1、Mathematical Finance Probability Review Tatiana Kirsanova Semester 1, 2014-15 Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 1 / 48Outline 1 Discrete Random Variables 2 Expectation of a Random Variable 3 Equivalence of Probability Measures 4 Variance of a Random Variable 5 Continuous R
2、andom Variables 6 Examples of Probability Distributions 7 Correlation of Random Variables 8 Conditional Distributions and Expectations Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 2 / 48Sample Space We collect the possible states of the world and denote the set by W. The states are c
3、alled samples or elementary events. The sample space W is either countable or uncountable. I A toss of a coin: W=fHead,Tailg=fH,Tg. I The number of successes in a sequence of n identical and independent trials: W=f0,1,. ,ng. I The moment of occurrence of the rst success in an innite sequence of iden
4、tical and independent trials: W=f1,2,.g. I The lifetime of a light bulb: W=fx2Rjx 0g. The choice of a sample space is arbitrary and thus any set can be taken as a sample space. However, practical considerations justify the choice of the most convenient sample space for the problem at hand. Discrete
5、(nite or innite, but countable) sample spaces are easier to handle than general sample spaces. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 3 / 48Discrete Random Variables Examples of random variables: I Prices of stocks. I Exchange rates. I Payos corresponding to portfolios. Denitio
6、n (Discrete Random Variable) A real-valued function X :W!R on a discrete sample space W=(w k ) k2I , where the set I is countable, is called a discrete random variable. Denition (Probability) A mapP :W7!0,1 is called a probability on a discrete sample spaceW if P.1. P(w k ) 0 for all k2 I, P.2. k2I
7、P(w k )= 1. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 4 / 48Probability Measure LetF = 2 W stand for the class of all subsets of the sample space W. The set 2 W is called the power set ofW. Note that the empty set also belongs to any power set. Denition (Probability Measure) A map
8、P :F7!0,1 is called a probability measure on (W,F) if M.1. For any sequence A i F , i = 1,2,. of events such that A i A j = for all i6= j we have P( i=1 A i )= i=1 P(A i ). M.2. P(W)= 1. Any set A2F is called an event. For any A2F the equalityP(WnA)= 1 P(A) holds. Kirsanova (Glasgow) Mathematical Fi
9、nance Semester 1, 2014-15 5 / 48Probability Measure on a Discrete Sample Space Note a probabilityP :W7!0,1 on a discrete sample space W uniquely species probabilities of all events A k =fw k g. It is common to writeP(fw k g)=P(w k )= p k . The theorem shows that any probability on a discrete sample
10、space W generates a unique probability measure on (W,F). Theorem LetP :W7!0,1 be a probability on a discrete sample space W. Then the unique probability measure on (W,F) generated byP satises, for all A2F, P(A)= w k 2A P(w k ). The proof of the theorem presents no di culties, since W is assumed to b
11、e discrete. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 6 / 48Example: Coin Flipping Let X be the number of headsappearing when a fair coin is tossed twice. We choose the sample space W to be W=f0,1,2g where a number k2W represents the number of times headhas occurred. The probabili
12、ty measureP onW is dened as P(k)= 1 4 , if k = 0,2, 1 2 , otherwise. We recognise here the binomial distribution with n = 2 and p = 1 2 . A single ip of a coin is a Bernoulli trial. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 7 / 48Example: Coin Flipping We now suppose that the coin
13、 is not a fair one. Let the probability of headbe p for some p6= 1 2 . Then the probability measureP is given by P(k)= 8 : q 2 , if k = 0, 2pq, if k = 1, p 2 , if k = 2, where q := 1 p is the probability of tailappearing. We deal here with the binomial distribution with n = 2 and 0 p 1. Kirsanova (G
14、lasgow) Mathematical Finance Semester 1, 2014-15 8 / 48Expectation of a Random Variable Denition (Expectation) Let X be a random variable on a discrete sample space W endowed with a probability measureP. The expectation (expected value or mean value) of X is dened to be E P (X)= m := k2I X(w k )P(w
15、k )= k2I x k p k . E P ( ) is called the expectation operator over the probabilityP. Note that the expectation of a random variable can be seen as the weighted average. Since it is impossible to know the exact event in the future, expectation could help one to make decisions. Kirsanova (Glasgow) Mat
16、hematical Finance Semester 1, 2014-15 9 / 48Expectation Operator Any random variable dened on a nite set W admits the expectation. When the setW is countable (but innite), we say that X is P-integrable wheneverE P (jXj)= w2W jX(w)jP(w). Then the expectationE P (X) is well dened (and nite). Theorem (
17、1.1) Let X and Y be twoP-integrable random variables andP be a probability measure on a discrete sample space W. Then for all a,b2R E P (aX +bY)= aE P (X)+bE P (Y) . HenceE P ( ) :X!R is a linear operator on the spaceX ofP-integrable random variables. Kirsanova (Glasgow) Mathematical Finance Semeste
18、r 1, 2014-15 10 / 48Expectation Operator Proof of Theorem 1.1. We note that E P (jaX +bYj)j ajE P (jXj)+jbjE P (jYj) so that the random variable aX +bY belongs toX. The linearity of expectation can be easily deduced from the denition: E P (aX +bY)= w2W (aX(w)+bY(w)P(w) = w2W aX(w)P(w)+ w2W bY(w)P(w)
19、 = a w2W X(w)P(w)+b w2W Y(w)P(w) = aE P (X)+bE P (Y) where a and b are arbitrary real numbers. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 11 / 48Expectation: Coin Flipping A fair coin is tossed three times. The player receives one dollar each time headappears and loses one dollar w
20、hen tailoccurs. Let the random variable X represent the player s payo. The sample space W is dened as W=f0,1,2,3g where k2W represents the number of times headoccurs. The probability measure is given by P(k)= 8 : 1 8 , if k = 0,3, 3 8 , if k = 1,2. This is the binomial distribution with n = 3 and p
21、= 1 2 . Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 12 / 48Expectation: Coin Flipping The random variable X is dened as X(k)= 8 : 3, if k = 0, 1, if k = 1, 1, if k = 2, 3, if k = 3. Hence the player s expected payo equals E P (X)= 3 k=0 X(k)P(k) = 3 8 + 3 8 + 3 8 + 3 8 = 0. Kirsanov
22、a (Glasgow) Mathematical Finance Semester 1, 2014-15 13 / 48Expectation of a Function of a Random Variable Let X be a random variable andP be a probability measure on a discrete sample space W. We dene Y = f(X) where f :R!R is an arbitrary function. Then Y is also a random variable on the sample spa
23、ce W endowed with the same probability measureP. Moreover, E P (Y)=E P (f(X)= w2W f(X(w)P(w). If a random variable X is deterministic thenE P (X)= X and E P (f(X)= f(X). Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 14 / 48Equivalence of Probability Measures LetP andQ be two probabili
24、ty measures on a discrete sample space W. Denition (Equivalence of Probability Measures) We say that the probability measuresP andQ are equivalent and we denoteP Q whenever for all w2W P(w) 0 , Q(w) 0. The random variable L :W!R + given as L(w)= Q(w) P(w) is called the Radon-Nikodym density ofQ with
25、 respect toP. Note that E Q (X)= w2W X(w)Q(w)= w2W X(w)L(w)P(w)=E P (LX) . Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 15 / 48Example: Equivalent Probability Measures The sample space W is dened as W=f1,2,3,4g. Let the two probability measuresP andQ be given by 1 8 , 3 8 , 2 8 , 2 8
26、 and 4 8 , 1 8 , 2 8 , 1 8 . It is clear thatP andQ are equivalent, that is,P Q. Moreover, the Radon-Nikodym density L ofQ with respect toP can be represented as follows 4, 1 3 ,1, 1 2 . Check that for any random variable X: E Q (X)=E P (LX) . Kirsanova (Glasgow) Mathematical Finance Semester 1, 201
27、4-15 16 / 48Risky Investments When deciding whether to invest in a given portfolio, an agent may be concerned with the risk“ of his investment. Example Consider an investor who is given an opportunity to choose between the following two options: The investor either receives or loses 1,000 dollars wi
28、th equal probabilities. This random payo is denoted by X 1 . The investor either receives or loses 1,000,000 dollars with equal probabilities. We denote this random payo as X 2 . Hence in both scenarios the expected value of the payo equals 0 E P (X 1 )=E P (X 2 )= 0. The following question arises:
29、which option is preferred? Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 17 / 48Variance of a Random Variable Denition (Variance) The variance of a random variable X on a discrete sample set W is dened as Var(X)= s 2 :=E P n (X m) 2 o , whereP is a probability measure on W. Variance i
30、s a measure of the spread of a random variable about its mean and also a measure of uncertainty. In nancial applications, it is common to identify variance of the price of a security with its degree of risk. Note that Var(X)= s 2 0. It equals 0 if and only if X is deterministic. Kirsanova (Glasgow)
31、Mathematical Finance Semester 1, 2014-15 18 / 48Variance of a Random Variable Example The variance of option 1 equals Var(X 1 )= (1000 0) 2 2 + ( 1000 0) 2 2 = 10 6 . The variance of option 2 equals Var(X 2 )= (10 6 0) 2 2 + ( 10 6 0) 2 2 = 10 12 . Therefore, the option represented by X 2 is more ri
32、sky than the option represented by X 1 . A risk-averse agent would prefer the rst option over the second. However, a risk-loving agent would prefer the second option over the rst. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 19 / 48Variance of a Random Variable Theorem (1.2) Let X be
33、 a random variable andP be a probability measure on a discrete sample spaceW. Then the following equality holds Var(X)=E P X 2 m 2 . Proof of Theorem 1.2. Var(X)=E P (X m) 2 =E P X 2 2mX +m 2 (linearity) =E P X 2 2mE P (X)+m 2 =E P X 2 m 2 . Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-
34、15 20 / 48Independence of a Random Variable Denition (Independence) Two discrete random variables X and Y are called independent if and only if for all x,y2R P(X = x,Y = y)=P(X = x)P(Y = y) whereP(X = x) is the probability of the eventfX = xg. A useful property of independent random variables X and
35、Y is E P (XY)=E P (X)E P (Y) . Theorem (1.3) Let X and Y be two independent discrete random variables. Then we have, for arbitrary a,b2R, Var(aX +bY)= a 2 Var(X)+b 2 Var(Y) . Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 21 / 48Independence of a Random Variable Proof of Theorem 1.3. L
36、etE P (X)= m X andE P (Y)= m Y . Theorem 1.1 yields E P (aX +bY)= am X +bm Y . Using Theorem 1.2, we obtain Var(aX +bY)=E P (aX +bY) 2 (am X +bm Y ) 2 = a 2 E P X 2 +2abE P (XY)+b 2 E P (Y) (am X +bm Y ) 2 = a 2 E P X 2 m 2 X +b 2 E P Y 2 m 2 Y +2ab(E P (X)E P (Y) m X m Y ) = a 2 Var(X)+b 2 Var(Y) .
37、 Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 22 / 48Continuous Random Variables Denitions (Continuous Random Variable) A random variable X on the sample space W is said to have a continuous distribution if there exists a real-valued function f such that f(x) 0, Z f(x)dx = 1, and for
38、 all real numbers a b P(a X b)= Z b a f(x)dx. Then f :R!R + is called the probability density function (pdf) of a continuous random variable X. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 23 / 48Continuous Random Variables Assume that X is a continuous random variable. Note that a p
39、robability density function need not satisfy the constraint f(x) 1. A function F(x) is called a cumulative distribution function (cdf) of a continuous random variable X if for all x2R F(x)=P(X x)= Z x f(y)dy. The relationship between the pdf f and cdf F F(x)= Z x f(y)dy , f(x)= d dx F(x). Kirsanova
40、(Glasgow) Mathematical Finance Semester 1, 2014-15 24 / 48Continuous Random Variables The expectation and variance of a continuous random variable X are dened as follows: E P (X)= m := Z xf(x)dx, Var(X)= s 2 := Z (x m) 2 f(x)dx, or, equivalently, Var(X)= Z x 2 f(x)dx m 2 =E P X 2 (E P (X) 2 The prop
41、erties of expectations of discrete random variables carry over to continuous random variables, with probability measures being replaced by pdfs and sums by integrals. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 25 / 48Discrete Probability Distributions Example (Binomial Distribution
42、) LetW=f0,1,2,. ,ng be the sample space and let X be the number of successes in n independent trials where p is the probability of success in a single Bernoulli trial. The probability measureP is called the binomial distribution if P(k)= n k p k (1 p) n k for k = 0,. ,n where n k = n! k!(n k)! . The
43、n E P (X)= np and Var(X)= np(1 p). Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 26 / 48Discrete Probability Distributions Example (Poisson Distribution) Let the sample space be W=f0,1,2,.g. We take an arbitrary number l 0. The probability measureP is called the Poisson distribution i
44、f P(k)= l k e l k! for k = 0,1,2,. . Then E P (X)= l= Var(X) . It is known that the Poisson distribution can be obtained as the limit of the binomial distribution when n tends to innity and the sequence np n tends to l 0. Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 27 / 48Discrete P
45、robability Distributions Example (Geometric Distribution) LetW=f1,2,3,.g be the sample space and X be the number of independent trials to achieve the rst success. Let p stand for the probability of a success in a single trial. The probability measureP is called the geometric distribution if P(k)=(1
46、p) k 1 p for k = 1,2,3,. . Then E P (X)= 1 p and Var(X)= 1 p p 2 . Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 28 / 48Continuous Probability Distributions Example (Uniform Distribution) We say that X has the uniform distribution on an interval (a,b) if its pdf equals f(x)= 1 b a , i
47、f x2(a,b) , 0, otherwise. It is clear that Z f(x)dx = Z b a 1 b a dx = 1. We have E P (X)= a+b 2 and Var(X)= (b a) 2 12 . Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 29 / 48Continuous Probability Distributions Example (Exponential Distribution) We say that X has the exponential dist
48、ribution on (0,) with the parameter l 0 if its pdf equals f(x)= le lx , if x 0, 0, otherwise. It is easy to check that Z f(x)dx = Z 0 le lx dx = 1. We have E P (X)= 1 l and Var(X)= 1 l 2 . Kirsanova (Glasgow) Mathematical Finance Semester 1, 2014-15 30 / 48Continuous Probability Distributions Example (Gaussian Distribution) We say that X has the Gaussian (normal) distribution with mean m2R and variance s 2 0 if its pdf equals f(x)= 1 p 2ps 2 e (x m) 2 2s 2 for x2R. We write X N(m,s 2 ). One can show that Z f(x)dx = Z 1 p 2ps 2 e (x m) 2 2s 2 dx = 1. We have E P (X
