!) X P P (A (!) ).

Here A (!) is the member of A containing! and P (A (!) ) = P 2 A (!) P. Notice thatE (X|{}) = E (X). This holds more generally. When iterating conditional expectations it is the smaller or sparser partition that determines the expected outcomes. A partition A is smaller than Bif B is a refinement of A. Proposition 1.6. If B is a refinement of A, thenE (X|A) = E (E (X| B) |A) = E (E (X|A) | B) Lecture Notes - MAP 5601 map 5601 LecNotes. tex 2 8/27/2003 for a random variable X: ! IR.

Exercise 1. Prove Proposition 1.6. Definition 1.7. A random variable X: !

IR is measurable with respect to a partition A ifX is constant on each set in A. So, for example, a function constant on is measurable with respect to any partition of. On the other hand if X assumes n distinct values on = {! 1, ! 2, ... , ! n} then X is measurable only with respect to the discrete partition {{! 1}, {! 2}, ... , {! n}}.

Proposition 1.8. The conditional expectation, E (X|A), is measurable with respect toA. Proof. The proof follows immediately from the definition of E (X|A). Proposition 1.9. If X is measurable with respect to the partition A, then E (X|A) = X. Moreover, if Y is another random variable, thenE (XY |A) = XE (Y |A).

Proof. The proof again follows immediately from the definition of E (X|A). We give an interpretation of Proposition 1.9. If X is measurable with respect to A, then the "best guess" of X, given A, is X itself.

Example. Let be the outcomes of a toss of a die and A = {{2, 4, 6}, {1, 3, 5}}. Define = 1 if the outcome is even and X = − 1 otherwise. ThenE (X|A) (!) = X (!) P (A (!) ) P (A (!) ) = 1 when the outcome is even and − 1 otherwise.

Notice here the result is the same independent of the probabilities of the individual outcomes. The following are immediate consequences of the definition of E (X|A). Proposition 1.10. Lecture Notes - MAP 5601 map 5601 LecNotes. tex 3 8/27/2003 Defninition 1.11. Given a random variable X on, we denote by p (X) the partition generated by X. This is the smallest (coarsest) partition of such that X is measurable with respect top (X). Notice it follows that X is measurable with respect to a partition A if and only if A isa refinement of p (X).

Defninition 1.12. A filtration (of partitions of) is a sequence of partitions 1, A 2, A 3, ... , AN whereAt+1 is a refinement of At for t = 1, 2, ... , N − 1. Example. If = {!

1, ! 2, ! 3, ! 4}, then A 1 = {}, A 2 = {{! 1, ! 2}, {!

3.! 4}}. A 3 = {{! 1}, {! 2}, {! 3}, {!

4}} is a filtration of. A sequence of random variables 1, X 2, ... , XN on is called a stochastic process or simply a process. We associate a process and a filtration in the following ways. Defninition 1.13.

A sequence of random variables {Xt}Nt = 1 is adapted to the filtration {At}Nt = 1 if Xt is measurable with respect to At for all t = 1, 2, ... , N. Example. Let = {! 1, ! 2, ! 3, ! 4}.

Let A 1 = {}, A 2 = {{! 1, ! 2}, {! 3, ! 4}} and A 3 = {{! 2}, {!

2}, {! 3}, {! 4}}. Then {At}3 t = 1 is a filtration. Let 1 (! i) = 1, i = 1, 2, 3, 4, X 2 (! i) = 1, i = 1, 22, i = 3, 4, X 3 (! i) = i, i = 1, 2, 3, 4. The process {Xt}3 t = 1 is adapted to {At}.

Defninition 1.14. A process {Xt} is predictable with respect to {At} if Xt is measurable with respect to At− 1 for all t. Lecture Notes - MAP 5601 map 5601 LecNotes. tex 4 8/27/2003 We remark that if {Xt} is adapted to {At}, then {Yt}, where Yt = Xt− 1, is predictable with respect to {At} (when suitably defined at the initial time). Defninition 1.15.

A process {Xt} is a martingale with respect to the filtration {At} ifE (Xt|As) = X sfor 1's t N. Notice this condition involves the probability measure P on. Example. Let X be a random variable on (, P). Define Yt = E (X|At) for t = 1, 2, ... , N. Then, using the law of iterated expectations, Proposition 1.6, we get (Yt|As) = E (E (X|At) |As) = E (X|As) = Ys for 1's t N. As such {Yt} is a martingale with respect to {At}. We remark that if {Xt} is a martingale with respect to {At}, then {Xt} is adapted to{At}.

Indeed, if we set's = t, then we haveE (Xt|At) = Xt, which implies that Xt is measurable with respect to At. Proposition 1.16. A process {Xt} is a martingale with respect to {At} if and only ifE (Xs|As− 1) = Xs− 1 for all's = 2, 3, ... , N. Proof. If {Xt} is a martingale, then the conclusion follows from Definition 1.15. Repeating this argument shows that {Xt} is a martingale. Defninition 1.17.

Given two processes {Xt}Nt = 1 and {Yt}Nt = 1, we define the optional process by [X, Y ] t = t Xs = 1 XsYsfor 1 t N. Here Xs = Xs − Xs− 1 and X 0 = Y 0 = 0. Exercise 2. We remark that the so-called optional quadratic variation process of {Xt}, { [X, X] t}is an increasing process with respect to t. Defninition 1.18. The predictable quadratic co variation process is defined by hX, Y it = t Xs = 1 E (XsYs|As− 1) where we also set A 0 = {}. We remark that since each term is conditioned on the previous partition, {hX, Y it} isa predictable process.

Proposition 1.19. Proof. It then follows from Proposition 1.16 that {Xt Yt− [X, Y ] t} is a martingale. To this purpose notice that [X, Y ] t = XtYt + [X, Y ] t− 1.

Now since martingales are in particular adapted, we haveE (XsYs|At− 1) = XsYsfor all 1's t − 1. It follows from linearity thatE ( [X, Y ] t− 1|At− 1) = [X, Y ] t− 1. Here we have used Proposition 1.9. The result follows. Exercise 3. Show that, under the above assumptions, {Xt Yt− hX, Y it} is a martingale.

Defninition 1.20. Two martingales {Xt}, {Yt} are orthogonal if hX, Y it = 0 all 1 t N. Theorem 1.21. Two processes {Xt} and {Yt} are orthogonal martingales if and only ifX 1 Y 1 = 0 and {Xt Yt} is a martingale. Proof. If we have two orthogonal martingales, then X 1 Y 1 = hX, Y i 1 = 0 and Theorem 1.19 shows that {Xt Yt} is a martingale. Conversely, if {Xt Yt} is a martingale, the nhX, Y it = Xt Yt − (Xt Yt − hX, Y it) Lecture Notes - MAP 5601 map 5601 LecNotes. tex 7 8/27/2003 is also a martingale since the conditions in linear.

However by construction hX, Y it is also predictable. Now any predictable martingale is necessarily constant in t. Here (hX, Y is|As− 1) = hX, Y is− 1, since hX, Y i is a martingale, andE (hX, Y is|As− 1) = hX, Y is since hX, Y i is predictable. Evidently hX, Y iN = hX, Y iN− 1 = ... = hX, Y i 1 = 0 by assumption. This shows that X and Y are orthogonal processes. We remark that in this proof, for the first time, we have denoted the process {Xt} is simply by X. The utility is worth the abbreviation.

We can distinguish the process From a single random variable X by the context. When an expression involving combinations of random variables is so abbreviated the tacit time is the same. For example 2 − [X, X] = {X 2 t − [X, X] t}. Exercise 4.

Two martingales X and Y are orthogonal if and only if X 1 Y 1 = 0 andE (Xt Yt|As) = E (Xt|As) E (Yt|As) for all 1's t. Example. We now consider the outcomes of the toss of a fair coin three times. As such the sample space contains 23 = 8 outcomes. Suppose that the toss of a head wins a dollar while an outcome of a tail looses a dollar. Furthermore, let Xt denote the sum of the winnings atti me t.

The following table lists these quantities. Lecture Notes - MAP 5601 map 5601 LecNotes. tex 8 8/27/20031 2 3 X 1 X 2 X 3! 1 H H H 1 2 3! 2 H H T 1 2 1! 3 H T H 1 0 1!

4 H T T 1 0 − 1! 5 T H H − 1 0 1! 6 T H T − 1 0 − 1! 7 T T H − 1 − 2 − 1! 8 T T T − 1 − 2 − 3 Next we define a filtration 1 = {{!

1, ! 2, ! 3, ! 4}, {!

5, ! 6, ! 7, ! 8}}A 2 = {{! 1, ! 2}, {!

3, ! 4}, {! 5, ! 6}, {! 7, ! 8}}, A 3 = {{!

1}, {! 2}, {! 3}, {! 4}, {! 5}, {!

6}, {! 7}, {! 8}}. Notice that {Xt}3 t = 1 is an adapted process with respect to {At}3 t = 1. Moreover, {Xt} is a martingale with respect to {At}. Exercise 5.

Show that X is a martingale. Exercise 6. Calculate X 2− [X, X] and X 2− hX, Xi and show that they are martingales. Definition 1.22. For any processes X and Y we define the stochastic sum of Y with respect to X (the discrete stochastic integral) as the process (Y. X) t = 8 } 2 F for all 2 IR, (ii) {x|f (x) } 2 F for all 2 IR, ( ) {x|f (x) } 2 F for all 2 IR, (iv) {x|f (x) } 2 F for all 2 IR, Moreover, any of (i) - (iv) implies (v) {x|f (x) = } 2 F for all 2 IR, Definition 2.15. A function f: !

IR, with domain a member of F, is measurable with respect to the -field F if (i) - (iv) holds in Proposition 2.14. Proof of Proposition 2.14. Assume (i). Then {x|f (x) } = T 1 n = 1 An where An = {x|f (x) + 1 n}. This proves (iv). Similarly {x|f (x) } = S 1 n = 1{x|f (x) − 1 n}, so that (ii) implies (i).

Also, the sets in ( ) and (iv) are complements of thos in (ii) and (i). Proposition 2.14 follows. Lecture Notes - MAP 5601 map 5601 LecNotes. tex 15 8/27/2003 INSERT 1 (NOT YET!) Example. Any real-valued function f: !

IR is measurable with respect to the measurable space (, P ). Example. Let = {! 1, ! 2, !

3, ! 4}, P (! i) = 14 for all i = 1, 2, 3, 4 and A = {{! 1, ! 2}, {!

3, ! 4}}. We define two random variables on (, (A), P).! 1!

2! 3! However X 2 is not. Notice {!

|X 2 (!) 0} = {! 1, ! 3}, which is not an element of (A). Proposition 2.1. A random variable X is measurable with respect to the measurable space (, (A) ) where is finite and A is a partition of, if and only if X is measurable with respect to the partition A (Definition 1.7).

Example. Let f (x) = x. Then {x|f (x) } is the open set {x|x }. As such f (x) is measurable with respect to (IR, B (IR), m).

Definition 2.17. A simple function, ', on (, F) is a real-valued function on that assumes a finite number of distinct nonzero values {a 1, a 2, ... , an}. Moreover, Ai = {! |' (!) = ai} isa member of F for all i = 1, 2, ... , n. The representation' (!) = n Xi = 1 aiA (!) is called the standard representation of '. Here A (!) = 1 if!

2 A and 0 otherwise. Notice that a single function on, F) is a measurable function on (, F). Definition 2.18. When ' is a simple function on the measure space (, F, u), we define the integral of ', with respect to u, as ' = Z'du = n Xi = 1 ain (Ai).

Moreover, if E 2 F we define ZE " du = n Xi = 1 ain (Ai E). Lecture Notes - MAP 5601 map 5601 LecNotes. tex 16 8/27/2003 Example. Let = {! 1, ! 2, ! 3, !

4} and P (! i) = 14. Define a random variable X on (, P, P) by X (! 1) = X (! 2) = 1, X (!

3) = X (! 4) = − 1. We calculate, ZXdP = 1. P ({! 1, ! 2}) + (− 1).

P ({! 3, ! 4}) = 1.12 − 1.12 = 0. Proposition 2.19. If X is a random variable on a finite probability space (, P, P), then ZXdP = E (X).

Exercise 10. Prove Proposition 2.19. Definition 2.20. Suppose that f is a nonnegative measurable function on the measure space (, F, u).

We defineZfu = sup'f simple Z 'du. Here the is over all simple functions ' with ' f. The following theorem allows us to approximate the integral of a nonnegative measurable by a sequence of simple functions increasing to f. Theorem 2.21 (Monotone Convergence Theorem). Suppose that {fn} is a sequence of nonnegative measurable functions, limn!

1 fn = f almost everywhere and fn f for all. Then Zfdu = limn! 1 Zfdu. Next, given a nonnegative measurable function f on, we construct simple functions'N such that limN!

1'N = f. Notice that the mensurability of f allows us to accomplish this Lecture Notes - MAP 5601 map 5601 LecNotes. tex 17 8/27/2003 approximation by partitioning the interval [0, N), N = 1, 2, 3, ... , into 2 N that converges to zero as N! 1. We define the simple functions'N by the first endpoint.

That is'N (!) = 2 N− 1 Xk = 1 kN 2 NAN, k. By the Monotone Convergence Theorem limN! 1 Z'Ndu = Zfdu. Exercise 11.

Show that limN! 1'N (!) = f (! ). We extend the integral a general: ! IR by the decomposition f = f+ − f− .

Here+ = max (0, f) and f− = max (0, − f). Exercise 12. Show that if f is measurable, then so is f+ and f− . Next, in preparation for constructing the conditional expectation, we need the following concepts. Definition 2.22. Given a measure space (, F, u), we say that a second measure on (, F), , is absolutely continuous with respect to u if the following condition holds.

Whenever E 2 F, with u (E) = 0, we have (E) = 0. In other words, sets of measure zero for u are always sets of measure zero for as well. When is absolutely continuous with respect to u, we write.