math stats review

7/26/2019 Math Stats Review

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Math/Stats Review

The following is a quick tour of the basic mathematical/statistical principles and algebra

needed in FNCE 370.1

Ba s i c P r i n c i p l e s

Many students will find it redundant. However, if you feel uncertain,study this review material carefully.

Tables and Graphs

Keep the following in mind when studying tables and graphs.

1. It is a good habit to look for the units. Are the units in $US, $CAN, Sterling, or in kg,

%, or numbers of people? Are units clearly stated and easily understood and compared?

Cereal boxes are, for example, notoriously ambiguous. (Did you get your daily allowanceof vitamins at breakfast?) How can you tell?

2. Look for the source of data and consider its possible reliability/unreliability.

3. In the case of Tables 1.3 and 1.4 (textbook, pp. 1819) there is a problem. Can youidentify the problem?

Answer

Table 1.3 is in units of billions of US dollars. Table 1.4 is in $billions. Does Table 1.4specify US or Canadian dollars? If we assume it is Canadian dollars, then Table 1.3 isdifficult or impossible to compare with Table 1.4. Moreover, currency values change with

inflation. What year are the tables for? Furthermore, Table 1.4 does not specify what

month (or in the beginning or end of the year) the data was collected.

Calculations with Brackets

Suppose you are doing the following calculation for Net Working Capital Turnover.

NWC Turnover = SALES/NWC= $2,311/($708 - $540)

From experience, you should know that the first thing to do in order to get NWC Turnover isto perform the operation withinthe brackets ($708 $540) = $168.

The second operation is to simplify the ratio $2,311/$168 = $13.756.The following sections contain some common algebraic principles that will get you on yourway to solving equations in this course.

Commutative Axioms

For Addition

a + b = b + a

For Multiplication

n . m = m . n

These two axioms state that changing the order in which two numbers are added or

multiplied will not alter the result.

For example2 + 3 = 3 + 2

and

2 . 4 = 4 . 2 (a dot or period may be used in place of x as a multiplication sign)

Division is not commutative; order matters in division.

2/3 does not equal 3/2

1Based on Auvil, D. L, & Poluga, C. (1984). Elementary algebra(2nd ed.). Don Mills, ON: Addison-Wesley Publishing.


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Associative Axioms

For Addition

a + (b + c) = (a + b) + c

For Multiplication

n . (m . p) = (n . m) . p

For example

3 + (2+ 4) = (3 + 2) + 4 = 9

3 . (2 . 4) = (3 . 2) . 4 = 24

Addition Axiom for Equality

If a = b, then a + c = b + c.

Multiplication Axiom for Equality

If a = b, then a . c = b . c

Distributive Axiom

a(b + c) = ab + ac

Compute 2 (10 + 5) in two ways

2(10 + 5) = 2 . 15 = 30

or, using the distributive axiom,

2 (10 + 5) = 2 . 10 + 2 . 5 = 20 + 10 = 30

Since multiplication is commutative,

(b + c)a = ba + ca

Distributive Axiom

Multiplication can be distributed over any finite number of terms.

a(b + c + d) = ab + ac + ad

The distributive axiom comes in handy when we must remove parentheses, such as theexample problem from Calculations with Brackets above.

Remove parenthesis from the expression 6(x + 4)

6(x + 4) = 6 . x + 6 . 4

= 6x + 24

The distributive axiom enables us to combine like terms

3(x2+ 3x2) = 3(4x2) = 12x2


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Example 2

Remove parentheses in

(3x + 5) (3x + 5) = -3x - 5

Example 3

Remove parentheses in 2(x + ) -2(x + ) = -2x 1

Example 4

Remove parentheses in 4(-3x + 1)

-4(-3x + 1) = 12x 4

Example 5

Remove parentheses in (x 5)

-(x-5) = (-1) . (x 5)

= (-1) . [x + (-5)]= (-1 . x + (-1) . (-5)= -x + 5

Example 6 Combine the like terms 6a+ 7a

6a+ 7a = (6 + 7) a

= 13a

Recall that the addition axiom for equalityallows us to add equal quantities to both sides ofan equation and the multiplication axiom for equalityallows us to multiply both sides of an

equation by equal quantities. Using these two properties, as well as the other axioms, we

shall change the formof an equation without changing its solution.

To solve the equation x +2 = 6, we shall isolate the variable on one side of the equation.

In order to isolate x on one side of the equation, we must remove the 2 from the left-hand side. This is done by adding the negative of 2, namely 2, to both sides.

x + 2 = 6x + 2 +(-2) = 6 + (-2)x + 0 = 4

x = 4

In the example above, the number 2 was removed from the left-hand side by adding thenegative of 2 (i.e., 2) to each side.

Example 7

Solve 5x = 10

Again we wish to isolate the variable x on one side of the equation. Therefore, we must

remove the coefficient 5. To do this we multiply both sides of the equation by the

reciprocal of 5, namely 1/5.5x = 10

1/5 . 5x = 1/5 . 10

1 . x = 10/5x = 2

In this last example, the number 5 was removed from the left-hand side by multiplying

each side by the reciprocal of 5, namely 1/5. The same result could have been achieved by

dividingeach side of the equation by 5.


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Ma t h R ev i e w : Le s so n s 3 t o 6

Exponential Functions2

Start by noting that

F(x) = 2x

and

g(x) = x2

are not the same function.

Whether a variable appears as an exponent with a constant base or as a base with a

constant exponent makes a big difference. The function gis a quadratic function. The

function fis called an exponential function. In general, the equation

f(x) = bx b>0, b 1

defines an exponential functionfor each different constant b, called the base. The domainof

fis the set of all positive real numbers.

Exponential Function Properties

Foraand b positive, , andx and y real:1. Exponent laws

2. ax= ay if and only ifx = y

Example: If 7 5t+1= 7 3t3, 5t + 1 = 3t 3 and t = -2

3.

For x 0, ax= bxif and only if a= b.

Example: if a4= 24then a = 2

Exponential functions are useful in many disciplines, not just in finance and economics. For

example, cholera, an intestinal disease, is caused by a cholera bacterium that multipliesexponentially by cell division as given approximately by

N = N0e 1.386t

where Nis the number of bacteria present after t hours and N0is the number of bacteria

present at the start (t = 0).

2This section draws extensively from Barnett, R. A., Ziegler, M. R., & Byleen, K. E. (2002). Finite mathematics for business,economics, life sciences, and social sciences(9th ed.). Upper Saddle River, NJ: Prentice Hall.

1a 1b

2424 54 34

54

( ) ( )

4Example: 4 4

4

xx xx y x y x y xy x y x x x

y x

a a aa a a a a a ab a b

ba b

+

= = = = =

= =


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7/21


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Example 9

Remove parentheses and simplify (4x/y2)3.

Definitions

Let a be any real number except zero. Then a0= 1

Let abe any real number except zero and nany natural number.

Then

We can observe certain patterns that occur with negative exponents. For example

Hence the rule

In other words, a negative exponent on a fraction may be replaced by the correspondingpositive exponent simply by inverting the fraction.

Also

Hence the rule

( )

3 3 3 3 3 3

2 3 6 62

22 2 0

2

4 4 4 64

55 5 1

5

x x x x

y y yy

= = =

= = =

1

1

2 2 2 02

1

Therefore,

1

5 5 5 15

n

na a

aa

=

=

= = =

1 1

/

n nn

n n n n

a b b

b aa b aa

b

= = = =

0 0

n na b

where a and bb a

=

1 1

1 /

n

n n a

a a= =

10

= n

n a where a

a


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Adding and Subtracting Expressions

Recall that to add two rational numbers we use the rule

a/c + b/c = (a + b)/c where c 0

That is, to add two rational numbers having the same denominator we add their numeratorsand place that sum over the common denominator.

Example 1

4/15 + 7/15 = 11/15.

On the other hand, to subtract two rational numbers we use the rule

a/c - b/c = (a - b)/c where c 0

These same two rules are used to add and subtract rational expressions.

Example 2

Find the sum 1/7x + 5/7x

Since the denominators are the same, we simply add the numerators

1/7x + 5/7x = 6/7x

Example 3

Find the sum

Least Common Denominator

The least common denominator(LCD) of a collection of rational expressions is the simplest

polynomial that is a polynomial multiple of each denominator in the collection. LCD canoften be determined simply by inspection.

Example 4

Determine the LCD of 1/8xand 1/12x

The simplest polynomial that is a multiple of both 8xand 12xis 24x. Note that

24x= 3 x 8xand

24x= 2 x12x

Therefore, the LCD is 24x.

Example 5

Determine the LCD of 5/x2y and 4/xy2. The LCD is x2y2sincex2y2= yx2y, and

x2y2= xxy2, and

x2y2is the simplestmultiple of both x2y and xy2.

( )( )3 23 6 3 6 3 6

32 2 2 2 2 2

xx x x

x x x x x x

+++ = + = = =

+ + + + + +


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Complex Fractions

A complex fractionis a fraction that contains other fractions in its numerator or denominator

(or both). A fraction that is not a complex fraction is called a simple fraction.

Because of numerous fractional parts, complex fractions may be difficult to read andinterpret. They may be simplified, however, using either of two basic methods. Both of

these methods are illustrated in Example 1.

Example 1

Simplify the complex fraction

Method 1: Simplify numerator and denominator separately, and then divide.

= 48

Method 2: Multiply numerator and denominator by the LCD of the fractions in thedenominator.

The Quadratic Formula

Common Factors

Suppose we wished to solve the quadratic equation x2 5x + 6 = 0.

If we were able to recognize that x2- 5x + 6 is actually the product of (x 2) and (x 3),we could write this equation as

(x 2) (x 3) = 0.

When the equation is in this form we can see that both 2 and 3 satisfy the equation, since 2

causes the first factor to be 0 and 3 causes the second factor to be 0.

The key to solving this equation lies in recognizing the fact that

x2- 5x + 6 = (x 2) (x 3)

When a polynomial is expressed as a product we say that the polynomial has been factored.

40

1 13 2

+

40 40 /1 40 /1

1 /3 1/2 2 /6 3 /6 5 /6= =

+ +

1 1Since the LCD of and is 6, we have

3 2

40 6 40 6 24048

1 1 6 2 3 53 2

= = =

++


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Completing the Square

Factoring provides us with an easy method for solving many quadratic equations.

Unfortunately, as we have already seen, not all polynomials are factorable using integersonly. For example, the left-hand side of the following quadratic equation cannot be factoredusing integers only.

x

2

6x + 4 = 0We can solve the equation above, however, using a method called completing the square.The basic idea of this method is to put one side of the equation in the form (x + d)2. This

side of the equation is then said to be in perfect square.

In order to put the equation in this form, we first observe that

(x + d)2= x2+ 2dx + d2.

Now, note that the term d2is the square of one-half the coefficient of x. That is, thecoefficient of x is 2d, and one-half of 2d squared is (2d/2)2= d2.

We shall now return to the original equation and solve it by completing the square.

x2 6x + 4 = 0

We begin by subtracting 4 from each side.

x2 6x = -4.We then complete the square on the left-hand side by adding the square of one-half thecoefficient of x to the left-hand side. That is, we add (-6/2)2= 9 to the left-hand side. Of

course, if we add 9 to the left-hand side we must also add 9 to the right-hand side. After

doing that, our equation becomes

x2 6x + 9 = -4 + 9 .

The left-hand side is now the perfect square (x 3)2. The equation can therefore be writtenas

(x 3)2= 5.

Solving this last equation by extraction of roots, we have

Example 1

Solve x2+ 4x 7 = 0

We begin by adding 7 to each side.

x2+ 4x 7 = 0

x2+ 4x = 7.

We now complete the square by adding to each side.

x2+ 4x + 4 = 7 + 4

The left-hand side is now the perfect square (x + 2)2.

(x + 2)2= 11.

The solutions are

x 3 5

x 3 5

Therefore the solutions are 3 5 and 3 5

=

=

+

24

42

=

-2 11 and -2 11+


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Algebraic Solutions to the Quadratic Formula

Consider the general quadratic equationas

Since the coefficient of x is b/a, the square of one-half the coefficient of x is (b/2a)2. Thus,we complete the square by adding (b/2a)2to each side.

We now have a formula for solving any quadratic equation.

The solutions of the quadratic equation

are given by

2

2

2

0

0

ax bx c

b c

x xa a

b cx x

a a

+ + =

+ + =

+ =

2 22

2 2

2 2

2 2

2

2

2

2

2

2

2 2

4

2 4 4

4

2 4

4

2 4

4

2 2

4

2 2

4

2

b b c bx x

a a a a

b ac b

x a a a

b b ac x

a a

b b ac x

a a

b b ac x

a a

b b ac x

a a

b b ac x

a

+ + = +

+ = +

+ =

+ =

+ =

=

=

2 0 ( 0)+ + = ax bx c a

2 4

2

b b ac x

a

=


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Example 1

Solve using the quadratic formula. We first put the equation in standard form

in order to identify a, b, and c.

a= 3, b= -4, c= 1

Substituting these values for a, b, and c into quadratic formula, we have

Therefore,

23 4 1 = x x

2

2

3 4 1

3 4 1 0

=

+ =

x x

x x

( ) ( ) ( ) ( )

( )

24 4 4 3 1

2 3

4 16 12

6

4 4

6

4 26

=

=

=

=

x

x

x

x

4 2 4 2 11 or

6 6 3

+ = = = =x x


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Ma t h R e v i e w : Le s so n s 7 a n d 8

Problems in Lessons 7 and 8 may involve solving two equations in two unknowns. (Wheredo the 2 lines cross?) Generally, one can substitute logic and an understanding of what one

is trying to do for mathematical savvy. But sometimes the mathematical sophisticationmakes things much easier.

Systems of Equations

For a graphical example of solving two equations in two unknowns in finance, see Figure 9.7on page 253 in the textbook.

Example 1

A paperboy has $11 extra in Christmas tips to divide between his two helpers, Mary and

Ray. Since Mary has helped more often than Ray, he decides that Mary should receive $3

more than Ray. How much should each helper receive?

This particular problem is easy to do simply by trial and error. Many problems of this type,

however, are too difficult to be solved using a trial-and-error approach. Our intent here is todevelop a systematic approach to solving problems of this type. Once developed, this

method will prove useful in many different situations in mathematics.We shall let x represent the number of dollars Mary is to receive and y the number ofdollars Ray is to receive. We can then write the two equations

x + y= 11

and

x y = 3

These two equations taken together are called a system of equations4

We know that all solutions to the first equation (x + y = 11) lie on a straight line. Also, all

solutions to the second (x y = 3) lie on some other straight line. If the two equations aregraphed using the same coordinate axes, we obtain two intersecting lines.

. To solve the system

means to find a pair of values x and y that satisfies the conditions of both equations. Forexample, the ordered pair (8,3) is a solution to the first equation, since 8 + 3 = 11; but it isnot a solution to the second equation, since 8 3 3.

4Since the two equations are to be true at the same time, they are also referred to as simultaneous equations.

y = -x + 11

y = x - 3

-6

-4

-2

0

2

4

6

8

10

12

14

0 2 4 6 8 10 12 14 16 18


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From a graph, the intersection point of the two lines appearsto be (7,4). Since the point(7,4) is the only point that lies on both lines, it is the only point that is a solution to both

equations. The point (7,4) is therefore called the solutionof the system.5

x + y= 11 or 7 + 4 = 11

x y= 3 or 7 4 = 3

Another way of writing the solution of the system above is to say that x = 7 and y= 4.

This method of solving a system of equations is called the graphical method.

If two linear equations that form a system represent two distinct lines, then the system issaid to be independent.If the system has at least one solution, it is said to be consistent.

Therefore, the system of Example 1 is independent and consistent, which we usually

shorten to just independent.

Example 2

Solve the system below using the graphical method.

x + y= 2

x + y = 4Graphing both equations, we see that the two lines are parallel.

Since the lines do not intersect, there is no solution to the system.

When the two lines of a system are parallel, the system is said to be independent and

inconsistent, which we usually shorten to just inconsistent.

Given a system of two linear equations in two variables,exactly one of three situations must

occur. Either the system has one solution, no solution, or infinitely many solutions.(The twolines lie on top of one another.)

Substitution Method

While solving the problems in the Systems of Equations section above, you probablydiscovered that the graphical method for solving a system is not always an accurate

method.

The graphical method gives us a clear picture of what is happening in the solving process,

but in order to be assured of obtaining an exact solution to a system, we must use an

algebraic method.

The substitution methodinvolves solving one of the equations for one of its variables (we

usually choose the easiest variable to solve for) and then substituting that expression intothe other equation. The resulting equation then contains only one variable.

5Auvil & Poluga (1984, p. 332).


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Example 1

Solve the system below using the substitution method.

i. x 2y = 6

ii. x + 2y = 10

We decide to solve the first equation for x. Adding 2yto both sides gives

x = 6 + 2y

Substituting the expressionx= 6+ 2yinto equation #2 we obtain a first-degree equation inone variable, and it is easily solved.

6 + 2y+ 2y= 10

6 + 4y = 10

4y= 4

y = 1

Finally, we substitute 1 for yin the equationx= 6 + 2y(actually this substitution could alsobe made in either of the original equations).

x = 6 + (2 x 1)

x= 8

Therefore the solution is (8,1).

Addition Method

The same quantity, when added to equals, produces equals. This axiom can be extended tostate

If a = band c= d, then a + c = b + d.

In other words, equal quantities when added to equals produce equals. Using this property,

we can add the corresponding sides of two given equations in two variables to produce oneequation in one variable. This method for solving a system of equations is called the

addition method.

Example 2

Solve the system below using the addition method.

x + y = 6

x y = 4

Adding the corresponding sides of the two equations, we see that the variable y drops out.

x + y = 6

2x = 10

x y = 4

Solving this last equation, we have2x= 10

x = 5

Finally, we substitute 5 for x in either of the original equations and solve for y. For example,using the first equation we have

5 + y = 6

y= 1


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Example 3

Solve the system below using the addition methods.

4x + 5y= 3

3x - 2y = 8

In this case, both equations must be multiplied by a suitable number in order to force either

the variable x or the variable y to drop out.Notice that we could have eliminated the variable x in Example 3 by multiplying the firstequation by 3 and the second equation by 4, and then adding. There are generally a

variety of ways in which the variable of our choice may be eliminated.

In all of the examples thus far, the terms involving like variables have been lined up on the

left-hand side of the equation, and the constants have been lined up on the right-hand side.

When a system of equations is so arranged, it is said to be in standard form.

System in Standard Form

2x+ 3 y= 2

4x + 9y= 9

System not in Standard Form

2x = 2 3y

4x+ 9y 9 = 0

When solving a system using the addition method, it is generally convenient to write the

system in standard form before proceeding with the solution.


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St a t i s t i cs R ev i e w : L e ss o n s 1 0 a n d 1 1

Standard Deviation

Standard deviation (SD) is important in finance since it directly measures risk. With SD onecan determine the

likelihood of a given level of return in a particular portfolio of investments

probability of returns on a given portfolio falling below a specified level

probability of one portfolio having higher returns than another.

From a students perspective, SD is important because it occurs on many if not most final

examinations and because many students have difficulty answering questions related to

SD. The textbook does not provide an elaborate treatment of SD, presumably because thesubject is assumed to have been covered in prerequisite courses.

Most descriptions of standard deviation start with the calculation. Here we will start with the

interpretation.

First, about 99.7% of all observations are found within 3 standard deviations of the mean.

In other words, only about 3 observations per 1000 are farther from the mean than 3standard deviations. Thus, 3 standard deviations aboveplus 3 standard deviations below

(that is, 6 in all) encompass about 997 observations in 1000. This 6 sigma criterionunderlies some quality control systems. If its quality is 3 sigma or more from the mean it

is not good enough.

Then, given 1000 observations of most biological and economic phenomena (e.g., weight ofmice, yield of crops, profitability of investments) the two extreme values (heaviest and

lightest weight, largest and smallest yield) are likely to be about 6 standard deviationsapart. This leads to an estimate of standard deviation.

Take 1000 observations and calculate:

(Largest value smallest value)/6.

This is an estimate of Standard Deviation.

For weights of hockey players in pounds this might be

(260 165)/6 = 15.83 pounds.(Note: There are about 800 players in the NHL. Their weights are published. A scanning of

this data at time of writing indicates: George Laroque (at 260 pounds) is the heaviest while

Paul St. Denis (at 165 pounds) is the lightest.

Some data sets are not normally distributed or bell shaped. Some, such as street addresses

(every number is equally likely) tend to be rectangular. But the Central Limit Theorem tellsus that means from sampling such distributions (indeed almost any distribution) will be

normally distributed or bell shaped . More importantly, and this is why the normaldistribution is so ubiquitousany data (observation) such as weight of hockey players that

is a consequence of many small and unrelated influences (e.g., heredity, childhood

nutrition, amount of exercise) is likely to be normally distributed.

This has been called the Fuzzy Central Limit Theorem. It turns out to be a boon to

researchers who can use statistical tests on many populations. Even when the population isnot normally distributed, Chebyshevs Theorem6

tells us that standard deviation can still be

used. For example, according to Chebyshev, at least 75% of a population will lie within the

interval . (That is, two standard deviations above, plus two below the mean.)

6See Triola, M. F., Goodman, W. M., & Law, R. (2002). Elementary statistics(2nd Canadian ed.). Toronto, ON: Addison WesleyLongman. The text uses (range)/4 instead of (range)/6 (see p. 75).

/- 2x +


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The standard score (or z-score) enables us to determine if an observation is unusual, and ifso, how unusual it is. It is independent of units used so that z-score for prices is a measure

of how unlikely is a particular price, while the z-score for weight provides a measure of how

unusual is that weight. The standard score (or z) is not tabulated in the textbook nor is itdescribed there, so analysis of z will be brief.

For any sample

where xiis an observation

is the sample means of observations and

s is an estimate of standard deviation.

For example, a stock price on the TSX may have a mean of $30 and an s (estimated

standard deviation) of 20. A particular stock (e.g., Royal Bank) may have a price of $60.The questionis Royal Banks price unusually high?

Answer z = ($60 - 30)/20 = 30/20 = 1.5.Since most observations (95%) fall within two standard deviations of the mean, 1.5 wouldnot be considered particularly unusual. Royal Banks price is not unusually high. If we lookup the 1.5 in the z table, we find a coefficient of 0.43 (out of 0.50). So Royal Banks price is

higher than about 43 firms out of 50higher than average, but not remarkable.

Highest observationlowest observation

To this point has been estimated as (range)/6. That is rough and ready. Moreover it isnot efficient since it discards the information contained in all but the highest and lowest

observations. Instead the general method involves the following steps.

1.

Take a sample of the subject phenomena and measure the interesting characteristic.

2. Find the mean or average of the sample characteristics.

3. Take the deviations of each characteristic from the sample mean characteristic.

4. Square each deviation, add them, and divide by N - 1. This is the variance.

5. Take the square root of the variance. This is the standard deviation.

The notation used in our text is = standard deviation, and = variance.

But since we are doing this on a sample and not on the whole population, the notation s isfrequently useds is an estimate of .

Suppose sample observations are 1, 2, and 3. 7

The mean is (1 + 2 + 3)/3 = 2

What is the standard deviation?

s2= {(1 - 2)2+ (2 - 2)2+ (3 - 2)2}/(3 - 1) = 2/(3 - 1) = 2/2 = 1

s = (2/2)1/2= 1

So our estimated standard deviation is 1. Are these numbers reasonable?

7While there are difficulties associated with such very small numbers of observations, it certainly simplifies the math.

( )iz x x s=

x

2


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The mean (2) plus and minus 2 standard deviations (2 + 2 to 2 - 2) = 4 to 0 should containabout 95% of observations.

The range 0 to 4 contains all our observations, thus this seems reasonable.

Suppose we have observations 2, 4, and 6.

= 4; deviations are 2, 0, +2, deviations squared are 4, 0, 4.

s2= 8/2 = 4

s = (4)1/2= 2

Is this realistic? Are 95% of observations between 4 + 2s and 4 - 2s, in other words,between +8 and 0?

The answer is yes.

Now suppose both of these sets of observations arise in one sample which contains (1, 2,

3,) (2, 4, 6).

In that case = 18/6 = 3 and

s2= {(1 - 3)2+ (2 - 3)2+ (3 - 3)2+ (2 - 3)2+ (4 - 3)2+ (6 - 3)2}/(6 - 1) = 16/5

And s is (16/5)1/2= 1.788.

Consider a sample 1, 2, 3, 1, 2, 3, 2, 4, 6. Since several observations occur more thanonce, is there a way to simplify the calculations of standard deviation?

Yes, there is. Examine the table below.

Relative Frequency Method8

Observed x Frequencyf f.x fx2

1 2 2 22 3 6 12

3 2 6 18

4 1 4 166 1 6 36

Sum 9 24 84

Using the formula

We can calculate the standard deviations of the sample data set {1, 2, 3, 1, 2, 3, 2, 4, 6}

as follows.

8Adapted from pages 74 and 75 in Triola, M. F., Goodman, W. M., & Law, R. (2002). Elementary statistics(2nd Canadian ed.).Toronto, ON: Addison Wesley Longman.

x

x

( )22 2 2

2 2

( )s

( 1) ( 1)

where x the sum of all x, x the sum of all x

= =

= =

n x x nfx fx

n n n n

29(84) (24)s 1.58113883

9(9 1)

= =


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Traditional Method

Sample data Mean Deviationfrom Mean

DeviationSquared

1 2.666667 -1.666667 2.77777782 2.666667 -0.666667 0.4444444

3 2.666667 0.3333333 0.1111111

1 2.666667 -1.666667 2.7777778

2 2.666667 -0.666667 0.4444444

3 2.666667 0.3333333 0.1111111

2 2.666667 -0.666667 0.4444444

4 2.666667 1.3333333 1.7777778

6 2.666667 3.3333333 11.111111

Sum = 24

Mean = 24/9 = 2.6667

Sum of Squared Deviations = 20

Estimated variance = 20/(9-1) = 2.5

Estimated standard deviation=

= 1.5811388

Note that the relative frequency method provides the same result as the traditional method.

More on Standard Deviation

On page 378 of the textbook, the standard deviation of the return on a portfolio consisting

of two stocks is given as . In the equation below, xiis weight and CORR

= correlation coefficient.

This is getting awkward. The correlation coefficient (CORR) is illustrated but not defined.

The correlation coefficient is calculated using the formula

Where

R1 = the series of return on stock 1

R2 = the series of return on stock 2, and

n = the number of data points in the series.

2.5

( )1 / 2

2 2p p p = =

1 2

2 2 2 21 1 2 2 1 2 1 22 CORR= + +p x xx x x x

( ) ( )

1 2 1 2

2 22 2

1 1 2 2

n R R R RCORR =

R R R R

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