QFFM ASSIGNMENT

QFFM Assignment

QFFM Assignment

QUESTION 1

The managing partner for Westwood One Investment Managers Inc. gave a public seminar in

which she discussed a number of issues, including investment risk analysis. In that seminar, she

reminded people that the coefficient of variation can often be used as a measure of risk of an

investment. To demonstrate her point of view, she used two hypothetical stocks as examples. She

let x equal the change in assets for a $1,000.00 investment in stock1 and y reflect the change in

assets for a $1,000.00 investment in stock2. She showed the seminar participants the following

probability distributions:

x

P(x)

y

P(y)

-$1,000

0.1

-$1,000

0.2

$0

0.1

$0

0.4

$500

0.3

$500

0.3

$1,000

0.3

$1,000

0.05

$2,000

0.2

$2,000

0.05

a. Compute the expected values for the random variables x and y. (2 marks)

b. Compute the standard deviations for the random variables x and y.(3 marks)

c. Compute the coefficient of variation for each random variable. (2 marks)

d. Referring to part c, suppose the seminar director said that the first stock was riskier since

its standard deviation was greater than the standard deviation of the second stock. How

would you respond to her assertion? (3 marks)

Solution:

EXPECTED VALUES FOR THE RANDOM VARIABLES 'X' & 'Y'

The managing partner for Westwood One Investment Managers Inc. used two hypothetical stocks to explain that the coefficient of variation can be used as a measure of risk of an investment. These two stocks are 'x' & 'y'. The random variables 'x' & 'y' as denoted by her is,

X= Change in assets for a $1000.00 investment in Stock 1

Y= Change in assets for a $1000.00 investment in Stock 2

According to the given probability distribution, the expected value of 'x' is given by,

E(X) = x. P(x)

= (-1000)(0.1) + (0)(0.1) + (500)(0.3) + (1000)(0.3) + (2000)(0.2)

E(X) = $750

Similarly, the expected value for the random variable 'y' is given by,

E(Y) = y. P(y)

= (-1000)(0.2) + (0)(0.4) + (500)(0.3) + (1000)(0.05) + (2000)(0.05)

E(Y) = $100

SOLUTION # 1(b): STANDARD DEVIATIONS FOR THE RANDOM VARIABLES 'X' & 'Y'

As we know that,

Variance(x) = = E

= E( - 2x + )

= E) 2.E(x) + = E) 2. + = E() 2 + = E()

Variance(x) = = E()

Therefore, standard deviation is the root of the variance of 'x' and can be written as,

= E()

We need to find E(),

E() = . = (0.1) + 0 + (0.3) + (0.3) + (0.2)

= 100000 + 0 + 75000 + 300000 + 800000

E() = $ 1275000

Now, = [1275000 ]

= (712500)

= $844.09

Similarly, standard deviation of the random variable 'y' is given by,

= E() And, E() = . = (0.2) + 0 + (0.3) + (0.05) + (0.05)

= 200000 + 0 + 75000 + 50000 + 200000

E() = $ 525000

Therefore, = [525000 = (515000)

= $ 717.63

COMMENTS: Stock 1 has greater variation because it is spread out more from its average value than Stock 2 having less variation than Stock 1.

SOLUTION# 1(c): COEFFICIENT OF VARIATION FOR EACH RANDOM VARIABLE

When we are having two or more than two stocks to be compared, we use coefficient of variation as a measure of risk involve in investment made for ...