CFA Level 1 (2009) - 1

Study Session 2

Cross-Reference to CFA Institute Assigned Reading #8 - Probability Concepts

Substituting, we get:

Var(Rp)= 0.552 (0.0576) + 0.252 (0.0324) + 0.202 (0.0225) + 2(0.55)(0.25) (0.0367) + 2(0.55)(0.20)(0.0108) + 2(0.25)(0.20)(-0.0041)

= 0.0174 + 0.0020 + 0.0009 + 0.0101 + 0.0024 - 0.0004

= 0.0324

The portfolio standard deviation is JO.0324 == 0.1800, or 18%.

C.You can answer this question any number of different ways, but here is the most obvious:

Event 1: The realized rerum is greater than rhe expecred return.

Event 2: The realized rerurn is equal ro the expecred return.

Event 3: The realized rerum is less [han the expecred rerum.

D.Wirh six assers in rhe portfolio, rhere wil.1 be 6 variance rerms and 15 unique covariance terms. The covariance marrix will have 6 x 6 = 36 celIs. The covariance of

a srock return with irself is irs variance; those will be rhe six entries on the diagonal. The other 30 cells are the covariance pairs, bur since each pair appears rwice in rhe matrix-Cov(A,B) is rhe same as Cov(B,A)-rhe number of unique covariance pairs is half of that, or 15. For any porrfolio of n assers, rhe portfolio variance calcularion would involve II variance rerms and n(n - 1) / 2 unique covariance rerms.

6.The prior probability thar rhe builder will increase irs orders is 25%.

P(increase) =0.25

P(no increase) = 0.75

There are four possible outcomes:

Builder increases its orders and starts new development.

Builder increases irs orders and does nor srarr new development. Builder does nor increase its orders and srarrs new development.

Builder does nor increase irs orders and does nor srarr new development.

The probabilities of each ourcome are as folIows: P(increase and development) = (0.25)(0.70) = 0.175.

P{increase and no development) = (0.25)(0.30) = 0.075. P(no increase and development) = (0.75)(0.20) = 0.15. P(no increase and no development) =(0.75)(0.80) =0.60.

We want ro updare rhe probability of an increase in orders, given the new informarion that the builder is starting rhe development. We can apply Bayes' formula:

From our assumptions, P(development I increase) = 0.70, and P(increase) = 0.25, so the numerator is (0.70)(0.25)

Study Session 2

Cross-Reference to CPA Institute Assigned Reading #8 - Probability Concepts

P(developmem) is the sum of P(development and increase) and P(development and no increase).

P(developmcnt) = 0.175 + 0.15 = 0.325

Professor's Note: J can never remember this formuLa, so J set these probLems up

~Like the probability modeL (tree) in the notes and focus on the probabilities of the

~new information-deveLopment in this case-which J have put in bold. TotaL

probability ofdeveLopment is 17.5 + 15 = 32.5. Ofthat probabiLity, 17.5/32.5, or 53.85% of the time, deveLopment is paired with an increase in saLes!

The following is a review of the Quantitative Methods: Application principles designed to address the learning outcome statements set forth by CFA Institute@. This topic is also covered in:

COMMON PROBABILITY DISTRIBUTIONS

Study Session 3

EXAM Focus

LOS 9.a: Explain a probability distribution and distinguish between discrete and continuous random variables.

LOS 9.b: Describe the set of possible outcomes of a specified discrete random variable.

A probability distribution describes the probabilities of all the possible outcomes for a random variable. The probabilities of all possible outcomes must sum to 1. A simple

probability distribution is that for the roll of one fair die; there are six possible outcomes and each one has a probability of 1/6, so they sum to 1. The probability distribution

of all the possible returns on the S&P 500 index for the next year is a more complex version of the same idea.

A discrete random variable is one for which the number of possible outcomes can be counted, and for each possible outcome, there is a measurable and positive probability. An example of a discrete random variable is the number of days it rains in a given month, because there is a finite number of possible outcomes-the number of days it can rain in a month is defined by the number of days in the month.

A continuous random variable is one for which the number of possible outcomes is infinite, even if lower and upper bounds exist. The actual amount of daily rainfall between zero and 100 inches is an example of a continuous random variable because the actual amount of rainfall can take on an infinite number of values. Daily rainfall can be measured in inches, half inches, quarter inches, thousandths of inches, or in even smaller increments. Thus, the number of possible daily rainfall amounts between zero and 100 inches is essentially infinite.

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #9 - Common Probability Distributions

The assignment of probabilities to the possible outcomes for discrete and continuous random variables provides us with discrete probability distributions and continuous probability distributions. The difference between these types of distributions is most apparent for the following properties:

• For a discrete distribution, p(x) = 0 when x cannot occur, or p(x) > 0 if it can. Recall that p(x) is read: "the probability that random variable X = x." For example, the probability of it raining on 33 days in June is zero because this cannot occur, but the probability of it raining 25 days in June has some positive value.

•For a continuous distribution, p(x) = 0 even though x can occur. We can only consider P(x 1 :S X :S x2) where XI and x2 are actual numbers. For example, the probability of

receiving two inches of rain in June is zero because two inches is a single point in an infinite range of possible values. On the other hand, the probability of the amount of rain being between 1.99999999 and 2.00000001 inches has some positive value. In the case of continuous distributions, it is interesting to note that P(x I :S X :S x2) = P(x, < X < x2) because p(x 1) = O.

In fin:lllce, some discrete distributions are treated as though they are continuous because the number of possible outcomes is very large. For example, the increase or decrease in the price of a stock traded on an American exchange is recorded in dollars and cents. Yet, the probability of a change of exactly $1.33 or $1.34 or any other specific change

is almost zero. It is customary, therefore, to speak in terms of the probability of a range of possible price change, say between $1.00 and $2.00. In other words p(price change = 1.33) is essentially zero, but p($ j < price change < $2) > O.

LOS 9.c: Interpret a probability function, a probability density function, and a cumulative distribution function, and calculate and interpret probabilities for a random variable, given its cumulative distribution function.

A probability function, denoted p(x), specifies the probability that a random variable is equal to a specific value. More formally, p(x) is the probability that random variable X takes on the value x, or p(x) = P(X := x).

The two key properties of a probability function are:

• o :S p(x) :S 1.

•:Lp(x) = 1, the sum of the probabilities for all possible outcomes, x, for a random variable, X, equals 1.

Example: Evaluating a probability function

Consider the following function: X ={I, 2, 3, 4}, p(x) = ...:.., elsep(x) = 0

10 .

Determine whether .this function satisfies the conditions for a probability function.

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #9 - Common Probabiliry Distributions

Answer:

Note that all of the probabilities are be~~~rM~~ofall prObabilities·

A probability density function (pdf) is a function, denoted f(x), that can be used to generate the probability that outcomes of a continuous distribution lie within a

particular range of outcomes. For a continuous distribution, it is the equivalent of a probability function for a discrete distribution. Remember, for a continuous distribution the probability of anyone particular outcome (of the infinite possible outcomes) is zero.

A pdf is used to calculate the probability of an outcome between two values (i.e., the probability of the outcome falling within a specified range). How that is actually done (it involves using calculus to take [he integral of the function) is. thankfully. beyond the scope of the material required for the exam.

A cumulative distribution function (cdf), or simply distributioll{ulletion. defines the probability that a random variable, X. takes on a value equal to or less than a specific value, x. It represents the sum. or cumulative value, of the probabilities for the outcomes up to and including a specified outcome. The cumulative distribution function for random variable, X, may be expressed as F(x) = P(X ::; x). For example, consider

the probability function defined earlier for X = {1, 2, 3, 4}, p(x) = x I 10. For this distribution, F(3) = 0.6 = 0.1 + 0.2 + 0.3, and F(4) = 1 = 0.1 + 0.2 + 0.3 + 0.4. This means that F(3) is the cumulative probability that outcomes 1, 2, or 3 occur, and F(4) is the cumulative probability that one of the possible outcomes occurs.

Figure 1 shows an example of a cumulative distribution function (for a standard normal distribution, described later in this topic). There is an 84.13% probability of a value less than 1 and a 15.87% probability of a value less than -1. Because this distribution is

continuous, the cdf for any value x is the proportion of the total area under the curve for all values less than or equal to x.

Figure 1: Cumulative Distribution Function

F(-l.O)=O.1587

F(l.O)=O.8413

©2008 Kaplan Schweser

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #9 - Common Probability Distributions

Cumulative:Distribution Function for X ... Uniform {2, 4.,6, 8,IO} ..

0.60

0.40

0.20

0' --------------

2 4 6 8 10 12

The Binomial Distribution

A binomial random variable may be denned as the number of "successes" in a givcn numbcr of rrials, whereby thc outcome can be either "success" or "failure." The probability of success, p, is constanr for each trial, and the rrials are independent. (A binomial random variable for which the number of rrials is 1 is called a Bernoulli random variable.) Think of a trial as a mini-experimcnt (or "Bernoulli trial"). The nnal outcome is the number of successes in a series of II rrials. Under these conditions, the binomial probabili t)' function den nes the probability of x successes in n rrials. Ir can be expressed using the following formula:

p(x) = P(X = x) = (number of ways ro choose x from n)px(l _ p)n-x

where:

(: J 0< "'ted ;: ~:;~~o"x"

p = the probability of "success" on each trial (don't confuse it with p(x))

So the probability of exactly x successes in n trials is:

Example: Binomial probability

Assuming a binomial distribution, compute the probability of drawing three black beans from a bowl of black and white beans if the probability of selecting a black bean in any given attempt is 0.6. You will draw five beans from the bowl.

Answer:

P(X = 3) =p(3) =~(0.6)3(0.4)2=(120/12)(0.216)(0.160) = 0.3456

213!

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #9 - Common Probability Distributions

Some inruition about these results may help you remember the calculations. Consider that a (very large) bowl of black and white beans has 60% black beans and that each time you select a bean, you replace it in the bowl before drawing again. We wanr to know the probability of selecting exacrly three black beans in five draws, as in the above problem. .

One way this might happen is BBBWW. Since the draws are independenr, the probability of this is easy to calculate. The probability of drawing a black bean is 60% and the probability of drawing a white bean is 1 - 60% = 40%. Therefore, the probability of selecting BBBWW, in order is, 0.6 x 0.6 x 0.6 x 0.4 x 0.4 = 3.456%. This is the p3( 1 - p)2 from the formula and pis 60%, the probability of selecting a black bean on any single draw from the bowl. BBBWW is not, however, the only way to choose exactly three black beans in five trials. Another possibility is BBWWB, and a third is BWWBB. Each of these will have exacrly the same probability of occurring as our initial outcome, BBBWW. That's why we need to answer the question of how many ways (different orders) there are for us to choose three black beans in five draws. Using the

10 x 3.456% = 34.56%, the answer we computed above.

The Expected Value of a Binomial Random Variable

For a given series of n trials. the expected number of successes or E(A) is given by the following formula:

expected value of X = E(X) = np

The inruition is srraightforward; if we perform 17 trials and the probability of success on each trial is p, we expect 17p successes.

Example: Expected value of a binomial random "Variable

Based on empirical data, the probability that the Dow Jones Industrial Average (DJIA) will increase on any given day has been determined to equal 0.67. Assuming that the only other outcome is that irdecreases, we can state p(UP) = 0.67 and p(DOWl\I) =: 0.33. Further, assume that movements in the DJlA are independenr (i.e., an increase in one day is independenr of what happened on another day).

Using the information provided, compute the expected value of the number of up days in a 5-day period.

5wdy Session 3

Cross-Reference to CFA Institute Assigned Reading #9 - Common Probability Distributions

. "',':.1

,::.

A Binomial Tree to Describe Stock Price Movement

..\ bi!lumi,,; n\(I,1<:1 can be: applied lO srock rrice nW"','!T1CIH.I. \\'e: rust need t(J define rh,·

or decrease in ','aluc by 1(Yo (the only two possible ourcomes). The probabiliry of an up- move (the up transition probabilit;·, u') is p and the probabilit\' of it down-move (rhe

A binomial tree i, consrrucred 0;' showing; all rhe possible combinations of upmoves and down-moves over a number of successive periods. For cwo periods,these

combinarions are UU, UO, DU, and DO. Imporranrly, UD and DU result in the same srock price S aErer rwo periods since S (1. 01)(1/1.01) = S and the order of rhe moves does not change rhe resulr. Each of rhe possible values along a binomial tree is a node. Figure::' illustrales a binomiaJ cree for rhree periods,

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #9 - Common Probabiliry Disuibutions

Figure 2: A Binomial Tree

Since a srock price of 50 can result from eicher ud or du moves, the probabiliry of a stock price of 50 after cwo periods (the middle value) is 2 x (0.6)(0.4) = 48%. A binomial

cree wich S = '50. U = 1.1. and Prob(U) = 0,'" is illustrated in Figure 3. Note [h3.r [he middJe value after two periods (50) is equal (0 the beginning value. The probbilit)' that the stOck price is down I -:'i0)after [\Va periods i~ simply the probabilir}· of two down movemems, (1 - 0.7)2 = '1%,