The probabilities P(I) and P(IC) are also called the priors. Bayes' formula now allows us to compute P(I given O) where this is the updated probability given new information about "I." We recall the following formulas:

Conditional probability: P(I given O) = P(IO) / P(O)

Joint probability: P(IO) = P(O given I) * P(I)

For this case, Bayes' formula becomes:

P(I given O) = {P(O given I) / P(O)} * P(I)

Use the total probability rule to compute P(O):

P(O) = P(O given I) * P(I) + P(O given IC) * P(IC)

P(O) = 0.6 * 0.3 + 0.2 * 0.7 = 0.32

P(I given O) = {.6 / .32} * .3 = .5625

If the new information of expand overseas is announced, the prior probability of P(I) = 0.30 must be updated with the new information to give P(I given O) = .5625.

u: Calculate the number of ways a specified number of tasks can be performed using the multiplication rule of counting.

The multiplication rule of counting applies when we have a list of k choices, and each choice "i" has ni ways of being chosen. The number of total choices is n1 * n2 *...*nk.

Example: An analyst is interested in whether a firm raises prices, lowers prices, or keeps prices the same. The analyst also is interested in whether the firm expands overseas. Each of these represent separate choices that can occur in different ways: n1 = 3 and n2 = 2. This gives n * n2 = 2*3 = 6 possible ways of arranging the choices. The list of possible pairs of choices is: (raise prices, expand), (raise prices, not expand), (lower price, expand), (lower prices, not expand), (keep prices the same, expand), (keep prices the same, not expand).

If a third item is up for consideration for which there are n3 choices, then there will be n1 * n2 * n3 ways of arranging the items.

v: Solve counting problems using the factorial, combination, and permutation notations.

Labeling is where there are n items of which each can receive one of k labels. The number of items that receive label "1" is n1 and the number that receive label "2" is n2, etc. Also, the following relationship holds:

n1 + n2 + n3 +...+nk = n.

The number of ways labels can be assigned is:

n! / (n1!) * (n2!) *...*(nk!).

On your TI financial calculator, factorial is [2nd], [x!].

Example: A portfolio consists of eight stocks. The goal is to designate four of the stocks as "long-term holds," designate three of the stocks as "short-term holds," and designate one stock a "sell." How many ways can these labels be assigned to the eight stocks? The answer is (8!) / (4!*3!*1!) = (40,320) / (24*6*1) = 280.

w: Distinguish between problems for which different counting methods are appropriate.

The multiplication rule of counting is used when there are two or more groups. The key is that we choose only one item from each group.

Factorial is used by itself when there are no groups - we are only arranging a given set of n items. Given n items, there are n! ways of arranging them.

The labeling formula applies to three or more sub-groups of predetermined size. Each element of the entire group must be assigned a place or label in one of the three or more sub-groups.

The combination formula applies to only two groups of predetermined size. Look for the word "choose" or "combination."

The permutation formula applies to only two groups of predetermined size. Look for a specific reference to "order" being important.

x: Calculate the number of ways to choose r objects from a total of n objects, when the order in which the r objects is listed does or does not matter.

A special case of this labeling is when k = 2. That is that n items can only be in one of two groups and n1 + n2 = n. In that case, we can let r = n1 and n2 = n - r. Since there are only two categories, we usually talk about choosing r items. Then (n - r) are not chosen. The general formula for labeling when k = 2 is called the combination formula (or binomial formula):

n! / (n - r)! * r!

Example: In a portfolio of eight stocks, we decide to sell three stocks. How many ways can we choose three of the eight to sell? The answer is 8! / (5! * 3!) = 56.

1.D: Common Probability Distributions

a: Explain a probability distribution.

A probability distribution specifies the probabilities of all the possible outcomes of a random variable. Those outcomes may be discrete or continuous.

b: Distinguish between and give examples of discrete and continous random variables.

A discrete random variable is one where we can list all the possible outcomes and for each possible outcome there is a measurable, positive probability. An example of a discrete random variable is the number of days it rains in a given month. A continuous distribution would define the probabilities for the actual amount of rainfall. A continuous random variable is one where we cannot list the possible outcomes because we can always list a third number between any two numbers on the list. The number of outcomes is essentially infinite even if lower and upper bounds exist. The number of points between the lower and upper bounds are essentially infinite.

c: Describe the range of possible outcomes of a specified random variable.

In the discrete case, such as the number of days of rain in a month, there are a finite number of outcomes as defined by the number of days in the month. In the continuous case, such as the amount of rainfall, the outcome can be recorded out to many decimal places. We say the probability of two inches of rainfall is essentially zero because it is a single point. We must talk about the probability of the amount of rain being between two and three inches. In other words:

For a discrete distribution p(x) = 0 when "x" cannot occur, or p(x) > 0 if it can.

For a continuous distribution p(x) = 0 even though "x" can occur, but we can only consider p(x1≤ X ≤ x2), where x1 and x2 are actual numbers.

d: Define a probabiliry function and state whether a given function satisfies the conditions for a probability function.

The probability function specifies the probability that the random variable takes on a specific value.

Example: The following is a probability function:

For X: (1, 2, 3, 4), p(x) = x / 10, else p(x) = 0.

The probabilities are p(1) = 0.1, p(2) = 0.2, p(x) = 0.3, and p(x) = 0.4, all of which are between zero and one. Also, 0.1 + 0.2 + 0.3 + 0.4 = 1.

e: State the two key properties of a probability function.

The two key properties of a probability function are:

0 ≤ p(x) ≤ 1

The sum of all the probabilities for mutually exclusive and exhaustive outcomes must equal one.

f: Define a cumulative distribution function and calculate probabilities for a random variable, given a cumulative distribution function.

The cumulative distribution function or just "distribution function" defines the probability that a random variable takes a value equal to or less than a given number: P(X ≤ x) or F(X). Using the probability function defined earlier: X:(1, 2, 3, 4), p(x) = x / 10, F(1) = 0.1, F(2) = 0.3, F(3) = 0.6, F(4) = 1. In other words, F(3) represents the cumulative probability that outcomes 1, 2, and 3 occur.

g: Define a discrete uniform random variable and calculate probabilities, given a discrete uniform probability distribution.

The discrete uniform random variable is one where the probabilities are equal for all possible outcomes. One example is X:(1,2, 3, 4, 5), p(x) = 0.2. In this case, the probabilities are equal for each possible outcome (20%). The probability of any one outcome is 0.2 and the probability of any "n" outcomes is n * 0.2. For example, p(2 ≤ X ≤ 4) = p(2) + p(3) + p(4) = 0.6, and F(2) = p(1) + p(2) = 0.4.

h: Define a binomial random variable and calculate probabilities, given a binomial probability distribution.

The binomial random variable is the number of "success" in a given number of "trial" where the outcome can either be "success" or "failure." The probability of success is constant for each trial, and the trials are independent. A trial is like a mini-experiment and the final outcome is the number of successes in the series of n-trials. Under these conditions, the probability of "x" success in "n" trials is calculated using the following formula:

p(x) = P(X = x) = [number of ways to choose x from n]px(1 - p)n - x

Where the term in brackets = n! / {(n - x)!x!}, and p = the probability of "success" on each trial.

i: Calculate the expected value and variance of a binomial random variable.

Expected value of X: E(X) = n * p

Variance of X: V(X) = n * p * (1 - p)

Example: Using empirical probabilities, for any given day the probability that the DJIA will increase is 0.67. We will assume that the only other outcome is that it decreases in a day. Hence, p(UP) = 0.67 and p(DOWN) = 0.33. We will assume that whether the DJIA increases in one day is independent of whether it decreases in anther day. What is the probability that the DJIA will increase three out of five days? What is the expected number of up days in a five-day period?

Here we define success as UP, so p = 0.67. The definition of success is critical to any binomial problem. The n items are the five days: n = 5. The number of successes we are computing the probability for is x = 3. The formula is:

p(3) = P(X = 3) = 5! / [(5 - 3)! * 3!] * (0.673) * (0.335 - 3)

p(3) = 10 * 0.301 * 0.109

p(3) = 0.328

Expected value of X: E(X I n = 5, p = 0.67) = 5 * 0.67 = 3.35

Variance of X: V(X) = n * p * (1 - p) = 5 * 0.67 * 0.33 = 1.106

j: Construct a binomial tree to describe stock price movement and calculate the expected terminal stock price.

Example: Assume the DJIA is at 10,000. It moves an average of 2 percent a day. Using the probabilities from our previous example in LOS 1.D.i, what is the expected value of the DJIA at the end of two days?

	Day 1 prob.	Day 1 values	Day 2 prob.	Day 2 values	p(x) * x
			P(UU) = 0.4489	DJIA = 10,404	4,670.36
	P(U) = 0.67	DJIA = 10,200
			P(UD) = 0.2211	DJIA = 9,996	2,210.12
10,000
			P(DU) = 0.2211	DJIA = 9,996	2,210.12
	P(D) = 0.33	DJIA = 9,800
			P(DD) = 0.1089	DJIA = 9,604	1,045.88
				Expected value =	10,136.46

Since the move in prices is 2% per day in either direction, the value of the DJIA after day one in the up state will be 10,000 * 1.02 = 10,200. After one day in the down state, the value will be 10,000 * (1 - .02) = 9,800. After two days in the up-up state, the value will be 10,000 * (1.02)2 = 10,404.

The probability of the up-up state occurring is (.67)2 = .4489. The probability of up-down is (.67)(.33) = .2211.

k: Describe the continuous uniform distribution and calculate probabilities, given a uniform probability distribution.

Note: We're pretty confident that there is a typo in your study guide here. LOS 1.D.k actually says to calculate probabilities using the binomial distribution, but we did that in LOS 1.D.h. Therefore, we will calculate probabilities using the uniform distribution here.

The continuous uniform distribution is defined over a range that spans between some lower limit "a" and some upper limit "b" which serve as the parameters of the distribution. Outcomes can only occur between a and b. Because the distribution is continuous, even if a < x < b, p(x) = 0. For a ≤ x1 ≤ x2 ≤ b,

P(X < a or X > b) = 0

P(x1 ≤ X ≤ x2) = (x2 - x1) / (b - a).

Example: Random variable X follows a continuous uniform distribution over 12 to 28, that is a = 12, and b = 28. The probability of an outcome between 15 and 25 is:

P(15 ≤ X ≤ 25) = (25 - 15) / (28 - 12)

= 10 / 16 = 0.625

l: Explain the key properties of the normal distribution.

The normal distribution has the following key properties:

It is completely described by its mean and variance, we write X ~ N(μ, σ2). In words, this says that "X" is normally distributed with mean μ and variance σ2.

Skewness = 0. This means the normal distribution is symmetric about its mean, so that P(X ≤ μ) = P(μ ≤ X) = 0.5, and mean = median = mode.

Kurtosis = 3; this is a measure of how "flat" the distribution is. Recall that excess kurtosis is measured relative to the number "3."

A linear combination of normally distributed random variables is also normally distributed.

The following general properties are also very important:

34% of the area falls between 0 and +1 standard deviations from the mean. So, 68% of the observations fall within +/- one standard deviation of the mean.

45% of the area falls between 0 and +1.65 standard deviations from the mean. So, 90% of the observations fall within +/- 1.65 standard deviations of the mean.

47.5% of the area falls between 0 and 1.96 standard deviations from the mean. So, 95% of the observations fall within +/- 1.96 standard deviations of the mean.

m: Construct confidence intervals for a normally distributed random variable.

Often we must use an approximation of µ and σ with the sample mean and sample standard deviation denoted as "X bar" and "s." These are point estimates. We often frame probability statements for a random variable using confidence intervals that are built around these point estimates. Some important examples of confidence intervals are below. We should note the similarity between these and the statements above using µ and σ.

P(X will be within X bar +/- 1.65 * s) = 90%. We say: The 90 percent confidence interval for X is X bar - 1.65 * s to x bar + 1.65 * s.

P(X will be within X bar +/- 1.96 * s) = 95%. We say: The 95 percent confidence interval for X is X bar - 1.96 * s to X bar + 1.96 * s.

P(X will be within X bar +/- 2.58 * s) = 99%. We say: The 99 percent confidence interval for X is X bar - 2.58 * s to X bar + 2.58 * s.

Example: Using a 20-year sample, the average return of a mutual fund has been 10.5 % per year with a standard deviation of 18 %. Using these two point estimates, what is the 95% confidence interval for the return next year?

10.5 - 1.96 * 18 or -24.78 to 10.5 + 1.96 * 18 = 45.78.

In symbols, P(-24.78 < return < 45.78) = 95%.

n: Define the standard normal distribution and explain how to standardize a random variable.

The standard normal distribution is a normal distribution that has been "normalized" so that it has a mean of zero and a standard deviation of one. To standardize an observation from any given normal curve, you must calculate the observation's Z-value. The Z-value tells you how far away the given observation is from the population mean in units of standard deviation. This is how we standardize a random variable.

Z = (observation - population mean) / standard deviation = (X - µ) / σ

Example: The EPS figures for a large group of firms are normally distributed with a mean of $6 and a standard deviation of $2. What is the Z-value given and EPS of $8, that is X = 8? How about X = 2?

If X = 8, then Z = (( X - µ) / σ = (8 - 6) / 2 = +1

If X = 2, then Z = ( X - µ) / σ = (2 - 6) / 2 = -2

The Z of +1.00 indicates that an EPS of $8 is one standard deviation above the mean, and a Z of -2.00 shows that the EPS of $2 is two standard deviations below the mean. Using the symmetric properties of the normal distribution, we can approximate that the probability of the EPS falling between $2 and $8 is 0.475 + 0.34 = 0.815.

o: Calculate probabilities using the standard normal probability distribution.

With the aid of a normal probability table, we can precisely compute the probability of a normally distributed random variable falling between any two values.

Example: The Z table tells us that F(2) = 0.9772, thus F(-2) = 1 - .09772 = 0.0228. This tells us that 0.0228 or 2.28% of the area falls below Z = -2 and an equal amount falls above Z = +2. Furthermore, P(-2 ≤ Z ≤ 2) = 1 - 0.0228 - 0.0228 = 0.9544. Another way to do this is to write:

P(-2 ≤ Z ≤ 2) = F(2) - F(2) = 0.9772 - 0.0228 = 0.9544.

p: Distinguish between a univariate and a multivariate distribution.

A multivariate distribution specifies the probabilities for a group of random variables. It is meaningful when the random variable in the group are not independent. A joint probability table describes the multivariate distribution between two discrete random variables. Multivariate relationships can exist between two or more continuous random variables, e.g., inflation, unemployment, interest rates, and exchange rates.

A multivariate normal distribution applies when all the random variables have a normal distribution. As mentioned earlier, one of the characteristics of a normal distribution is that a linear combination of normal random variables is normal as well. For example, if the return of each stock in a portfolio is normally distributed, then the return on the portfolio will be normally distributed.

q: Explain the role of correlation in the multivariate normal distribution.

The relationships between the individual returns can be succinctly and completely described with a specific set of parameters. Just like a univariate normal distribution is completely described by its mean and variance, the multivariate normal distribution for two random variables is completely described by their means, variances, and the correlation between the two. More generally, and using the stock returns as an example, the multivariate normal distribution for the returns on n stocks is completely defined by the following three sets of parameters.

The n means of the n series of returns (µ1, µ2,..., µn).

The n variances of the n series of returns (σ1, σ2, ...,σn).

The (n * (n - 1)) / 2 pair-wise correlations.

If there are only two random variables, n = 2, then 2 * (2 - 1) / 2 = 1 and there is only one correlation. If there are three random variables, n = 3, then 3 * (3 - 1) / 2 = 3 and there are three correlations. The existence of correlation is the feature that distinguishes a multivariate normal distribution from a univariate normal distribution. These correlations indicate the strength of the linear relationships between each pair of variables.

r: Define shortfall risk.

Shortfall risk is the risk that a portfolio will fall below a particular value over a given horizon and is the focus of safety-first rules.

s: Select an optimal portfolio using Roy's safety-first criterion.

Roy's safety-first criterion states that the optimal portfolio minimizes the probability that the return of the portfolio falls below some minimum acceptable level. This minimum acceptable level is called the "threshold" level.

Rp = portfolio return

RL = threshold level return

Minimize P(Rp < RL)

SFRatio = (E[Rp] - RL /σp

Example: A college endowment is $120 million. Over the next year, the managers of the endowment plan to withdraw $1.6 million for operations and have set a minimum acceptable goal at the end of the year of $122 million for the endowment. They are considering three portfolios for the endowment with the following expected returns and variances:

Portfolio A: E(Rp) = 9%, σp = 12%

Portfolio B: E(Rp) = 11%, σp = 20%

Portfolio C: E(Rp) = 6.6%, σp = 8.2%

The threshold is RL = (1.6 + 122 - 120) / 120 = 0.030. The SFR's and corresponding F(-SFRatios) are below.

Portfolio A: SFRatio = (9 - 3) / 12 = 0.5 and F(-SFRatio) = 0.3085

Portfolio B: SFRatio = (11 - 3) / 20 = 0.4 and F(-SFRatio) = 0.3446

Portfolio C: SFRatio = (6.6 - 3) / 8.2 = 0.44 and F(-SFRatio) = 0.3300

The best choice is Portfolio A because its value for F(-SFRatio) = 0.3085 is the lowest of the three choices. Note, choosing the portfolio with the largest SFRatio gives the same result as choosing based on the smallest F(-SFRatio). Hence, you really don't have to look up -SFRatios in the normal table! Just pick the highest SFRatio.

t: Explain the relationship between the lognormal and normal distributions.

A lognormal distribution is simply the distribution of a random variable Y = eX, where X ~ N(μ, σ2) and e = 2.718, the base of the natural logarithms.

What you need to know for the exam:

The lognormal distribution is skewed to the right.

The lognormal distribution is bounded from below by zero.

The random variable Y = eX. Or alternatively, from the properties of logarithms, ln[Y] = X and X is normally distributed. To prove this, remember the following fun fact about logarithms: ln[eX] = X.

u: Distinguish between discretely and continuously compounded rates of return.

A discretely compounded rate of return is based on the relationship of the change of an asset's price and its starting price. It is denoted in the readings as "R." If the beginning price and ending price of an asset are S0 and S1 respectively, R = S1 / S0 - 1. Note that this is the same thing as the holding period return that we calculated in Chapter 3. The only difference is that there are no periodic cash flows considered here.

A continuously compounded rate of return measures the change over the average of all numbers between S0 and S1. This gives a useful perspective on the return earned over a period. This is done with the use of the natural logarithm (ln). Given a holding period return of R, the continuously compounded rate of return is:

ln(1 + R) = ln(S1 / S).

v: Calculate a continuously compounded return, given a specific holding period return.

Example: A stock increases from $100 to $110 during a year. From a discretely compounded return perspective, we would say that the return was R = ($110 / $100) - 1 = 0.10, and this is the holding period return. Yet, if the stock falls from $110 to $100, using the same technique, the return is R = ($100 / $110) - 1 = -0.0909. This could be misleading because the stock ends at the same value it started at $100m, but the average of the discretely compounded returns is (0.1000 + [-0.0909]) / 2 = 0.00455.

A continuously compounded return measures the nominal return with respect to the average of all values between the beginning and ending value: r = ln($110 / $100) = 0.0953 and it is also true that r = ln($100 / $110) = -0.0953. The fact that the stock ends at the same value it started at and (0.0953 - 0.0953) = 0 more accurately depicts the true movement of the stock.

w: Explain Monte Carlo simulation and describe its major implications.

Monte Carlo simulation is used to find approximate solutions to complex problems. The procedure usually uses a random number generator from a computer to generate sets of realized random variables from specified distributions. The results combine to form a new series of variables for which the true distributions are too mathematically complex to define. After generating a large number of sets of realizations, the statistics of the generated numbers can be used as estimates of the true parameters of the complex distribution.

Major applications:

Projecting the interaction of pension plan assets and liabilities.

Developing estimates of value at risk.

Estimating the potential success of a given trading strategy.

Estimating the distribution of the return of a portfolio composed of assets that do not have normally distributed returns.

Estimating the distribution of an asset that has features such as embedded options, call features, and parameters that change as market conditions change.

x: Explain historical simulation and describe its limitations.

Historical simulation uses historical data to generate the sets of realized random variables (as opposed to a random number generator as in Monte Carlo simulation).

Limitations:

Historical simulation cannot take into account the effect of significant events that did not occur during the sample period. For example, if a particular security only came into existence after 1987, we do not have observations for its behavior during a "market crash."

It cannot perform "what if" analysis. The source of the sample data is a fixed set, and we cannot investigate the effects of changing parameters in certain ways.

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