1、Investment Tools Probability,SASF CFA Quant. Review,天馬行空官方博客： ；QQ:1318241189；QQ群：175569632,2,Probability,A random variable is a quantity whose outcome is uncertain. Two defining properties of Probability. Probability of any event E is a number between 0 and 1, p(E). Sum of the probabilities of any l

2、ist of mutually exclusive and exhaustive events equals 1. Mutually exclusive = one and only one event can occur at any time. Exhaustive = one of the events must occur, jointly cover all possible outcomes. Empirical probability - probability of an event occurring is estimated from data, usually in th

3、e form of a relative frequency. A priori probability - probability of an event is deduced by reasoning about the structure of the problem itself. Subjective probability - probability of an event is based on a personal assessment without reference to any particular data.,3,Visualizing Sample Space,1.

4、Listing S = Head, Tail 2.Contingency Table 3.Decision Tree Diagram,4,2,nd,Coin,1,st,Coin,Head,Tail,Total,Head,HH,HT,HH, HT,Tail,TH,TT,TH, TT,Total,HH,TH,HT,TT,S,Contingency Table,Experiment: Toss 2 Coins. Note Faces.,5,Tree Diagram,Outcome,S = HH, HT, TH, TT,Sample Space,Experiment: Toss 2 Coins. No

5、te Faces.,T,H,T,H,T,HH,HT,TH,TT,H,6,Second Die,First Die,Probabilities for Two Dice,7,Forming Compound Events,1.Intersection Outcomes in Both Events A and B AND Statement Symbol (i.e., A B) 2.Union Outcomes in Either Events A or B or Both OR Statement Symbol (i.e., A B),8,Answer: Count the combinati

6、ons that yield 7 and add the individual probabilities to find the probability of the event, A, “Rolling a seven” P(A) = P(A B1) + P(A B2) + P(AB3) + P(A B6) = 6/36 = 1/6,Second Die,First Die,Joint Probabilities for Two Dice,What is the Probability of throwing a Seven with the two Dice?,9,Addition Ru

7、le,1. Used to Get Compound Probabilities for Union of Events: P(A OR B) = P(A B) = P(A) + P(B) - P(A B) 2. For Mutually Exclusive Events: P(A OR B) = P(A B) = P(A) + P(B),10,Answer: Count the combinations that have one die with 1 and add the individual probabilities OR use the Addition Rule for prob

8、abilities. P(AB) = P(A) + P(B) - P(A B) = 1/6 + 1/6 -1/36 = 11/36,Second Die,First Die,Addition Rule for Probabilities,What is the Probability of throwing at least one die showing 1 with the two Dice?,11,Answer: Count the combinations that have one die with 4 or greater & add the individual probabil

9、ities OR use the Addition Rule for probabilities. P(AB) = P(A) + P(B) - P(A B) = 1/2 + 1/2 -1/4 = 3/4,Second Die,First Die,Addition Rule - Again,What is the Probability of throwing a four or higher on either the first or second die?,12,Conditional Probability,Conditional Probability: Probability tha

10、t one Event will occur Given that Another Event has already occurred. To calculate a Conditional probability Revise Original Sample Space to Account for New Information Eliminates Certain Outcomes Result: P(A | B) = P(A and B) P(B),13,Second Die,First Die,Conditional Probability,What is the Probabil

11、ity of throwing at least an 8 if the first die is a 4?,Answer: Note we know we are in the row associated with a 4 for the first die. There are only 3 ways of the possible six outcomes for the second die that yield a sum 8. Hence P(A|B) = = P(AB)/P(B) = (3/36)/(6/36) =.5,14,1.Event Occurrence Does No

12、t Affect Probability of Another Event Toss first die, what does its outcome tell you about the likely results in tossing the second die? Answer: Nothing. The two die are independent 2.Causality Not Implied 3.Tests For Statistical Independence P(A | B) = P(A) P(A and B) = P(A)*P(B),Statistical Indepe

13、ndence,15,Multiplication Rule,1.Used to Get Compound Probabilities for Intersection of Events Called Joint Events 2.P(A and B) = P(A B)= P(A)*P(B|A) = P(B)*P(A|B) 3. For Independent Events:P(A and B) = P(A B) = P(A)*P(B),16,Answer: The result on the first die is independent of the second die. We use

14、 The multiplication rule to find prob. of A and B Hence P(A B) = = P(A)*P(B) = (1/6)*(1/6) = (1/36),Second Die,First Die,Multiplication Rule with Independence,What is the Probability of throwing “snake eyes” i.e. a 1 on each die?,17,Discrete R.V.s - Summary Measures,Expected Value Mean of Probabilit

15、y Distribution Weighted Average of All Possible Values X = E(X) = Xi P(Xi) = p(X1) X1+ p(X2) X2+ + p(Xn) Xn Variance Weighted Average Squared Deviation about Mean X2 = E (Xi-X(Xi-XP(Xi) = p(X1)(X1- X)2 + p(X2 )(X2- X)2 + + p(Xn) )(Xn- X)2 Standard deviation Square root of the variance. Measure of ri

16、sk shows dispersion of possible outcomes around expected level of outcomes.,18,Mean/Variance Calculation Table,X,i,P(X,i,),X,i,P(X,i,),X,i,-,(X,i,-,),2,(,X,i,-,),2,P(,X,i,),Total,X,i,P(X,i,),(,X,i,-,),2,P(,X,i,),19,Asset Expected Return & Risk,Calculate the Expected Returns of alternative assets. Rk

17、i= Return to Asset k if Event i occurs P(Rki) = Probability Event i occurs.,20,Calculating ER and Risk,Can use the previous table to calculate ER and variance (risk) for each asset. Show calculations for Asset A below.,A = 3.56%,ERA = 9.1%,ERA/A = 2.55,21,Covariance,Covariance between two random var

18、iables X and Y is defined as A negative covariance between X and Y means that when X is above its mean its is likely that Y is below its mean value. If the covariance of the two random variables is zero then on average the values of the two variables are unrelated. A positive covariance between X an

19、d Y means that when X is above its mean its is likely that Y is above its mean value. Covariance of a random variable with itself, its own covariance, is equal to its variance.,22,Correlation,Correlation between two random variables X and Y measured as: Correlation takes on values between 1 and +1.

20、Correlation is a standardized measure of how two random variables move together, i.e. correlation has no units associated with it. A correlation of 0 means there is no straight-line (linear) relationship between the two variables. Increasingly positive (negative) correlations indicate an increasingl

21、y strong positive (negative) linear relationship between the variables. When the correlation equals 1 (-1) there is a perfect positive (negative) linear relationship between the two variables.,23,Portfolio Probability Calculations,Portfolio consisting of two assets A and B, wA invested in A. Asset A

22、 has expected return rA and variance s2A. Asset B has expected return rB and variance s2B. The correlation between the two returns is rAB. Portfolio Expected Return: E(rp) = wA rA + (1-wA )rB Portfolio Variance: or,24,Exam Questions on Probability,28.The probability that two or more events will happ

23、en concurrently is: A.joint probability. B.multiple probability. C.concurrent probability. D.conditional probability.,25,Exam Question on Expected Values,24.An analyst developed the following probability distribution of the rate of return for a common stock: Scenario ProbabilityRate of return 10.25

24、0.08 20.50 0.12 30.25 0.16 The standard deviation of the rate of return is closest to: A.0.0200. B.0.0267. C.0.0283. D.0.0400.,To calculate standard deviation must first calculate expected value, mean, of the rate of return. = .25 x .08 + .5 x .12 + .25 x .16 = 0.12 Calculate variance as probability

25、-weighted squared deviations of values from expected value. s2 = .25.08 - .122 + .5.12 - .122 + .25.16 - .122 = 0.0008 Finally calculate standard deviation as square root of variance. s = SQRT(0.0008) = 0.02828,Common Probability Distributions,27,Probability Distributions,Probability distribution sp

26、ecifies the probabilities of the possible outcomes of a random variable. Discrete random variable can take on at most a countable number of possible values, such as coin flip or rolling dice. Continuous random variable can take on an uncountable (infinite) number of possible values, such as asset re

27、turns or temperatures. Probability function specifies the probability that the random variable takes on a specific value: P(X = x). Two Key Properties of a Probability Function. 0 p(x) 1 because a probability lies between 0 and 1. The sum of probabilities p(x) over all values of X equals 1.,28,Discr

28、ete Probability Distributions,Discrete Uniform Random Variable: The uniform random variable X takes on a finite number of values, k, and each value has the same probability of occurring, i.e. P(xi) = 1/k for i = 1,2,k. Bernoulli random variable is a binary variable that takes on one of two values, u

29、sually 1 for success or 0 for failure. Think of a single coin flip as an example of a Bernoulli r.v. Binomial random variable: X B(n, p) is defined is the number of successes in n Bernoulli random trials where p is the probability of success on any one Bernoulli trial. For 4 = X B(10, 0.5) think “wh

30、ats probability of 4 heads in ten coin flips?” Probability distribution for a Binomial random variable is given by:,29,Uniform Distribution,Population Distribution,Summary Measures,.0,.1,.2,.3,1,2,3,4,Example of distribution of a uniform random variable.,30,.0,.1,.2,.3,0,2,4,6,8,10,X,P(X),Binomial D

31、istn Coin Flips,N = 10 p = .5,Probability of exactly 4 heads in 10 tosses = Bar Height,31,Binomial Probability Distn Function,P(X= x | n,p) = probability that X = x given n & p n=sample size =3 in previous example p=probability of success Pr(Price up)=.75 x=number of successes in sample (X = 0, 1, 2

32、, ., n),32,Binomial Distn - Characteristics,n = 5 p = 0.1,n = 5 p = 0.5,Mean,Standard Deviation,x,x,E,X,np,np,p,(,),(,),1,.0,.2,.4,.6,0,1,2,3,4,5,X,P(X),.0,.2,.4,.6,0,1,2,3,4,5,X,P(X),33,Week 1,Week 2,Week 3,Week 4,S=$100,Assuming:,1. Share Price goes up or down by $10 each week.,2. Probability that

33、 Share Price goes down by $10 is 25% each week. i.e. pdown = .25,3. Probability that Share Price goes up by $10 is 75% each week. pup = .75,4. Probability that Share Price goes up this week is NOT affected by what happened before.,P = .75,Evolution of Share Price,34,Normal Distribution,Normal distri

34、bution is a continuous, symmetric probability distribution that is completely described by two parameters: its mean, , and its variance, 2. General Normal random variable X N(, 2) The normal distribution is said to be bell-shaped with the mean showing its central location and the variance showing it

35、s “spread”. A linear combination of two or more Normal random variables is also normally distributed. Standard Normal distribution Z N(0, 1). is a Normal distribution with mean =0, and variance 2=1.,35,Effect of Varying Parameters (x & x),X,f(X),C,A,B,mA mC, sA = sC,mA = mB, sA sB,36,Normal Distribu

36、tion,General Normal random variable X N(, 2) X can be standardized to a Standard Normal random variable. Resulting variable has mean zero and variance equal to 1. Calculating probabilities for a normal random variable X N(, 2) taking on a range of specified values, say a X b, directly as the area un

37、der the normal curve using the cumulative Normal distribution function as: N(a X b| , 2) = N(X b| , 2) - N( X a| , 2) . You should be able to show what this looks like using a diagram of the Normal distribution.,37,Confidence Intervals,90% level,95% level,99% level,x+1.65x x+2.58x,X,x+1.96x,x-2.58x

38、x-1.65x,x-1.96x,x,X= x ZX,38,Questions on Probability Distributions,30. A normal distribution would least likely be described as: A.asymptotic. B.a discrete probability distribution. C.a symmetrical or bell-shaped distribution. D.a curve that theoretically extends from negative infinity to positive

39、infinity. 31.An investment strategy has an expected return of 12 percent and a standard deviation of 10 percent. If investment returns are normally distributed, the probability of earning a return less than 2 percent is closest to: A.10%. B.16%. C.32%. D.34%.,2% is 1 st. dev. (s=10%) away from the m

40、ean of 12%. We know that for a Normal distribution there is a 68% chance of being within 1 st. dev. of the mean. We are 1 st.dev below, so there is (1-.68)/2 = 16% chance of this occurring.,39,Question on Confidence Intervals,32.Based on a normal distribution with a mean of 500 and a standard deviat

41、ion of 150, the Z-value for an observation of 200 is closest to: A.-2.00. B.-1.75. C. 1.75. D. 2.00. 34.If the standard deviation of a population is 100 and a sample size taken from that population is 64, the standard error of the sample means is closest to: A.0.08. B.1.56. C.6.40. D.12.50.,40,Hypothesis Testing Process,Is,X,Population,41,Null Hypot