Chapter 2 - Probability

ESSENTIAL MATH AND STATISTICS FOR FINANCE AND RISK MANAGEMENT

Alan Anderson, Ph.D. Western Connecticut State University ECI Risk Training http://www.ecirisktraining.com

CHAPTER 2: PROBABILITY THEORY Probability theory is based on the notion of a random experiment, which is a process that generates outcomes in such a way that: • •

all possible outcomes are known in advance the actual outcome is not known in advance

Some examples of a random experiment are: • • •

spinning a roulette wheel choosing a card from a deck rolling a pair of dice

EXAMPLE Suppose that a die is rolled twice; assume that the variable of interest is whether the number that turns up is odd or even. There are four possible outcomes of this random experiment: {even, even}, {even, odd}, {odd, even}, {odd, odd}. The set of all possible outcomes of a random experiment is known as the sample space. The sample space for this experiment consists four sample points: S = {EE, EO, OE, OO}. A subset of the sample space is known as an event. EXAMPLE Based on the die-rolling experiment, suppose that the event F is defined as follows: “at most one even number turns up”. The event F is a set containing the following sample points: F = {EO, OE, OO}. COMPUTING PROBABILITIES OF EVENTS For a random experiment in which there are n equally likely outcomes, each outcome has a probability of occurring equal to 1/n. In the die-rolling example, each of the four sample points is equally likely, giving it a probability of 1/4. The probability of event F is therefore: P(F) = P{EO} + P{OE} + P{OO} = 1/4 + 1/4 + 1/4 = 3/4 Alternatively, the probability of an event E can be computed as P(E) = #E/#S

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where:

#E = number of sample points in event E #S = number of sample points in the sample space

In this example, the probability of event F would be calculated as P(F) = #F/#S = 3/4. AXIOMS OF PROBABILITY An axiom is a logical statement that is assumed to be true without formal proof; all further results are derived using axioms as a starting point. In probability theory, the three fundamental axioms are: Axiom 1:

The probability of any event A is non-negative; P(A) ≥ 0

Axiom 2:

The probability of the entire sample space is one; P(S) = 1

Axiom 3:

For any collection of disjoint (non-overlapping) events: n

P(A1 ∪ A2 ∪ ... ∪ An ) = ∑ P(Ai ) i =1

RULES OF PROBABILITY The rules of probability are derived from these axioms. Three of the most important rules are the Addition Rule, Multiplication Rule and the Complement Rule. ADDITION RULE For two events A and B, the addition rule is: P(A ∪ B) = P(A) + P(B) – P(A ∩ B) EXAMPLE Referring to the die-rolling experiment, suppose that the following events are defined: A = “the first roll is even” B = “exactly one roll is odd What is the probability that the first roll is even or at exactly one roll is odd? Events A, B and A ∩ B contain the following sample points:

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A = {EE, EO} B = {EO, OE} A ∩ B = {EO} Since the sample space consists of four equally likely sample points, P(A) = 1/2 P(B) = 1/2 P(A ∩ B) = 1/4 P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 1/2 + 1/2 – 1/4 = 3/4 ADDITION RULE FOR MUTUALLY EXCLUSIVE (DISJOINT) EVENTS If events A and B cannot both occur at the same time, they are said to be mutually exclusive or disjoint; in this case, P(A ∩ B) = 0. Therefore, the addition rule for mutually exclusive events is: P(A ∪ B) = P(A) + P(B). CONDITIONAL PROBABILITY The conditional probability of an event is the probability that it occurs given that another event occurs. For two events A and B, the probability that B occurs given that A occurs is written as: P(B|A). This can be computed as:

P(B|A) = P(A ∩ B) / P(A) = P(B ∩ A) / P(A)

Equivalently, the probability that A occurs given that B occurs is computed as: P(A|B) = P(A ∩ B) / P(B) = P(B ∩ A) / P(B) EXAMPLE For the die-rolling experiment, where: A = “the first roll is even” = {EE, EO} B = “exactly one roll is odd” = {EO, OE} Since (B ∩ A) = {EO}, P(B ∩ A) = 1/4 Since A = {EE, EO}, P(A) = 1/2 Since B = {EO, OE}, P(B) = 1/2

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P(B|A) = P(B ∩ A) / P(A) = (1/4)/(1/2) = 1/2 P(A|B) = P(B ∩ A) / P(B) = (1/4)/(1/2) = 1/2 MULTIPLICATION RULE For two events A and B the multiplication rule is: P(B ∩ A) = P(B|A)P(A) = P(A|B)P(B) EXAMPLE Suppose that a card is chosen from a standard deck without being replaced; a second card is then chosen. Define: A = “the first card is a club” B = “the second card is a club” The probability that both cards are clubs is computed as follows: P(B ∩ A) = P(B|A)P(A) Since there are thirteen clubs in a standard deck of cards and fifty-two cards in the deck, P(A) = 13/52 = 1/4 The probability of B given A is computed as follows: if the first card is a club (event A), then there will be 51 remaining cards in the deck when the second card is chosen. Of these, 12 of 13 clubs remain in the deck. Therefore, P(B|A) = 12/51. Therefore, the probability that both cards are clubs is: P(B ∩ A) = P(B|A)P(A) = (12/51)(1/4) = 12/204 = 0.0588 INDEPENDENT EVENTS Two events A and B are said to be independent if the occurrence of A does not affect the probability of B occurring, and vice versa. EXAMPLE Referring to the die-rolling experiment, suppose that the following events are defined:

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C = “the first roll is even” D = “the second roll is odd” Intuitively, events C and D are independent since the two rolls of the die have no influence on each other. INDEPENDENT EVENTS Events A and B are independent only if both of the following conditions are true: P(B | A) = P(B) P(A | B) = P(A) EXAMPLE For the die-rolling experiment, where: A = “the first roll is even” = {EE, EO} B = “exactly one roll is odd” = {EO, OE} P(A) = 1/2 P(B) = 1/2 P(B ∩ A) = 1/4 P(A|B) = P(B ∩ A) /P(B) = (1/4)/(1/2) = 1/2 P(B|A) = P(B ∩ A) /P(A) = (1/4)/(1/2) = 1/2 Since P(A|B) = P(A) and P(B|A) = P(B), A and B are independent events. MULTIPLICATION RULE FOR TWO INDEPENDENT EVENTS For two independent events, the multiplication rule for independent events is: P(B ∩ A ) = P(B)P(A) = P(A)P(B) COMPLEMENT RULE Two events A and B are said to be complements if: P(A ∪ B) = 1 P(A ∩ B ) = 0

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The complement of A is written AC. The complement rule is: P(A) = 1 - P(AC) or:

P(AC) = 1 - P(A)

BAYES’ THEOREM Bayes’ Theorem can be used to determine the conditional probability of an event. The general formula for computing conditional probabilities is:

P(Ai | B) =

P(B | Ai )P(Ai ) n

! P(B | A )P(A ) i

i

i=1

The sample space, or set of all possible events, is partitioned into n events: A1, A2, ..., An; the denominator of the formula represents the total probability of event B. EXAMPLE For the die-rolling experiment, the following events are defined: A = “the first roll is even” = {EE, EO} B = “exactly one roll is odd” = {EO, OE} In this example, AC = "the first roll is odd" = {OE, OO}. Using Bayes’ Theorem, the probability that the first roll is even given that exactly one roll is odd is determined as follows: A = {EE, EO} B = {EO, OE} AC = {OE, OO} (B ∩ A) = {EO} (B ∩ AC) = {OE}

P(A) = 1/2 P(B) = 1/2 P(AC) = 1/2 P(B ∩ A) = 1/4 P(B ∩ AC) = 1/4

P(B|A) = P(B ∩ A)/P(A) = (1/4)/(1/2) = 1/2 P(B|AC) = P(B ∩ AC)/P(AC) = (1/4)/(1/2) = 1/2 In this case, the formula for computing P(A|B) is:

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P(A | B) =

P(B | A)P(A) P(B | A)P(A) + P(B | AC )P(AC )

P(A | B) =

(1 / 2)(1 / 2) 1 = (1 / 2)(1 / 2) + (1 / 2)(1 / 2) 2

RANDOM VARIABLES A random variable is a function that assigns numerical values to the outcomes of a random experiment. EXAMPLE Referring to the die-rolling experiment, suppose that a random variable X is defined as “the number of times an even number turns up during the experiment”. X assigns a number to each element of the sample space, as shown in the following table:

SAMPLE POINT EE EO OE OO

X 2 1 1 0

From the table, it can be seen that: P(X = 0) = 1/4 P(X = 1) = 2/4 P(X = 2) = 1/4 Since X can only assume a finite number of different values, it is said to be a discrete random variable. If X can assume an infinite number of different values, it is said to be a continuous random variable. CUMULATIVE DISTRIBUTION FUNCTION (CDF)

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The cumulative distribution function (cdf), designated F(x), shows the probability that a random variable X assumes a value that is less than or equal to a constant x: F(x) = P(X ≤ x) where: X = a random variable x = a realization (value) of X For a discrete random variable, the cumulative distribution function is:

F(x) = P(X ≤ x) = ∑ P(X = x) x

For a continuous random variable, the cumulative distribution function is: x

F(x) = P(X ≤ x) =

∫

f (u)du

−∞

EXAMPLE Referring to the die-rolling experiment, the following table illustrates the cdf of X: (X ≤ x) {} {OO} {OO, EO, OE} {OO, EO, OE, EE} {OO, EO, OE, EE}

x <0 0 1 2 >2

F(x) = P(X ≤ x) 0 1/4 3/4 1 1

PROBABILITY MASS FUNCTION (PMF) For a discrete random variable, a table or a function showing the probability of each possible value is known as a probability mass function (pmf), designated p(x): p(x) = P(X = x) where:

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∑ p(x ) = 1 i

x

COMPUTING PROBABILITIES WITH A CDF For a discrete random variable, probabilities can be derived from the cdf as follows: p(xi) = F(xi) – F(xi-1) EXAMPLE For the die-rolling experiment, the probability mass function of X is shown in the following table:

x <0 0 1 2 >2

F(x) 0 1/4 3/4 1 1

p(x) 0 (1/4 – 0) = 1/4 (3/4 – 1/4) = 1/2 (1 – 3/4) = 1/4 (1 – 1) = 0

This table shows that: P(X = 0) = 1/4 P(X = 1) = 2/4 P(X = 2) = 1/4 PROBABILITY DENSITY FUNCTION (PDF) For a continuous random variable, a function showing the probability of any range of possible values is known as a probability density function (pdf), designated f(x): b

∫ f (x)dx = P(a < X < b) a

where: ∞

∫

f (x)dx = 1

−∞

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The pdf is related to the cdf as follows: f (x) =

dF(x) dx

Equivalently, the cdf is related to the pdf as follows: x

F(x) =

∫

f (u)du

−∞

COMPUTING PROBABILITIES WITH A CDF For a continuous random variable, probabilities can be derived from the cdf as follows: p(a < X < b) = F(b) – F(a) PROBABILITY FUNCTION Both probability mass functions and probability density functions are sometimes known more simply as probability functions. JOINTLY DISTRIBUTED RANDOM VARIABLES For two jointly distributed discrete random variables, X and Y, the joint probability mass function is defined as: p(x, y) = P(X = x, Y = y) EXAMPLE Suppose that a census is taken for a small town; the distribution of the number of children among the families in this town is given as follows:    

20% have no children 30% have one child 40% have two children 10% have three children

Define: X = “number of boys in a family” Y = “number of girls in a family”

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The following table shows the joint probability mass function for X and Y:

i,j

0

1

2

3

0 1 2 3 Row Sum P(Y = j)

0.20 0.15 0.10 0.0125 0.4625

0.15 0.20 0.0375 0 0.3875

0.10 0.0375 0 0 0.1375

0.0125 0 0 0 0.0125

Row Sum P(X = i) 0.4625 0.3875 0.1375 0.0125 1.0000

This table shows the joint probabilities for every possible value of X and Y. For example, the joint probability that a family has 2 boys and 1 girl (X = 2, Y = 1) is 0.0375. MARGINAL PROBABILITY For two jointly distributed random variables, X and Y, the marginal probability of X is the probability that X assumes a given value for all possible values of Y. Equivalently, the marginal probability of Y is the probability that Y assumes a given value for all possible values of X. Marginal probabilities are computed as follows:

p X (X = x) = ∑ p(x, y) y

This is the marginal probability mass function of X. The marginal probability mass function of Y is:

pY (Y = y) = ∑ p(x, y) x

In the census example, the marginal probabilities of X are given by the row sums of the joint pmf; the marginal probabilities of Y are given by the column sums. EXAMPLE (c) ECI Risk Training 2009

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The marginal probability that a family has one boy can be determined as follows: P(one boy) = P(one boy and no girls) + P(one boy and one girl) + P(one boy and two girls) + P(one boy and three girls) = P(X = 1, Y = 0) + P(X = 1, Y = 1) + P(X = 1, Y = 2) + P(X = 1, Y = 3) = 0.15 + 0.20 + 0.0375 + 0 = 0.3875 EXAMPLE The (marginal) probability that a family has two girls can be determined as follows: P(two girls) = P(no boys and two girls) + P(one boy and two girls) + P(two boys and two girls) + P(three boys and two girls) = P(X = 0, Y = 2) + P(X = 1, Y = 2) + P(X = 2, Y = 2) + P(X = 3, Y = 2) = 0.10 + 0.0375 + 0 + 0 = 0.1375 UNCONDITIONAL PROBABILITY A marginal probability is also known as an unconditional probability. CONDITIONAL PROBABILITY A conditional probability P(X = x|Y = y) is the probability that X assumes a specific value x given that Y assumes a specific value y. CONDITIONAL PROBABILITY MASS FUNCTION The probability mass function of X given that Y = y is known as a conditional probability mass function:

p X |Y (x | y) = P(X = x | Y = y) =

p(x, y) pY (Y = y)

The probability mass function of Y given that X = x is:

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pY | X (y | x) = P(Y = y | X = x) =

p(x, y) p X (X = x)

EXAMPLE The probability that a family has one boy given that it has two girls is computed as follows:

P(X = x | Y = y) =

p(x, y) pY (Y = y)

P(X = 1 | Y = 2) =

p(1, 2) pY (Y = 2)

= 0.0375 / 0.1375 = 0.2727 JOINT PROBABILITY DENSITY FUNCTION For two jointly distributed continuous random variables, X and Y, the joint probability density function is defined as: b d

∫ ∫ f (x, y)dx dy = P(a < X < b, c < Y < d) a c

The marginal probability density function of X is: ∞

f X (x) =

∫

f (x, y)dy

−∞

The marginal probability density function of Y is: ∞

fY (y) =

∫

f (x, y)dx

−∞

CONDITIONAL PROBABILITY DENSITY FUNCTION (c) ECI Risk Training 2009

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For jointly distributed continuous random variables, the conditional probability density function is:

f X |Y (x | y) =

f (x, y) fY (y)

JOINT CUMULATIVE DISTRIBUTION FUNCTION For two jointly distributed random variables, X and Y, the joint cumulative distribution function is defined as: F(x, y) = P(X ≤ x, Y ≤ y) If X and Y are discrete random variables and a and b are constants:

F(a,b) = P(X ≤ a,Y ≤ b) = ∑ ∑ p(x, y) x ≤a y≤b

If X and Y are continuous random variables and a and b are constants: a b

F(a,b) = P(X ≤ a,Y ≤ b) =

∫∫

f (x, y)dx dy

−∞ −∞

EXAMPLE Using the census example, the probability that a family has two boys or less and one girl or less can be computed as follows: P(two boys or less and one girl or less) = P(no boys, no girls) + P( no boys, one girl) + P(one boy, no girls) + P(one boy, one girl) + P(two boys, no girls) + P(two boys, one girl) = P(0, 0) + P(0, 1) + P(1,0) + P(1,1) + P(2,0) + P(2, 1) = 0.20 + 0.15 + 0.15 + 0.20 + 0.10 + 0.0375 = 0.8375 INDEPENDENCE OF RANDOM VARIABLES If X and Y are discrete random variables and p(X=xi ,Y = yj) = p(X=xi)p(Y = yj) then X and Y are independent. For continuous random variables, if: (c) ECI Risk Training 2009

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f(x,y) = fX(x)fY(y) then X and Y are independent. Some of the consequences of independence are: 1) E(XY) = E(X)E(Y) 2) COV(X, Y) = 0 3) Var(X, Y) = Var(X) + Var(Y) MOMENTS OF A RANDOM VARIABLE A random variable can be characterized by its moments. These are summary measures of the behavior of a random variable. The most important of these are: • • • •

Expected Value Variance Skewness Kurtosis

EXPECTED VALUE The first moment of a random variable X is known as its expected value; this is the average or mean value of X. For a discrete random variable X, the expected value is computed as follows: n

E(X) = ∑ xi P(X = xi ) i =1

where: xi = a possible value of X i = an index n = the number of possible values of X Σ = “sigma”; this is the summation operator EXAMPLE For the die-rolling experiment, where X is the number of even numbers that turn up after rolling a die twice, the expected value is computed as follows: E(X) = (0)(1/4) + (1)(1/2) + (2)(1/4) = 1 This shows that on average, there will be one even number each time a die is rolled twice.

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For a continuous random variable X, the expected value is computed as follows: ∞

E(X) =

∫ xf (x)dx

−∞

PROPERTIES OF THE EXPECTED VALUE For two random variables X and Y and two constants a and b: 1) E(a) = a 2) E(aX+ b) = aE(X) + b 3) E(X + Y) = E(X) + E(Y) CONDITIONAL EXPECTATION For two random variables X and Y, the conditional expectation of X given that Y assumes a specific value y is written as: E[X|Y = y] If X and Y are discrete random variables:

E[X | Y = y] = ∑ xP(X = x | Y = y) x

If X and Y are continuous random variables: ∞

E[X | Y = y] =

∫ xf

X |Y

(x | y)dx

−∞

∞

=

∫x

−∞

f (x, y) dx fY (y)

EXAMPLE Using the census example, the expected number of boys in a family given that there are two girls in the family is computed as: E[X|Y = 2] =

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(0)P(X = 0 | Y = 2) + (1)P(X = 1 | Y = 2) + (2)P(X = 2 | Y = 2) + (3)P(X = 3| Y = 2)

= (0)

p(0, 2) p(1, 2) p(2, 2) p(3, 2) + (1) + (2) + (3) pY (Y = 2) pY (Y = 2) pY (Y = 2) pY (Y = 2)

= (0)(0.10/0.1375) + (1)(0.0375/0.1375) + (2)(0/0.1375) + (3)(0/0.1375) = 0 + 0.2727 + 0 + 0 = 0.2727 EXPECTED VALUE OF A PRODUCT OF RANDOM VARIABLES The expected value of the product of two discrete random variables is:

E(XY ) = ∑ ∑ xyp(x, y) x

y

The expected value of the product of two continuous random variables is: ∞ ∞

E(XY ) =

∫ ∫ xyf (x, y)

−∞ −∞

VARIANCE The second central moment of a random variable X is known as its variance. This indicates the degree of dispersion or spread of X around its expected value. The variance of random variable X is computed as follows:

σ X2 = E[(X − E(X))2 ] = E[X 2 ] − (E[X])2 For a discrete random variable, the variance can also be expressed as:

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n

σ X2 = ∑ [ xi − E(X)] P(X = xi ) 2

i =1

EXAMPLE For the die-rolling experiment, the variance of X is computed as follows: σx2 = [0-1]2(1/4) + [1-1]2(2/4) + [2-1]2(1/4) = (1)(1/4) + (0)(2/4) + (1)(1/4) = 1/2 For a continuous random variable X, the variance is computed as follows:

σ X2 =

∞

∫ [ x − E(X)]

2

f (x)dx

−∞

PROPERTIES OF VARIANCE For two random variables X and Y and two constants a and b: 1) Var(a) = 0 2) Var(aX + b) = a2Var(X) 3) Var(aX + bY) = a2Var(X) + b2Var(Y) + 2abCov(X,Y) CONDITIONAL VARIANCE For two random variables X and Y, the conditional variance of X given that Y assumes a specific value y is written as: Var[X|Y] = E[[X-E(X|Y)]2|Y] This can be re-written as: Var[X|Y] = E[X2|Y] - (E[X|Y])2 where:

E[X 2 | Y ] = ∑ x 2 p(x, y) x

EXAMPLE

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Using the census example, the conditional variance of the number of boys in a family given that there are two girls is computed as follows:

p(0, 2) p(1, 2) p(2, 2) p(3, 2) + (1)2 + (2)2 + (3)2 pY (Y = 2) pY (Y = 2) pY (Y = 2) pY (Y = 2) 2 2 2 E[X |Y] = (0) P(X = 0 | Y = 2) + (1) P(X = 1 | Y = 2) + (2)2P(X = 2 | Y = 2) + (3)2P(X = 3 | Y = 2) = (0)2

= (0)(0.10/0.1375) + (1)(0.0375/0.1375) + (4)(0/0.1375) + (9)(0/0.1375) = 0 + 0.2727 + 0 + 0 = 0.2727 Var[X|Y] = E[X2|Y] - (E[X|Y])2 = (0.2727) - (0.2727)2 = 0.2727 - 0.0744 = 0.1983 STANDARD DEVIATION One of the drawbacks to using variance is that it is measured in squared units. Since these are difficult to interpret, the standard deviation is often used instead. For any random variable X, the standard deviation of X equals the square root of the variance of X. EXAMPLE For the die-rolling experiment, the standard deviation is computed as follows:

σ X = σ X2 =

1 = 0.7071 2

COVARIANCE Covariance is a measure of dependence between random variables: • • •

If X and Y tend to move in the same direction, Cov(X,Y) > 0 If X and Y tend to move in opposite directions, Cov(X,Y) < 0 If X and Y are unrelated to each other (i.e., independent ), Cov(X,Y) = 0

Covariance is defined as:

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Cov(X,Y) = E([X - E(X)][Y - E(Y)]) This can also be written as: Cov(X,Y) = E(XY) - E(X)E(Y) If X and Y are discrete random variables, the covariance can also be expressed as: n

n

∑ ∑ (x i =1 j =1

i

− E(X))(y j − E(Y ))P(X = xi ,Y = y j )

where: i, j are indexes P(X=xi, Y=yj) = the joint probability mass function of X and Y For two continuous random variables, covariance is computed as: ∞ ∞

∫ ∫ (x − E(X))(y − E(Y )) f (x, y)dxdy

−∞ −∞

where: f(x,y) = the joint probability density function of X and Y EXAMPLE Using the following joint probability mass function for two random variables X and Y, the covariance is computed as follows:

i, j

0

1

0 1 Column Sum P(Y = j)

0.40 0.10 0.50

0.30 0.20 0.50

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Row Sum P(X = i) 0.70 0.30 1.00

21

The first step is to compute E(X) and E(Y). Using the row sums from the joint probability mass function, the probability mass function for X is:

x 0 1

P(x) 0.70 0.30 n

E(X) = ∑ xi P(X = xi ) i =1

= (0)(0.70) + (1)(0.30) = 0.30 Using the column sums from the joint probability mass function, the probability mass function for Y is:

y 0 1

P(y) 0.50 0.50 n

E(Y ) = ∑ yi P(Y = yi ) i =1

= (0)(0.50) + (1)(0.50) = 0.50 n

n

COV (X,Y ) = ∑ ∑ (xi − E(X))(y j − E(Y ))P(X = xi ,Y = y j ) i =1 j =1

= (0 - 0.3)(0 - 0.5)(0.40) + (0 - 0.3)(1 - 0.5)(0.30) + (1 - 0.3)(0 - 0.5)(0.10) + (1 - 0.3)(1 - 0.5)(0.20) = 0.06 - 0.045 + 0.035 + 0.07 = 0.12 PROPERTIES OF COVARIANCE For two random variables X and Y and two constants a and b: 1) Cov(X, Y) = Cov(Y, X) 2) Cov(X + a,Y + b) = Cov(X, Y)

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3) Cov(aX, bY) = abCov(X, Y) 4) Cov(X, X) = Var(X)

CORRELATION A related measure of dependence is correlation. The correlation coefficient is defined as:

ρ=

Cov(X,Y ) σ Xσ Y

where: ρ = “rho”; this is the correlation coefficient σX = the standard deviation of X σY = the standard deviation of Y The correlation coefficient always assumes a value between negative one and positive one and is unit-free: -1 ≤ ρ ≤ 1 EXAMPLE Using the data from the covariance example, the correlation is computed as follows. First, the variance of X is computed as: n

σ X2 = ∑ [ xi − E(X)] P(X = xi ) 2

i =1

σ2X = [0 - 0.3]2(0.70) + [1 - 0.3]2(0.30) = 0.21 The standard deviation of X is:

σ X = 0.21 = 0.4583 Second, the variance of Y is computed as:

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n

σ Y2 = ∑ [ yi − E(Y )] P(Y = yi ) 2

i =1

σ2Y = [0 - 0.5]2(0.50) + [1 - 0.5]2(0.50) = 0.25 The standard deviation of Y is:

σ Y = 0.25 = 0.5 The correlation between X and Y is:

ρ=

Cov(X,Y ) 0.12 = = 0.5237 σ Xσ Y (0.4583)(0.5)

SKEWNESS The third central moment of a random variable X is known as its skewness. This indicates the degree of asymmetry in the values of X. Skewness is computed as follows:

α3 =

E[(X − E(X))3 ] σ3

where: α3 is the skewness coefficient σ3 is the standard deviation cubed EXAMPLE For the die-rolling experiment, the numerator of the skewness formula is computed as follows: = [0-1]3(1/4) + [1-1]3(2/4) + [2-1]3(1/4) = (-1)(1/4) + (0)(2/4) + (1)(1/4) = 0

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The denominator equals (σ2)3/2 = (1/2)3/2 = 0.3536 Therefore, α3 = 0/0.3536 = 0

KURTOSIS The fourth central moment of a random variable X is known as its kurtosis. This refers to the likelihood that X will assume an extremely small or large value. Kurtosis is computed as follows:

α4 =

E[(X − E(X))4 ] σ4

where: α4 is the kurtosis coefficient σ4 is the standard deviation raised to the fourth power EXAMPLE For the die-rolling experiment, the numerator of the kurtosis formula is computed as follows: = [0-1]4(1/4) + [1-1]4(2/4) + [2-1]4(1/4) = (1)(1/4) + (0)(2/4) + (1)(1/4) = 1/2 The denominator equals (σ2)2 = (1/2)2 = 0.25 Therefore, α4 = 0.5/0.25 = 2 COEFFICIENT OF VARIATION The coefficient of variation is a measure of relative variation; it is defined as the ratio of the standard deviation to the mean (expected value) of X:

CV =

(c) ECI Risk Training 2009

σX µX

25

EXAMPLE For the die-rolling experiment, the coefficient of variation is:

CV =

σ X 0.7071 = = 0.7071 µX 1

CHEBYSHEV’S INEQUALITY Chebyshev’s inequality gives an upper limit for the probability that X will be k or more standard deviations away from its expected value. Chebyshev’s inequality is written as:

P { X − µ X ≥ kσ X } ≤

1 k2

EXAMPLE Suppose that X is a random variable with an expected value of 2, and a standard deviation of 5. For k = 3: P { X − µ X ≥ kσ X } ≤

P { X − 2 ≥ 15} ≤

1 k2

1 9

In other words, P(X ≥ 17) and P(X ≤ -15) are both less than or equal to 1/9.

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