Question

The stocks in companies A and B both cost 100 USD today. We let

X

and

Y

denote

the stock price in the two companies one year from now. We assume that X and Y are independent

random variables with

E

[

X

] = 110

,

E

[

Y

] = 110

,V ar

[

X

] = 100

,V ar

[

Y

] = 400. We want to invest 10

million USD in the two stocks. Let p denote the fraction invested in company A.

a. Write down the equation for the value of the portfolio one year from now.

b. What is the expected value of portfolio and its variance one year from now?

Answer 1

Given stocks in company A and B both cost 100 USD today , here let X

and Y denote the stock price in the two companies one year

assume that X and Y are independent

random variables with

E[X] = 110

E[Y] = 110

,Var[X] = 100

,Var[Y] = 400.

Now E[aX] = aE(X) , E(bY) = bE(Y)..........................(1)

E(aX + bY) = aE(X) + bE(Y)................................(2)

V(aX) = a²V(X); V(bY) = b²V(Y); V(aX + bY) = a²V(X) + b²V(Y) + 2abC(X,Y)...……..(3)

When X and Y are Statistically Independent then

V(aX + bY) = a²V(X) + b²V(Y) ………………….………………………....................... (4)

(a)

Amount to be invested = $10 million

Stock cost = $100

Number of stocks to be invested = 10 million/100 = 0.1 million

Fraction to be invested in Company A = p

Number of stocks to be invested in Company A = 0.1 p million

Stock price of Company A one year from now = X

Therefore price of stocks to be invested in Company A one year from now = $0.1pX million

Similarly, price of stocks to be invested in Company B one year from now = $0.1(1 – p)Y million.

Then the  equation for the value of the portfolio one year from now

= ${(0.1pX) + 0.1(1 – p)Y} million

= $(0.1){pX + (1 – p)y} million

(b)

Vide (1) and (2), E[(0.1){pX + (1 – p)y}]

= 0.1[{pE(X) + (1 – p)E(y)}]

= 0.1{110p + 110(1 - p)} [given E(X) = E(Y) = 110]

= $11 million

Vide (4),

V[(0.1){pX + (1 – p)y}]

= (0.01){p²V(X) + (1 – p)²V(Y)}

= (0.01){100p² + 400(1 – p)²}

= {p² + 4(1 – p)²}

= $(5p² - 8p + 4) million²