Chapter 17 Question: 3, 4 & 5

[DMF]

This is a general question with regards to the approaches used in 17.3, 17.4 and 17.5.

In questions 17.3 and 17.4(ii) the solutions use the formula on page 18 of tables (truncated moments) but the solution to 17.5 uses the formula on page 14 (i.e. the moments are not truncated). Is that because we are told that X = S^0.5 - K^0.5 - i.e. r = 0.5 hence the reliance on the formula for E[E^r] on pg14 of the tables. Whereas in the other two questions we do not know necessarily know r - although I thought it was 0 and 1 for 4(ii)(a) and (b) respectively?

Thanks,

 

[Lolgabby]

The key is note the possible values your payoff can take.
Do you agree with me that your payoff will only be valid for values of (ST> =0 to infinity) so your intergral will be from 0 to infinity. The formula on page 18 of the tables will give us a problem because log of 0 is invalid, although some clever people may make a certain assumption about the invalid Ln (0) and continue with the formula on pg18 of the tables but that assumption is quit confusing I won't talk about it here.

So after the invalid solution we see if we can use the formula on page 14 of the tables. Note that the values of x our lognormal distribution can take is (x>0) which match with (ST>0) and ST follows a lognormal distribution so we can use the rth moment of the lognormal distribution to evaluate E (ST^(1/2)).

 

[Lolgabby]

Another thing :
Please note that we do know the value of r or k in the other questions

 

[Anna Bishop]

Hello

The difference is whether the payoff on the derivative is based on the full range of values of the share price at expiry (ie ST from 0 to infinity) or just a truncated range (ie ST>K only). I'll try and explain below with a couple of examples.

All three questions (17.3, 17.4, 17.5) involve using the result of the five-step proof:

Vt = exp{-r(T-t)}EQ[XT|Ft]

where XT is the payoff on the derivative at time T.

In Q17.3(i) the payoff is: XT = ST - K if ST > K
In Q17.4(ii)(a) the payoff is: XT = 1 if ST < K
In Q17.4(ii)(a) the payoff is: XT = ST if ST < K
In Q17.5(i)(a) the payoff is: XT = ST^0.5 - K^0.5 for all values of ST

We can see that the range of values of T is truncated for Q17.3 and Q17.4 but not for Q17.5.

We can see above that the five-step proof result involves working out the expectation of the payoff. Typically, we calculate expectations of continuous random variables, by integrating, eg:

E[X] = ∫xf(x)dx

E[X^0.5] = ∫[x^0.5]f(x)dx

In Questions 17.3, 17.4, 17.5, the random variable is ST and hence we integrate over the range of values that ST can take.

In Q17.3(i), we would be integrating from K to ∞.
In Q17.4(ii)(a) and (ii)(b), we would be integrating from 0 to K.
In Q17.5(i)(a), we would be integrating from 0 to ∞.

In all three questions, ST|Ft has a logN(logSt + (r - 0.5σ^2)(T-t), σ^2)(T-t)) distribution.

This is the case, since the questions tell us that the share price follows geometric BM or the BS assumptions (of which geometric BM is one of the assumptions).

All three of the questions could be done using the formulae on page 18 of the Tables.

In Q17.3(i), we would take U = ∞ and L = K.
In Q17.4(ii)(a), we would take U = K and L = 0.
In Q17.4(ii)(a), we would take U = K and L = 0.
In Q17.5(i)(a), we would take U = ∞ and L = 0.

However, for Q17.5(i)(a), since we know we are integrating from 0 to ∞, ie over the full range of values, rather than over a range truncated at K, it is a lot quicker just to use the short-cut formulae for E[X^0.5] for the lognormal distribution on Page 14 of the Tables.

Does this make sense?

It's a useful exercise to try Q17.5(i)(a) using the Page 18 formulae instead, ie with k=0.5. Note that, in the calculation of Uk and Lk we would assume that log ∞ = ∞ and log 0 = -∞. (It's useful to draw the graph of log x vs x to see why this is case.)

Anna

 

[DMF]

Thanks Anna, that is a clear explanation and very helpful.

Above you mentioned that in Q17.5(i)(a) the payoff is: XT = ST^0.5 - K^0.5 for all values of ST. I guess this partly through me off - as I didn't appreciate that it was for all values of ST (although it is mentioned in the question). To my mind, a requirement for the pay off (XT) was that ST must be greater than K but I guess this can be ignored given the wording in the question.

Thanks again!

 

[Anna Bishop]

Hi again DMF, you are welcome!

You should find that the requirement for ST > K or ST < K is linked to a call or put option respectively or alternatively will be explicitly specified in the question.

For example, the payoff on a call option is max(ST - K, 0). In other words the payoff is ST - K if ST > K or 0 if not.
For example, the payoff on a put option is max(K - ST, 0). In other words the payoff is K - ST if ST < K or 0 if not.

If nothing is specified and there is no mention of a call or put then I would assume ST can take any (positive) value.

Anna