Net income of a business reflects residual income remaining after all [...] costs have been paid.
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the [...] between two random variables is the probability-weighted average of the cross-products of each random variable’s deviation from its own expected value.
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\(n!\over n1!n2!…nk!\) is called [...]
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General Formula for Labeling Problems
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The number of ways that n objects can be labeled with k different labels, with n_{1} of the first type, n_{2} of the second type, and so on, with n_{1} + n_{2} + … + n_{k} = n,
\(n!\over n1!n2!…nk!\)
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In calculations of variance we do not know whether large deviations are likely to be positive or negative, hence the [...] in return distributions.
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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.
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Using the notation in the formula below, the number of objects with the first label is r = n_{1} and the number with the second label is n − r = n_{2} (there are just two categories, so n_{1} + n_{2} = 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 _{n}C_{r} and (nr) are shorthand notations for n!/(n − r)!r! (read: n choose r, or n combination r).
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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”):
_{5}C_{3 }= 5! /(5−3)!3! =(5)(4)(3)(2)(1)/(2)(1)(3)(2)(1) = 120 / 12 = 10 ways
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Answering the following questions may help you apply the counting methods we have presented in this section.
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.
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.
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.
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.
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.
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
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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 S_{1}, S_{2}, …, S_{n} are mutually exclusive and exhaustive scenarios or events, then P(A) = P(A | S_{1})P(S_{1}) + P(A | S_{2})P(S_{2}) + … + 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 | S_{1})P(S_{1}) + E(X | S_{2})P(S_{2}) + … + E(X
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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.
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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!
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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)!
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