Welcome to the hotseat. We've prepared a guide if you'd like to read more about how it works.

Tut 12 question 5(b)

0 votes
asked Oct 18 in BUS 2016H - Financial Mathematics by Ishrko (130 points)

Can someone please help for this question. I understand it is an application of the dividend yield case but I don't quite understand the logic behind it.

2 Answers

0 votes
answered Oct 31 by Muzammil_Jable (140 points)

Question 5 considers two cases regarding how income on the security is earned. 

(a) At two distinct time points, and, 

(b) Annually in arrear throughout the term of the contract. The question states that this dividend is earned on the market price of the security at that time. As such, dividend income is calculated on the growth of the security's value over time (which growing at 3%).  

Note that this dividend yield is at 2% per annum effective and so,

The forward price at the outset of the contract is 

K0 = 95×1.03^(12) ×1.02^(−12) 

Why is the income factor raised to the negative exponent, you might ask?

Intuitively, you can think of this as the income the short party receives until maturity which you as the buyer won't get. Thus the growth factor and hence forward price payable by the buyer is offset. 

The forward price now is:

K5 = 145×1.03^(7) ×1.02^(−7) 

Therefore the value of the forward contract now is:

(K5 − K0) v^(7) = (145 × 1.03^(7) × 1.02^(−7) − 95 × 1.03^(12) × 1.02^(−12) )  * v^ (7) = 145 × 1.02^(−7) − 95 × 1.03^(5) × 1.02^(−12) = 126.23 − 86.84 = R39.39.

commented Nov 17 by Ishrko (130 points)

Thank you. 
The income factor was the confusing part.

0 votes
answered Nov 3 by Kyle J van Vuuren (150 points)
The security pays dividends in arrears (not continuously), with a dividend yield D=2% p.a. The same logic applies as if it were the continuous case, however instead of using \(e^{nD}\) we use \((1+D)^n\). This is similar to interest paid continuously versus interest paid yearly.

In General we have:
At time 0 we borrow \(S(1+D)^{-(T-t)}\), and then sell at time T at price K.
Now with no arbitrage we have \( K- S(1+D)^{-(T-t)}(1+i)^{(T-t)} = 0\).
\( K=\frac {S(1+i)^{(T-t)}}{(1+D)^{(T-t)}} \)

Now to answer Question 5(b) it is just a matter of substituting in the relevant amounts.