on 12-Aug-2017 (Sat)

In calculations of variance we do not know whether large deviations are likely to be positive or negative, hence the degree of symmetry in return distributions.

If we know both the set of all the distinct possible outcomes of a random variable and the assignment of probabilities to those outcomes—the probability distribution of the random variable—we have a complete description of the random variable, and we can assign a probability to any event that we might describe

The unconditional variance of EPS is the sum of two terms:

1) the expected value (probability-weighted average) of the conditional variances (parallel to the total probability rules) and

2) the variance of conditional expected values of EPS.

The second term arises because the variability in conditional expected value is a source of risk. Term 1 is σ²(EPS) = P(declining interest rate environment) σ²(EPS | declining interest rate environment) + P(stable interest rate environment) σ²(EPS | stable interest rate environment) = 0.60(0.004219) + 0.40(0.0096) = 0.006371.

Term 2 is σ²[E(EPS | interest rate environment)] = 0.60($2.4875 − $2.34)² + 0.40($2.12 − $2.34)² = 0.032414. Summing the two terms, unconditional variance equals 0.006371 + 0.032414 = 0.038785.

Regarding counting, there can never be more combinations than permutations for the same problem, because permutations take into account all possible orderings of items, whereas combinations do not.

For if it is true that no production of knowledge in the human sciences can ever ignore or disclaim its author's involvement as a human subject in his own circumstances, then it must also be true that for a European or American studying the Orient there can be no disclaiming the main· circumstances of his actuality: that he comes up against the Orient as a European or American first, as an individual second.

A number of experiments have demonstrated the effectiveness of the KWM. One of the most common research designs has been to test the KWM against other methods of learning vocabulary. The KWM has been compared with learning words in context (e.g., immersion) or learning words with no given strategy. For example, McDaniel and Pressley (1989) found the KWM to be significantly more facilitative to learning over the context method. It has also been shown that the KWM is superior over systematic teaching (King-Sears et al., 1992). Additionally, this method has been shown to be an effective tool in learning many words in a short amount of time. Atkinson and Raugh (1975) found that students were able to learn 120 new words in three days (40 words a day). In sum, a good deal of research has shown that the KWM of vocabulary learning is highly effective.

In all the problems to which we apply n! method, we must use up all the members of a group (sampling without replacement).

The empirical rule can be broken down into three parts: 68% of data falls within the first standard deviation from the mean. 95% fall within two standard deviations.

the sample mean presents a central value for a particular set of observations as an equally weighted average of those observations.

To summarize, the contrast is forecast versus historical, or population versus sample.

If we have the probability distribution of a random variable, we can assign a probability to any event that we might describe.

A mutual fund guide ranked 18 bond mutual funds by total returns for the year 2014. The guide also assigned each fund one of five risk labels: high risk (four funds), above-average risk (four funds), average risk (three funds), below-average risk (four funds), and low risk (three funds); as 4 + 4 + 3 + 4 + 3 = 18, all the funds are accounted for. How many different ways can we take 18 mutual funds and label 4 of them high risk, 4 above-average risk, 3 average risk, 4 below-average risk, and 3 low risk, so that each fund is labeled?

The answer is close to 13 billion. We can label any of 18 funds high risk (the first slot), then any of 17 remaining funds, then any of 16 remaining funds, then any of 15 remaining funds (now we have 4 funds in the high risk group); then we can label any of 14 remaining funds above-average risk, then any of 13 remaining funds, and so forth. There are 18! possible sequences. However, order of assignment within a category does not matter. For example, whether a fund occupies the first or third slot of the four funds labeled high risk, the fund has the same label (high risk). Thus there are 4! ways to assign a given group of four funds to the four high risk slots. Making the same argument for the other categories, in total there are (4!)(4!)(3!)(4!)(3!) equivalent sequences. To eliminate such redundancies from the 18! total, we divide 18! by (4!)(4!)(3!)(4!)(3!). We have 18!/(4!)(4!)(3!)(4!)(3!) = 18!/(24)(24)(6)(24)(6) = 12,864,852,000. This procedure generalizes as follows.

Using the notation in the formula below, the number of objects with the first label is r = n₁ and the number with the second label is n − r = n₂ (there are just two categories, so n₁ + n₂ = n). Here is the formula:

• Combination Formula (Binomial Formula). The number of ways that we can choose r objects from a total of n objects, when the order in which the r objects are listed does not matter, is

  \(_nC_r=(\frac{n}{r})= \frac{n!}{(n-r)!r!}\)
  
  Here _nC_r and (nr) are shorthand notations for n!/(n − r)!r! (read: n choose r, or n combination r).

If we label the r objects as belongs to the group and the remaining objects as does not belong to the group, whatever the group of interest, the combination formula tells us how many ways we can select a group of size r. We can illustrate this formula with the binomial option pricing model. This model describes the movement of the underlying asset as a series of moves, price up (U) or price down (D). For example, two sequences of five moves containing three up moves, such as UUUDD and UDUUD, result in the same final stock price. At least for an option with a payoff dependent on final stock price, the number but not the order of up moves in a sequence matters. How many sequences of five moves belong to the group with three up moves? The answer is 10, calculated using the combination formula (“5 choose 3”):

₅C₃ = 5! /(5−3)!3! =(5)(4)(3)(2)(1)/(2)(1)(3)(2)(1) = 120 / 12 = 10 ways

Answering the following questions may help you apply the counting methods we have presented in this section.

1. Does the task that I want to measure have a finite number of possible outcomes? If the answer is yes, you may be able to use a tool in this section, and you can go to the second question. If the answer is no, the number of outcomes is infinite, and the tools in this section do not apply.

2. Do I want to assign every member of a group of size n to one of n slots (or tasks)? If the answer is yes, use n factorial. If the answer is no, go to the third question.

3. Do I want to count the number of ways to apply one of three or more labels to each member of a group? If the answer is yes, use the multinomial formula. If the answer is no, go to the fourth question.

4. Do I want to count the number of ways that I can choose r objects from a total of n, when the order in which I list the r objects does not matter (can I give the r objects a label)? If the answer to these questions is yes, the combination formula applies. If the answer is no, go to the fifth question.

5. Do I want to count the number of ways I can choose r objects from a total of n, when the order in which I list the r objects is important? If the answer is yes, the permutation formula applies. If the answer is no, go to question 6.

6. Can the multiplication rule of counting be used? If it cannot, you may have to count the possibilities one by one, or use more advanced techniques than those presented here.23

Does the task that I want to measure have a finite number of possible outcomes? If the answer is yes, you may be able to use a tool in this section, and you can go to the second question. If the answer is no, the number of outcomes is infinite, and the tools in this section do not apply.

Can the multiplication rule of counting be used? If it cannot, you may have to count the possibilities one by one, or use more advanced techniques than those presented here

In this reading, we have discussed the essential concepts and tools of probability. We have applied probability, expected value, and variance to a range of investment problems.

• A random variable is a quantity whose outcome is uncertain.

• Probability is a number between 0 and 1 that describes the chance that a stated event will occur.

• An event is a specified set of outcomes of a random variable.

• Mutually exclusive events can occur only one at a time. Exhaustive events cover or contain all possible outcomes.

• The two defining properties of a probability are, first, that 0 ≤ P(E) ≤ 1 (where P(E) denotes the probability of an event E), and second, that the sum of the probabilities of any set of mutually exclusive and exhaustive events equals 1.

• A probability estimated from data as a relative frequency of occurrence is an empirical probability. A probability drawing on personal or subjective judgment is a subjective probability. A probability obtained based on logical analysis is an a priori probability.

• A probability of an event E, P(E), can be stated as odds for E = P(E)/[1 − P(E)] or odds against E = [1 − P(E)]/P(E).

• Probabilities that are inconsistent create profit opportunities, according to the Dutch Book Theorem.

• A probability of an event not conditioned on another event is an unconditional probability. The unconditional probability of an event A is denoted P(A). Unconditional probabilities are also called marginal probabilities.

• A probability of an event given (conditioned on) another event is a conditional probability. The probability of an event A given an event B is denoted P(A | B).

• The probability of both A and B occurring is the joint probability of A and B, denoted P(AB).

• P(A | B) = P(AB)/P(B), P(B) ≠ 0.

• The multiplication rule for probabilities is P(AB) = P(A | B)P(B).

• The probability that A or B occurs, or both occur, is denoted by P(A or B).

• The addition rule for probabilities is P(A or B) = P(A) + P(B) − P(AB).

• When events are independent, the occurrence of one event does not affect the probability of occurrence of the other event. Otherwise, the events are dependent.

• The multiplication rule for independent events states that if A and B are independent events, P(AB) = P(A)P(B). The rule generalizes in similar fashion to more than two events.

• According to the total probability rule, if S₁, S₂, …, S_n are mutually exclusive and exhaustive scenarios or events, then P(A) = P(A | S₁)P(S₁) + P(A | S₂)P(S₂) + … + P(A | S_n)P(S_n).

• The expected value of a random variable is a probability-weighted average of the possible outcomes of the random variable. For a random variable X, the expected value of X is denoted E(X).

• The total probability rule for expected value states that E(X) = E(X | S₁)P(S₁) + E(X | S₂)P(S₂) + … + E(X

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A random variable is a quantity whose outcome is uncertain.

Probability is a number between 0 and 1 that describes the chance that a stated event will occur.

An event is a specified set of outcomes of a random variable.

Mutually exclusive events can occur only one at a time. Exhaustive events cover or contain all possible outcomes.

The two defining properties of a probability are, first, that 0 ≤ P(E) ≤ 1 (where P(E) denotes the probability of an event E), and second, that the sum of the probabilities of any set of mutually exclusive and exhaustive events equals 1.

A probability estimated from data as a relative frequency of occurrence is an empirical probability. A probability drawing on personal or subjective judgment is a subjective probability. A probability obtained based on logical analysis is an a priori probability.

A probability of an event E, P(E), can be stated as odds for E = P(E)/[1 − P(E)] or odds against E = [1 − P(E)]/P(E).

Probabilities that are inconsistent create profit opportunities, according to the Dutch Book Theorem.

A probability of an event not conditioned on another event is an unconditional probability. The unconditional probability of an event A is denoted P(A). Unconditional probabilities are also called marginal probabilities.

A probability of an event given (conditioned on) another event is a conditional probability. The probability of an event A given an event B is denoted P(A | B).

The probability of both A and B occurring is the joint probability of A and B, denoted P(AB).

P(A | B) = P(AB)/P(B), P(B) ≠ 0.

The multiplication rule for probabilities is P(AB) = P(A | B)P(B).

The probability that A or B occurs, or both occur, is denoted by P(A or B).

The addition rule for probabilities is P(A or B) = P(A) + P(B) − P(AB).

When events are independent, the occurrence of one event does not affect the probability of occurrence of the other event. Otherwise, the events are dependent.

The multiplication rule for independent events states that if A and B are independent events, P(AB) = P(A)P(B). The rule generalizes in similar fashion to more than two events.

According to the total probability rule, if S₁, S₂, …, S_n are mutually exclusive and exhaustive scenarios or events, then P(A) = P(A | S₁)P(S₁) + P(A | S₂)P(S₂) + … + P(A | S_n)P(S_n).

The expected value of a random variable is a probability-weighted average of the possible outcomes of the random variable. For a random variable X, the expected value of X is denoted E(X).

The total probability rule for expected value states that E(X) = E(X | S₁)P(S₁) + E(X | S₂)P(S₂) + … + E(X | S_n)P(S_n), where S₁, S₂, …, S_n are mutually exclusive and exhaustive scenarios or events.

The variance of a random variable is the expected value (the probability-weighted average) of squared deviations from the random variable’s expected value E(X): σ²(X) = E{[X − E(X)]²}, where σ²(X) stands for the variance of X.

Variance is a measure of dispersion about the mean. Increasing variance indicates increasing dispersion. Variance is measured in squared units of the original variable.

Standard deviation is the positive square root of variance. Standard deviation measures dispersion (as does variance), but it is measured in the same units as the variable.

Covariance is a measure of the co-movement between random variables

The covariance between two random variables R_i and R_j is the expected value of the cross-product of the deviations of the two random variables from their respective means: Cov(R_i,R_j) = E{[R_i − E(R_i)][R_j − E(R_j)]}. The covariance of a random variable with itself is its own variance.

Correlation is a number between −1 and +1 that measures the co-movement (linear association) between two random variables: ρ(R_i,R_j) = Cov(R_i,R_j)/[σ(R_i) σ(R_j)].

To calculate the variance of return on a portfolio of n assets, the inputs needed are the n expected returns on the individual assets, n variances of return on the individual assets, and n(n − 1)/2 distinct covariances.

Portfolio variance of return is \(σ^2(Rp) = \displaystyle\sum_{i=1}^{n}\displaystyle\sum_{j=1}^{n}w_iw_jCov (R_i,R_j)\)

The calculation of covariance in a forward-looking sense requires the specification of a joint probability function, which gives the probability of joint occurrences of values of the two random variables.

When two random variables are independent, the joint probability function is the product of the individual probability functions of the random variables.

Bayes’ formula is a method for updating probabilities based on new information.

Bayes’ formula is expressed as follows: Updated probability of event given the new information = [(Probability of the new information given event)/(Unconditional probability of the new information)] × Prior probability of event.

The multiplication rule of counting says, for example, that if the first step in a process can be done in 10 ways, the second step, given the first, can be done in 5 ways, and the third step, given the first two, can be done in 7 ways, then the steps can be carried out in (10)(5)(7) = 350 ways.

The number of ways to assign every member of a group of size n to n slots is n! = n (n − 1) (n − 2)(n − 3) … 1. (By convention, 0! = 1.)

The number of ways that n objects can be labeled with k different labels, with n₁ of the first type, n₂ of the second type, and so on, with n₁ + n₂ + … + n_k = n, is given by n!/(n₁!n₂! … n_k!). This expression is the multinomial formula.

A special case of the multinomial formula is the combination formula. The number of ways to choose r objects from a total of n objects, when the order in which the robjects are listed does not matter, is

nCr=(nr)=n!(n−r)!r!nCr=(nr)=n!(n−r)!r!

The number of ways to choose r objects from a total of n objects, when the order in which the r objects are listed does matter, is

nPr=n!(n−r)!nPr=n!(n−r)!