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# INTRO FINANCIAL MATH MAP 5601

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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 Spaces 1 2 Elements of Continuous Probability Theory 12 3 Differential Equations 21 Lecture Notes MAP 5601 map5601LecNotesteX i 8272003 1 Finite Probability Spaces The toss of a coin or the roll of a die results in a nite 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 PH PT likely we may write P1 P2 P3 P Similarly if each face of the die is equally 4 P5 P6 g Defninition 11 A finite probability space is a pair QP where Q is the sample space set and P is a probability measure If 9 w1w2 w then i 0 ltPwi S 1 for alli 1n ii 2PM 1 In general given a set of A we denote the power set of A by 73A By definition this is the set of all subsets of A For example if A 1 2 then 73A 0 12 1 Here as always 0 is the empty set By additivity a probability measure on 9 extends to 739 if we set P 0 Example For the toss of a fair die P1 P2 P3 P4 P5 P6 while P toss is even P toss is odd and P246 P2 P4 P6 3 The division of 12 6 into even and odd 246 and 13 5 is an example of a partition Defninition 12 A partition of a set 9 of arbitrary cardinality is a collection of nonempty disjoint subsets of 9 whose union is Q If the outcome of a die toss is even then it is an element of 246 In this way partitions may provide information about outcomes Defninition 13 Let A be a partition of Q A partition 8 of Q is a refinement of A if every member of B is a subset of a member of A For example 8 123456 is a refinement of 123456 Notice Lecture Notes MAP 5601 map5601LecNotesteX 1 8272003 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 9 is called a random variable This is because the value of such a function depends on the random occurrence of a point of 9 However without this interpretation a random variable is just a function Given a finite probability space 9 P and the real valued random variable X Q a IR we define the expected value of X or expectation of X to be the weighted probability of its values Definition 14 The expectation of the random variable X Q a R is by definition EX ZXWDPW 2391 where Q w1w2 wn 77 We see in this definition an immediate utility of the property 2 Pwi 1 If X is 2391 identically constant say X C then C When a partition of Q is given giving more informaiton in general than just 9 we define a conditional expectation Definition 15 Given a finite probability space QP and a partition of Q A we define the conditional expectation of the random variable X Q a R with respect to the partition A at the point w E Q by X 70PM WEAW Here Aw is the member of A containing w and PAw 2 PM WEAW Notice that EXl9 EX 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 B if B is a refinement of A Proposition 16 HE is a re nement of A then EXl A EEXl Bl A EEXl Al 8 Lecture Notes MAP 5601 map5601LecNotesteX 2 8272003 for a random variable X Q a IR Exercise 1 Prove Proposition 16 Definition 17 A random variable X Q a IR is measurable with respect to a partition A if X is constant on each set in A 0 for example a function constant on Q is measurable with respect to any partition of 9 On the other hand if X assumes n distinct values on Q w1w2wn then X is measurable only with respect to the discrete partition wl tug Proposition 18 The conditional expectation A is measurable With respect to A Proof The proof follows immediately from the de nition of A Proposition 19 le is measurable With respect to the partition A then A X Moreoever if Y is another random variable then Eom 4 mm 4 Proof The proof again follows immediately from the de nition of A We give an interpretation of Proposition 19 If X is measurable With respect to A then the best guess77 of X given A is X itself Example Let Q be the outcomes of a toss of a die and A 246135 Define X 1 if the outcome is even and X 71 otherwise Then XwPAw E X A w 1 lt l l PAw when the outcome is even and 71 otherwise Notice here the result is the same indepen dent of the probabililties of the individual outcomes The following properites are immediate consequences of the definition of A Proposition 110 Let X and Y be random variables on QP a If ab 6 IR then EaX bYl A aEXl A bEYl A b IfX Y then EGG1 EYl A Lecture Notes MAP 5601 map5601LecNotestex 3 8272003 Defninition 111 Given a random variable X on 9 we denote by pX the partition generated by X This is the smallest coarsest partition of 9 such that X is measurable with respect to Notice it follows that X is measurable with respect to a partition A if and only if A is a refinement of Defninition 112 A filtration of partitions of Q is a sequence of partitions A17A27A377AN where At1 is a re nement of At fort 12N 71 Example Q 141171412714131414 then A1 A2 w1w2w3w4 A3 wl tag tag w4 is a filtration of Q A sequence of random variables X1X2XN On Q is called a stocastic process or simply a process We associate a process and a ltration in the following ways Defninition 113 A sequence of random variables Xt v1 is adapted to the filtration At il if Xt is measurable with respect to At for all t 12 N Example Let Q w1w2w3w4 Let A1 A2 w1w2w3w4 and A3 wgw2w3w4 Then At 1 is a filtration Let X1wi17 i17273747 1 i 12 X2wi2 34 X3wZ i i1234 The process Xt 1 is adapted to At Defninition 114 A process Xt is predictable with respect to At if Xt is measurable with respect to 451 for all 5 Lecture Notes MAP 5601 map5601LecNotestex 4 8272003 We remark that if Xt is adapted to At then Yt where Yt Xt11 is predictable with respect to At when suitably de ned at the initial time Defninition 115 A process Xt is a martingale with respect to the filtration At if Eth As X5 for 1 S s S t S N Notice this condition involves the probability measure P on 9 Example Let X be a random variable on QP Define Yt EXl4t fort 12N Then using the law of iterated expectations Proposition 16 we get EYtl As EEXl Atlls EXl As Y5 for 1 S 5 3153 N As such 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 5 t then we have Eth At th which implies that Xt is measurable with respect to At Proposition 116 A process Xt is a martingale with respect to At if and only if EGG1571 X571 for ails 23N Proof If Xt is a martingale then the conclusion follows from Definition 115 Conversely Eth 4172 At72l 4171 EltEltth 4t71llt72 Lecture Notes MAP 5601 map5601LecNotesteX 5 8272003 by the laW of iterated expectations Hence if Eth 451 Xt11 then EXti 4172 EXt71i 4172 Xt72 Repeating this argument shows that Xt is a martingale Defninition 117 Given two processes Xt il and we define the optional quadratic covariation process by t X Ylt Z AXSAY 51 for l gth HereAXs X57X511 and X0Y00 Exercise 2 Prove the polarization identity 1 7 Yt Yt 7 7 Xt 7 if7 Yb We remark that the so called optional quadratic variation process of Xt XXt is an increasing process With respect to t Defninition 118 The predictable quadratic covariation process is defined by t ltXYgtt ZEltAXSAYsi AH 51 where we also set A0 We remark that since each term is conditioned on the previous partition ltXYgtt is a predictable process Proposition 119 If Xt and are martingales for QPAt then so are th 7 X7 and 7 ltX Proof We show that EXth 7 Xleti 4171 Xtilytil 7 X7 Yit71 It then follows from Proposition 116 that Xth 7 X Ylt is a martingale To this purpose notice that inyit AXtAYt X7 Yit71 Lecture Notes MAP 5601 map5601LecNotesteX 6 8272003 Furthermore XYlt1 X1Y1 AXZAYZ AXt1Alg1 Now since martingales are in particular adapted we have EltAXsAmAt1gt AXSAYS for all 1 S s S t 7 1 It follows from linearity that EltlXaYlt71l 4171 leYlteL We calculate EXth nyltl Atil EXth AXtAYt nyltiil Atil EXt71Yl XthA Xtilytill 4171 Xletii Xt71EYl l4t71 1471511th 4171 Xtilytil X7 Ylt71 Here we have used Proposition 19 Now since Xt and are martingales 4171 K571 and Eth 4171 Xt71 The result follows Exercise 3 Show that under the above assumptions Xthe ltX Ygtt is a martingale Defninition 120 Two martingales Xt are orthogonal if ltXYgtt 0 all 1 S t S N Theorem 121 Two processes Xt and are orthogonal martingales if and only if X1Y1 0 and Xth is a martingale Proof If we have two orthogonal martingales then X1Y1 ltXYgt1 0 and Theorem 119 shows that Xth is a martingale Conversely if Xth is a martingale then ltX7Ygtt Xth 7 Xth 7 ltX7Ygtt Lecture Notes MAP 5601 map5601LecNotesteX 7 8272003 is also a martingale since the conditions in linear However by construction ltX Ygtt is also predictable Now any predictable martingale is necessarily constant in 15 Here EltltX7YgtslA571 ltX7Ygtsih since ltXYgt is a martingale and EltltX7Ygtsl A571 ltX7Ygts since ltXYgt is predictable Evidently ltXYgtN ltXYgtN1 ltXYgt1 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 X 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 X2 i XX X3 7 XXt Exercise 4 Two martingales X and Y are orthogonal if and only if X1Y1 0 and As Eth As forall 1333i Example We now condsider the outcomes of the toss of a fair coin three times As such the sample space 9 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 at time t The following table lists these quantities Lecture Notes MAP 5601 map5601LecNotesteX 8 8272003 Next we de ne a ltration A1 wlzw27w37w47 w57w67w7zw8 A2 W1yw27w37w47 Maw w7aw87 A3 wlh Win wsh Min Main wash Win W8 Notice that Xt 1 is an adapted process With respect to At 1 Moreover Xt is a martingale With respect to At Exercise 5 Show that X is a martingale Exercise 6 Calculate X27 X X and X27 ltX Xgt and show that they are martingales Definition 122 For any processes X and Y we define the stochastic sum of Y with respect to X the discrete stochastic integral as the process 0 ift0 t YXt7 ZYSAXM 51 when the sum is defined Proposition 123 For processes X and Y AltYXgtt ltYXgtrltYXgtt1 YtAXt Proof The proof is immediate by de nition Lecture Notes MAP 5601 map5601LecNotesteX 9 8272003 Theorem 124 le is a martingale and Y is a predictable process then Y X is a martingale Proof First EY 39 XM A1571 EOQAth A1571 EY 39 Xt1l 4171 Since X is a martingale and Y is predictable in particular both are adapted to the given filtration As such EY Xt1l At1 Y Xt1 Furthermore since Y is predictable Y is measurable on 451 and so EYtAth At1 Y5EAth At1 Finally since X is a martingale EAth At1 0 All together EY XM At1 Y Xt1 proving the assertion Proposition 125 For processes Y W and X we have Y W Proof Y w X inmw or 51 where we have used Proposition 123 Proposition 126 For processes Y W and X we have YX7Wl Y Xanl Proof By de nition Z YSAXSAWS 51 On the other hand t 51 Lecture Notes MAP 5601 map5601LecNotesteX 10 8272003 However AX Wls X Wls 7 X W51 AXSAWS Proposition 127 For processes Y W and X assuming that Y is predictable ltYX7Wgt Y ltX7Wgt Exercise 7 Prove Proposition 127 Proposition 128 De ne the process X as the right shift of the process X that is X Xt1 With X0 0 Then or XX27XX Proof One calculates t t t 7 X5271 7 51 81 51 The rst term on the right telescopes to X and the result follows Exercise 8 Given two processes X and Y use the polarization identity of Exercise 2 and Proposition 128 to show X Y XY7Y X eXY This is the stocastic summation by parts formula Lecture Notes MAP 5601 map5601LecNotestex 11 8272003 2 Elements of Continuous Probability Theory We rst de ne U elds Definition 21 Given an arbitrary set 9 a collection of subsets of 9 f is a U field if the following properties hold 0 45 e f ii ifA E f then AC 6 f and iii if is a countable sequence of sets in f then 611 E f Here AC QA is the complement of A We remark that be Q E f and so a5 9 is the smallest U eld for any set 9 On the other hand notice that the power set 739 is the largest U eld of Q We remark that U elds are used to de ne integration theory based on measure theory In probability theory we also use U elds to track information as partitions do in the nite case Combining properties ii and iii and using de Morgan7s law we see that if AZL9 C f then A 0113 6 r So U elds are closed under complementation countable unions and countable intersec 38 239 1 tions Notice in the nite case only nite unions and intersections need be considered In the nite case a U eld is often called simply a eld Definition 22 When 9 is a nonempty collection of subsets of 9 we write 09 for the U field generated by 9 By definition this is the smallest U field of 9 containing 9 Exercise 7 Show that 09 is unique Example Let Q w1w2w3w4 and g w1w2w3w4 Then 09 7w17w27w37w479 H w17w27ws7w4y the 07 0 739 Notice that the collections 9 and H in the above example are partitions of Q In the nite case all U elds arise this way Theorem 23 Let Q be nite and let f be a U eld on There is a unique partition A on so that 0A f Lecture Notes MAP 5601 map5601LecNotesteX 12 8272003 Proof Partially order f by inclusion We claim that A minimal elements of f is a partition of Q and moreover 0A f By definition A E f is minimal if there is no proper subset of A in f In other words if B E f and B C A then B A Now if A1 and A2 are minimal then A1 A2 b otherwise A1 A2 or one contains a nonempty k proper set A1 A2 6 f Next let A A1A2Ak then U AZ 9 Otherwise 2391 k C lt U AZ is a nonempty set containing a minimal element of f Hence A is a partition 2391 on Exercise 8 Assume that Q is finite f is a a field and A is the partition of minimal elements of f under inclusion Show that 0A f Moreover if f1 and f2 are two a fields of Q with corresponding partitions A1 and A2 and if f1 2 then A1 A2 80 in the finite case there is a one to one correspondence between partitions of Q and U fields However not all U fields of infinite sets arise in this way An important U field of the real numbers is the Borel sets Definition 24 The Borel subsets of the real line R is the U field generated by the collection of all open sets of R We denote this U field by 8R We remark that since every open subset of R is a countable union of open intervals the open intervals also generate BUR Notice that the closed interval 1 b is a countable intersection of open intervals lndeed ab a Wilma 711 Hence 1 b E BUR Definition 25 The pair Qf where Q is a set and f is a U field of subsets is called a measurable space Definition 26 A measure M defined on the measurable space Q is an extended real valued set function defined on sets in f and satisfying 0 ult gt o and ii if is a disjoint collection of subsets of f then M 21 21 Lecture Notes MAP 5601 map5601LecNotesteX 13 8272003 We call De nition 26ii countable additivity Example If 9 is finite and P is a probability function on Q then setting P 0 QPQP is a measure space Proposition 27 lfA C B are elements of f then MA S MB Proof Notice B A U B AC is a disjoint union so that MA MB 7 MB AC S MB We now give a brief outline of the construction of Lebesgue measure on the real line IR We wish to de ne a measure on subsets of IR that extend the natural euclidean length of an interval To this purpose we define outer measure Definition 28 For E C IR we define the outer measure as follows i m 7 E1231 1 I Here the infimum is over all countable collections of open intervals I with E C U We have the following properties Proposition 29 i m is de ned on 73IR ii mE 2 0 all E C IR iii ifE1 C E2 then mE1 S mE2 iv mE y mE all E C IR y 6 IR v mA 0 for any countable set A C IR vi mUEZ 3 Z for any countable collection Moreover one can show that vii mI 1 for any interval I C R Exercise 9 Prove Proposition 29 i Vi It turns out that even when is a disjoint collection that equalith in vi may fail Because of this inquot is not a measure on 73IR However remarkable inquot is a measure on a suitable sub U eld of 73IR which is in fact very large Lecture Notes MAP 5601 map5601LecNotesteX 14 8272003 Definition 210 Carath odory A set E C R is Lebesgue measurable if mA mA E mA EC for all sets A C R We write 5 for the collection of Lebesgue measurable sets Theorem 211 The collection 5 is a U eld of IR It is easy to see that b and Q are in 5 From the symmetry in E and Ec if E 6 5 then EC 6 5 follows immediately What is not so clear but true is that 5 is closed under countable unions See Theorem 212 The Borel subsets 8R are contained in 5 One only needs to show that an interval is in 5 Again see Theorem 213 The space IR 5 m where m ml is a measure space In particular 7quot is countably additive on the U eld 5 See We now discuss integration theory on a general measure space Proposition 214 Let f Q a IR be a real valued space 25 The following are equivalent i gt a E f for all a 6 IR ii 2 a E f for all a 6 IR iii lt a E f for all a 6 IR iv S a E f for all a 6 IR Moreover any of i7iv implies v a E f for all a 6 IR Definition 215 A function f Q a IR with domain a member of f is measurable with respect to the U field f if i iv holds in Proposition 214 Proof of Proposition 214 Assume Then S a oo An where A7 lt a I This proves iv Similarly lt a Uoon1xlfac S a 7 so that ii implies Also the sets in iii and iv are complements of thos in ii and i Proposition 214 follows Lecture Notes MAP 5601 map5601LecNotesteX 15 8272003 INSERT 1 NOT YET Example Any real valued function f Q a R is measurable with respect to the measurable space 9 Example Let Q w1w2w3w4 Pw i for all i 1234 and A w1w2w3w4 We define two random variables on QUAP i Q i W1 i 2 i w3 i w4 i 1 X1 1 1 i 1 i 1 i 1 i 1 X2 1 1 i 1 i 1 i 1 i Then X1 is measurable with respect to QUA However X2 is not Notice wiX2w gt 0w1w3 which is not an element of 0A Proposition 21 A random variable X is measurable With respect to the measurable space 9 UA Where 9 is nite and A is a partition of 9 if and only ifX is measurable With respect to the partition A De nition 17 Example Let at an Then gt a is the open set gt a As such ne is measurable with respect to IRBRm Definition 217 A simple function go on Q is a real valued function on 9 that assumes a finite number ofdistinct nonzero values a1a2 17 Moreover AZ wig0w ai is a member offfor all i 12n The representation 80W Z aiXAw 21 is called the standard representation of go Here XAW 1 if w E A and 0 otherwise Notice that a sinple function on 9 f is a measurable function on 9 f Definition 218 When go is a simple function on the measure space Qfu we define the integral of go with respect to u as so sodu Z MMw 0 21 Moreover if E E f we define sodu Z aMAi NE E 39 1 1 Lecture Notes MAP 5601 map5601LecNotesteX 16 8272003 Example Let Q w1w2w3w4 and PW Define a random variable X on 1 Z QPQP by Xw1 Xw2 1 Xw3 Xw4 71 We calculate X PdPanQD7DPQwampMR 0 This holds more generally Proposition 219 IfX is a random variable on a nite probability space QPQP then JXWE Exercise 10 Prove Proposition 219 Definition 220 Suppose that f is a nonnegative measurable function on the measure space Q i We define f u sup sodu wa 2 simple Here the supremum is over all simple functions go with go 3 f The following theorem allows us to approximate the integral of a nonnegative measur able by a sequence of simple functions increasing to f Theorem 221 Monotone Convergence Theorem Suppose that fn is a sequence of nonnegative measurable functions lim fn f almost everywhere and fn S f for all ngtltxgt n Then fdp lim fndp n 2 Next given a nonnegative measurable function f on 9 we construct simple functions goN such that lim goN f Notice that the measurability of f allows us to accomplish this Lecture Notes MAP 5601 map5601LecNotesteX 17 8272003 approximation by partitioning the interval 0N N 1 23 into 2N that converges to zero as N a 00 Because f is measurable N k7 2N AM wl fwltk1 is in f for all N 1 23 and all k 0 1 2N 71 We de ne the simple functions cm by the first endpoint That is 2N 71 N 501Vw Z kWXANJc k1 By the Monotone Convergence Theorem Alim g0NdMfdp 2 2 Exercise 11 Show that Alfim cpNw fw We extend the integral a general f Q a IR by the decomposition f f e f Here F max0 f and f7 max0 if 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 222 Given a measure space 977M We say that a second measure on 9 f 1 is absolutely continuous with respect to M if the following condition holods Whenever E E f with ME 0 we have 1E 0 In other words sets of measure zero for M are always sets of measure zero for 1 as well When 1 is absolutely continuous with respect to p we write 1 ltlt 11 Examples Given Q M and D E CHO let 1 011 Then 1 ltlt 11 and 11 ltlt 1 Let C be coounting measure on IRBR That is CE number of elements in E when E is finite and 00 otherwise If m is the Lebesgue measure on RBR then H ltlt 0 NH Lecture Notes MAP 5601 map5601LecNotestex 18 8272003 3 Let f be a nonnegative measurable function on Q M and set M fdu E for E E f then 1 is a measure on Q and moreover 1 ltlt M The fact that 1 is a measure can be proved from the Monotone Convergence Theorem The absolute continuity of 1 With respect to M is clear since integration of f over a set of M measure zero equals zero It turns out that in U nite cases the third case is typical Recall that a measure space 9 f M is finite if MiOmega More generally if Q U ooi1QZ Q E f and for all i then Q M is called U nite For example RBRm is a U nite measure 00 space since IR U 7NN N1 Theorem 223 RadonNikodym Theorem Suppose that Q M is a U nite measure space and 1 is another measure de ned on f With 1 ltlt M Then there exists a nonnegative measurable function f such that M fdu E for all E E f Moreover ifg is any other such function then f g ae With respect to M We denote the above function f by 37 and refer to it as the Radon Nikodym derivative of 1 With respect to M This function has the following properties 1 If 1 ltlt M and f is a nonnegative measurable function then d fdufldu dM 2 WW mr dM dM dM 3 lf1 ltlt M ltlt A then d1 7 d1 dM dA dM dA 4 lf1 ltlt M and M ltlt 1 then d1 1 67 Lu Exercise 13 Prove the above assertions 1 4 Lecture Notes MAP 5601 map5601LecNotesteX 19 8272003 Definition 224 A measure space Q P is a probability space if PQ 1 As such probability spaces are nite measure spaces Examples 1 Q finite P probability function and QPQP 2 01l m 3 RA where f is nonnegative and measurable with f fdm 1 R Definition 225 Given a probability space Q P and a random variable X we define the eXpectation of X as EX Q XdP Recall that this reduces to De nition 14 When 9 is finite and f Definition 226 We are given a measure space Q P a random variable X and a sub U field g C f The conditional eXpectation of X with respect to g is a random variable that we denote by g It has the following properties 1 g is measurable with respect to g 2 fXdP fEXlgdP for a G E g G G g is unique up to sets of measure zero Lecture Notes MAP 5601 map5601LecNotesteX 20 8272003 3 Differential Equations Example Assume that money deposited in a bank increases with continuously compounded interest This means that the rate of growth is proportional to the amount present Let Mt amount of money present at time t and let 7 be the proportionality constant Then 7 is the constant interest rate and Mt satisfies the differential equation dM 7 M dt T To solve this equation we relate the differentials using the chain rule dM dt dt rM dt In this case the variables M and t can be isolated on separate sides of the equation and d lnMrtO the equation can be integrated Or As such M A6 where A 60 As such Mt grows exponentially Notice the initial amount M0 A Consider the question How much must be invested today to achieve the amount E at the future time T To answer this one sets the amount M evaluated at time T equal to the desired future value E AeTT Solving for A A Ee TT This is referred to as discounting the future value E Example Let us denote by S the price or value of an asset at time t For a given 151 and 152 we write the change in value by AS 5052 75051 over the time interval At 152 7151 The Lecture Notes MAP 5601 map5601LecNotesteX 21 8272003 relative price change or return compares the absolute price change to the initial value If the relative price change is proportional to the time interval then AS 47 At 5 M Here M is a constant called the drift We can rewrite this as AS 7 S At M Letting At a 0 we obtain an equation of the same form Namely dSi 4 s w M These equations are examples of a special type of first order equations Definition 31 We here consider y as a function of x Equations of the form dyi ggi mmm are called separable Here at is a function only of x and gy is a function only of y The variables can be separated and integrated provided the integrals exist as indicated fxd Another often occurring type of first order differential equation is the linear equation In these equations y and y are present however no other functions of them occur We may express such an equation in the form ymmy m Here Mac and 110 are arbitrary functions of x We pause to remark that these equations are separable when 110 E 0 We call these equations homogeneous Indeed in this case ich dac Otherwise we solve these equations by multiplying by a so called integrating factor Let 10 efz pmdt Lecture Notes MAP 5601 map5601LecNotesteX 22 8272003 We remark that integrating factors are in general not unique In particular here any multiple of 10 Will work As such the lower limit in the definite integral fm pt dt is irrelevant and may be presupposed conveniently chosen Multiplying by I gives Iy Ipu The key to this procedure is that the left hand side is the derivative of the product of the solution y and the integrating factor lndeed lmy 1W ydiim However gm exp m pt dt pm exp m plttgt dt by the chain rule As such d QUWM WW plt96gt1lt96gty lntegrating With respect to x gives and dividing by 10 obtains the solution Example Let 1 2 y 7y 96 at Then 0 l 10 exp Edi explnac at We have actually made a choice of multiple here Multiplying by 10 gives any y 3 Lecture Notes MAP 5601 map5601LecNotesteX 23 8272003 So This gives d d tummy w 4 at any I 0 7 3 O yi 4 Example Suppose the price of an asset S grows at a constant rate 0 Then This equation is separable and d5 E41 SOt01 Now lets suppose that there are two contributions to the growth of S The first is pro portional to the asset price and the second a constant rate 0 Then OI This is a linear equation and So Lecture Notes MAP 5601 15 ds E 7 MS O 15 exp ipdS 6 d amp CTAS 06 O S 6M i ei it 01 E Oleit M map5601LecNotesteX 24 8272003 Now the initial amount is 50 01 Op so that 01 is larger than 50 Notice that SOS asymtotically approaches 016 as t a 00 We now consider solutions to equations of the form y We New We These equations are second order meaning that y is the highest derivative that is present They are linear meaning that the solution y and its derivatives 2 and y occur only to first power Otherwise 1915 qx and Mac are arbitrary functions of x We also consider the so called associated homogeneous equation y We New 0 Theorem 32 i If yl and y2 are two solutions to the homogeneous equation then so is their general linear combination namely y Oiyi O2y2 REVISION STOPS HERE for arbitrary constants 01 and 02 ii lfy is a solution to 21 and if yh is a solution to 22 then the sum y yh is a solution to 21 iii If u and v are solutions to 21 then the difference is a solution to 22 Proof 0 01m 02m Plt96gtltcw1 c2y2 19601y1 cm Ody1 WW1 qlt96gty1l 02 Wen2 qlt96gty2l 0 ii y yh WOW W mm W W my WM W pmyt qlt96gtyhl me 0 Lecture Notes MAP 5601 map5601LecNotesteX 25 8272003 iii u 7 v mm 7 v qltgtltu v in WOW WM 7 W We WM Notice that it follows from Theorem 23iii that the most general solution to 21 is any solution to 21 plus the general homogeneous solution B We now consider solutions to homogeneous equations with constant real coef cients 24 y 12 by 0 The following technique works for such equations of any order However here we restrict our attention to second order equations and because of the homogeneity may assume that the coef cient of y is one One can prove that there are two solutions to 24 which are linearly independent Two functions are linearly independent when they are not multipliers of each other or what is the same their ratio is not constant Solutions to 24 are linear combinations of solutions of the form y 6 where A is in general a complex constant As such every solution 24 is a linear combination of the two linearly independent solutions lndeed upon substitution into 24 one obtains AZCMC 1 aka 1 be 0 which can only be satisfied if 25 A2aAb 0 In fact 6 74 0 for any A or any The solutions to 25 are given by iaixaZ 74b A 2 Of course in some cases 25 can be factored by inspection Since the coef cients a and b are assumed real the solutions to 25 and as such to 24 fall into three cases Lecture Notes MAP 5601 map5601LecNotesteX 26 8272003 Case 1 The roots to 25 are real and distinct say A1 and A2 In this case CANE and 6A are solutions Since 24 is linear linear combinations of solutions are again solutions So y 016219 7 026229 is a solution for any constants 01 and 02 Case 2 The solutions to 25 are real and repeated This happens when the left hand side of 25 is a perfect square If the solution is A then 6 is a solution The other linearly independent solution to the first is 6 This can be derived from a reduction of otder argument As such y 016 7 025mm A90 is the general solution We warn you that a solution of the form x6 exists only when A is a repeated root Case 3 The solutions are complex numbers All complex numbers can be expressed in the form a ib where a and b are real numbers and i2 71 When A1 and A2 are complex solutions to 25 they necessarily are complex conjugates By this we mean if A1 1 ib then A2 a 7 ib One can see this from the quadratic formula since with real coef cients 7a2i l2m are complex conjugate numbers when 12 74b lt 0 One can also see this by conjugating equation 25 Often the complex conjugate of A a ib is denoted A a 7 ib It is easy to see that conjugation distributes over addition and multiplication of complex numbers As such m 0 or A2 aAb 0 This show that if A is a solution then so is A It follows that 6 ibm and CQ ibm are solutions to 24 It turns out that general linear combinations of the complex exponentials can be rewritten equivalently as general linear combinations of functions which are real valued Euler7s equation expresses complex exponentials as complex linear combinations of the sinusodial functions Namely for any real number 6 26 Big cos l isin l As such 27 eibm cos baa 7 isin bat and because the cosine is even and the sine is odd 28 6 me cos bit 7 isin bx Lecture Notes MAP 5601 map5601LecNotesteX 27 8272003 Using 27 and 28 general linear combinations of the complex exponentials can be rewritten as linear combinations of sine and cosine Try itl In the general solution of equation 24 the real exponentials from the term 1 remain as factors As such solutions have the form y 016 cos baa 026 sin bx Example 29 The equation y 5y6y0 gives rise to the characteristic equation A2 7 5A 6 0 The roots are A 72 and 73 As such y 016 026 is the general solution Example 210 Case 2 y 6y 9y 0 Here A 32 0 and A 73 is a double real root As such the general solution is y 0167 Cg eism Example 211 As an example of Case 3 we solve y y 0 Now A2 1 0 and A ii We can express the general result in terms of real valued functions y 01 cosac Cgsinac Suppose we compare the solutions to y y y 0 In this case A2 A 1 0 and A 7 i The solution now has real exponentials multiplying the sinusoidial solutions y016 E 2 Vi 1 Vi cos7x02e 2sin715 Lecture Notes MAP 5601 map5601LecNotestex 28 8272003 3 The Dirac Delta Function The Dirac delta function is a linear mapping which associates a real number to each real valued continuous function Such a mapping is called a linear functional lf IR denotes the real line and CIR the continuous functions de ned on IR then the 6 function maps CIR onto IR Given go 6 CIR we define That is 6 maps each continuous function to its value at the origin We can construct the action of 6 by integrating a continuous function go against a family of functions 65 and taking a limit This gives rise to the view of 6 as a function Translations of the 6 function arise naturally in this formalism and the continuous functions are evaluated in this way at all real values To this purpose we define 1 i hen is lt x lt 6 6 287 W E gt 0 otherwise for 8 gt 0 Notice 65 die 1 for all 5 Theorem 31 Let go 6 CIR Then cm gg WWMW Proof Notice femwwmjiwmw 25 On the other hand Lecture Notes MAP 5601 map5601LecNotesteX 29 8272003 Hence 5s5096d96 500 We esoltogtgtdx E 1 lt i 7 0 d e 26 We M M x is Since go is continuous at 0 given 7 gt 0 there exists 8 gt 0 such that M96 7 MOM lt n Whenever lt 5 This proves 32 D It is useful in calculations to write 610 in the integrand and so 55096d sow Notice if a lt 0 and b gt 0 in particular and Notice that it follows that 5a d soltagt for all a 6 IR Equivalently 6c e agtsoltxgtdx sac and one may think of 6 7 a as translating the concentration of the delta function from the origin to the point 1 Lecture Notes MAP 5601 map5601LecNotesteX 30 8272003 Example 33 Suppose at time t 0 the value of an asset is V0 and this value increases instantaneously at t l by an amount A The differential equation 34 A605 7 1 describes this Integrating 34 with respect to 15 gives A6T7ldT 0 i 0 0lttlt1 7 A tgt1 aclta acgta then we can write Vt Auac 7 a V0 In this sense we symbolically write d E7105 7 a 615 7 1 Lecture Notes MAP 5601 map5601LecNotesteX 31 8272003 4 The Laplace Transform Given a function t for t gt 0 we de ne its Laplace transform 41 fs ms dt 0 Notice that the improper integral 41 is de ned for a wide range of functions ft because when 5 gt 0 6 is a decaying exponential In most applications of the Laplace transform which we will encounter restricting the domain of the variable 5 will not cause dif culty We offer the following table and discuss some of the calculations Here and throughout this section we denote the Laplace transform of ft by When the context permits we will also represent the Laplace transform of functions denoted by lower case roman letters by their capitals 42 A Table of transforms Here the constant a is in general a complex nunmber 1 05 F 8 a t b 6 Sid c cos at 5212 d sin at 52112 e cosh at f sinh at g 7105 a With some practice one remembers this table without having to recompute We em phasize that the Laplace transform is linear As such af by MU bug for functions ft gt and constants a and b The functions 42 c d e and f are linear combinations of functions of the form 6 Because the Laplace transform is linear the Lecture Notes MAP 5601 map5601LecNotesteX 32 8272003 1 571139 transforms of c d e and f are linear combinations of This helps remember the pattern of transforms in Table 42 The Laplace transform transforms linear ordinary differential equations into algebraic equations of the transforms This is the key property that we here exploit Theorem 43 So that the Laplace transforms exist we assume that there exists an s gt 0 such that tlim lfte 5tl 0 Also we assume that ft exists except perhaps for a nite H00 number of points in any bounded interval Then 44 f8 7 sums 7 f0 Under appropriate conditions one can iterate 44 to get ltf gtltsgt same 7 No 44 7 82 f8 7 sf0 7 NW Notice that the initial conditions f0 and fO are in these formulas As such solutions to differential equations using the Laplace transform satisfy prescribed initial conditions which gives values for the integration constants in the general solutions This aspect of the Laplace transform is often convenient Example 46 Solve the differential equation y 75y6y0 with initial conditions 240 l and yO 0 Using 44 45 and the linearity of 5 we get My 7 5 Ad MW 7 50 Lecture Notes MAP 5601 map5601LecNotesteX 33 8272003 5 The Fourier Transform Lecture Notes MAP 5601 map5601LecNotesteX 34 8272003 6 The Diffusion Equation The one dimensional diffusion equation or heat equation is a partial differential in two variables x and t In most applications of the diffusion equation the solution Mac 15 is a function in the spatial variable x and the variable of time t The diffusion equation Bu 821i 61 E w is a parabolic partial differential equation Problem 62 Solve 61 over the real line foo lt x lt 00 for u uxt with prescribed initial condition Mac 0 71015 When the integral exists the solution to Problem 62 is given by new ut2 u0vgi 4t d1 The so called fundamental solution to Problem 62 is the above kernel 1 m 64 u at 76 4t lt 5lt 2x7rt 0 In this case ggr us at 576 in the distributional sense 32 As such from 63 1 0 few lutilgmuove 4t dv oo u0v67v dv u0ac We give two derivations of 63 Because the domain is the entire real line we can directly apply the Fourier transform Lecture Notes MAP 5601 map5601LecNotesteX 35 8272003 1 Applying the Fourier transform to 61 we get a A 2 2A 65 a ust 47139 s ust Solving 65 st 054628 and 178 0 C On the other hand 66 653 25 54625 Hence 67 as t E0s 56915 The convolution theorern gives uxt u0vu5x 7 mt dv no u5ac t This is 63 M It is useful to realize that 63 can also be derived from the superposition of so called separated solutions This technique is prevalent When building solutions to linear equations To this purpose we set uxt Then substitution into 61 gives FxGt F Gt or 68 Since the left and right hand sides of 68 depend separately on the independent variables and If they must be constant We choose this constant to be 7 The solutions in this case Will decay as t a 00 As such 69 F x p2Fx 0 and 610 Gt p2Gt 0 Lecture Notes MAP 5601 map5601LecNotesteX 36 8272003 Solving these equations one gets in general Acospx Bsinpx Gt Cap As such for general A and B ux t A cospac B sinpace p2t is a solution to 61 Now since 61 is linear a superposition of solutions is also a solution The integral is a limit of Riemann surns which are finite superpositions of solutions Here we can pass to the limit obtaining 611 U t Ap cospac 1 31 sinpace p2t dp 0 as a solution to 61 Notice that 611 is a Fourier integral In particular Um 0 ltAltpgt cospx Bltpgt shim dp 0 will equal no if we choose 1 0 Ap 7 u0v cospv dv 7r and 00 1 Bp i u0v sinpv dv 7r lnserting the formulas for the coe icients Ap and 31 into 611 we obtain 0 1 0 1 0 7 2 u0v cospv dv cospac E u0v sinpv dv sinpac e p dp 0 0 1 7 u0vcospv cospx sinpv sinpacle th dv dp 7r 0 0 Lecture Notes MAP 5601 map5601LecNotesteX 37 8272003 Using the summation formula for the cosine an equation worth remembering cosAB cos A cosB 7 sinA sin B we collapse the above integral into 1 7 2 612 7 u0v cospv 7 p6 p dv dp 7r 0 700 We next interchange the order of integration and change variables as follows stp and 2bxtac7v Then 612 becomes 6 52 cos 2bs ds dv xE 613 7 uov O8 Since e 52 cos 2bs ds 6472 0 we again obtain 63 We next show that certain equations of interest can be solved by solving the diffusion equation Proposition 614 Solutions to 81 821 81 615 7 7 7 b i 8t 82a8 where a and b are constants are of the form a2 616 vxt e mb77tluxt Here ux t solves the diffusion equation 61 Proof If v has the form 616 then substitution into 615 shows that ux t solves 61 2 Conversely if v is a solution to 615 then 6 QT WU solves 61 Proposition 617 For w wXt suppose that with A 7 0 8w 2 82w 8w There are solutions to 618 of the form vt wXt where em and 1 satis es 615 with a B 7 AA and b 0 Lecture Notes MAP 5601 map5601LecNotesteX 38 8272003 7 The BIackScholes Equation Lecture Notes MAP 5601 map5601LecNotesteX 39 8272003 8 An Introduction to Measure and Integration The concept of measure gives a unifying theory for many mathematical situations which may at rst seem quite different The discussion below shows how it arises naturally in the theory of the Lebesgue integral Let us rst consider the example of a nite probability space Here X is a set of outcomes called a sample space A probability function P is de ned on the set of all subsets of X This set is called the power set of X and denoted By de nition a probability function is a function which associates a nonnegative real number to a given subset and which satis es two conditions 81 PX 1 82 PA U B PA PB when A B Q ie A and B are disjoint Notice that these conditions imply that PA S l for any subset A E PX since PX PA PAC and in particular PQJ 0 Example 83 Consider the outcomes of tossing a coin H for heads T for tails The sample space is X H T The power set of X is PX 07 H TEX If the coin is fair then PltHgt PltTgt g and PX PH U T PH PT Notice that the power set PX of any set X nite or not trivially satis es the properties of an algebra of subsets Lecture Notes MAP 5601 map5601LecNotesteX 40 8272003 Definition 84 A set of subsets of a set X A is an algebra of subsets of X satisfying two properties 85 If A E A then AC XA E A 86 If AB E A then A U B E A Notice A U AC X and so X E A and so XC Q E A Also de Morganls laW gives A B ACUBCC and so A B E A Notice that X 0 is the smallest algebra of subsets of X In the case of a finite probability space X and therefore 73X is finite In other cases X is infinite and it is often necessary to consider sets of subsets of X Which are smaller than 73X but still retain the properties of an algebra Since When X is infinite algebras of subsets may also be infinite it is necessary to extend 83 as follows Definition 87 A U algebra f of subsets of X is an algebra of subsets such that 88 If AZL9 is a sequence of sets in f then their union U 14 is also in f 2391 Example 89 Let X R and A set of finite unions of half open intervals That is finite unions of sets of the form la b Where foo lt a S b 3 00 along With the intervals foo 1 Then A is an algebra of subsets of R Example 810 Given a collection of subsets C C 73X there is a srnallest U algebra con taining C We call this U algebra the U algebra generated by C and denote it by 0C For X R and A as in Example 89 0A is called the Borel subsets of IR and denoted by B The Borel subsets B is also the U algebra generated by the open sets of IR since each open set is a countable union of open intervals Example 811 On the other hand consider the collection of intervals of IR C 7000 0 Notice the intervals partition IR that is they are disjoint and IR is their union In this case 0C 9 700 0 0 00 R Lecture Notes MAP 5601 map5601LecNotestex 41 8272003 Also notice that a similar result occurs when C is any partition of IR We next discuss the concept of a measure on sets A probability function on a proba bility space is an important example We motivate the concepts by outlining the construction of the Lebesgue integral which requires Lebesgue measure To this purpose we first review the Riemann integral The Riemann integral when it exists is de ned to be a limit of so called Riemann sums We now describe these sums Suppose that x is defined on a closed interval 1 b We partition this interval into subintervals labeled x0 a lt x1 lt x2 lt lt x b In each subinterval xi1xZ we choose sample a functional value where xFl S 3 xi This functional value is used as an approximation to the functional values of x over the subinterval xi1xZ If the function f actually was constant on these subintervals the following Riemann sum would be the Riemann integral of x over 1 b In general however each such sum is an approximation to the Riemann integral of x over 1 b n 812 2 x2 AxZ 2391 Here is the sampled functional value from the subinterval xi1xZ and Axi xi 7 xFl is the length of the ith subinterval As such the Riemann sum 812 is a finite sum of areas of rectangles A sequence of such Riemann sums is created by subdividing a previous partition and creating a new Riemann sum The Riemann integral is by definition the limit of these Riemann sums as n the number of subintervals tends to infinity with the lengths Ax tending to zero One way to do this is to use a uniform partition with Axi 6 7 for all n and all i Elementary calculus texts often use this approach It turns out that when such a sequence of Riemann sums has a limit it is independent of the partitioning or the sampling We define the Riemann integral of x over ab as b 71 813 m dx A1330 2 x2 AxZ a 21 We remark that this limit exists in particular if f is continuous on 1 b or bounded with a finite number of discontinuities In summary to define the Riemann integral one partitions the domain of the functions by functional values The Lebesgue integral reverses this point of view In essence one partitions the range of the function and approximates the function Lecture Notes MAP 5601 map5601LecNotesteX 42 8272003 by a constant functional value over inverse images of the intervals partitioning the range Definition 8 Given a probability space Q P and a function F Q a R a random variable we de ne the expectation or average value of F as s EF FwdPw 2 Notice that PQ 1 We also need conditional expectation of F This in general is another function Definition 8 Given a subUalgebra g C f we de ne g the conditional expectation of F given 9 This is any function G such that 1 G is measurable with respect to the Ualgebra g 2 deP deP for all A E Q We remark that in general the smaller the oalgebra the less functions are measurable This is because less sets are available to allow the inverse images of measurable sets to be measurable Definition 8 Let x x on 01 Now 01 O1m where 01 are the Lebesgue measurable sets of 01 is a probability space Notice that the function is measurable with respect to the Ualgebra g Ulti0 00101 but that x is not For example at 6 01 is not measurable with respect to g Ks A 0 R V V Moo we Lecture Notes MAP 5601 map5601LecNotesteX 43 8272003 9 A Discussion of Brownian Motion Lecture Notes MAP 5601 map5601LecNotesteX 44 8272003 10 Derivation of the BIackScholes Equation and Applications Lecture Notes MAP 5601 map5601LecNotesteX 45 8272003 References Ikeda N and Watanabe 5 Stochastic Di erential Equations and Di u sion Processes NorthHolland Publishing Company Amsterdam New York Oxford October 28 1981 Durrett R Brownian Motion and Martingales in Analysis Wadsworth Inc Belmont California 1984 Hida T Brownian Motion SpringerVerlaf New York November 11 1980 Karatzas I and Shreve SE Brownian Motion and Stochastic Calculus Springer Verlag New York November 11 1988 Baxter M and Rennie A Financial Calculus Lecture Notes MAP 5601 map5601LecNotesteX 46 8272003

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