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## Saturday, December 14, 2013

### a contract paying log S_T or S^2

Background - It’s easy to learn BS without knowing how to price simpler contracts. As show in the 5 (or more) examples below, there are only a few simple techniques. We really need to step back and see the big picture.

Here’s a very common pricing problem. Suppose IBM stock price follows GBM with u and σ. Under RN, the drift becomes r, the bank account’s constant IR (or probably the riskfree rate), therefore, Given a (not an option) contract that on termination pays log ST, how much is the contract worth today? Note the payoff can be negative.

Here’s the standard solution --
1) change to RN measure, but avoid working with the discounted price process (too confusing).
2) write the RN dynamics as a BM or GBM. Other dynamics I don’t know how to handle.

Denote L:= log St and apply Ito’s

dL = A dt + σ dW ... where A is a time-invariant constant. So log of any GBM is a BM.

I think A = r – 0.5σ2 but the exact formula is irrelevant here.

3) so at time T, L ~ N(mean = L0 + A*T, std = ...)
4) so RN-expectation of time-T value of L is L0 + A*T
5) discount the expectation to PV

Note L isn’t the price process of a tradable, so below is wrong.
E (LT/ BT) = L0/ B0   ... CANNOT apply martingale formula
-- What if payout = ST2 ? By HW4 Q3a the variable Jt:=St2 is a GBM with some drift rate B and some volatility.  Note this random process Jt is simply derived from the random process St. As such, Jt is NOT a price of any tradable asset .
Expectation of J’s terminal value = J0 exp(B*T)

I guess B = 2r + σ2 but irrelevant here.

 if Jt were a price process, then the discounted value of it would be martingale i.e 0 drift rate. Our Jt isn't martingale. It has a drift rate, but this drift rate isn't equal to the risfree rate. Only a tradable price process has such a drift rate. To clear the confusion, there are common 3 cases
1) if Jt is a price process (GBM or otherwise), then under RN measure, drift rate in it must be r Jt. See P5.16 by Roger Lee
2) if Jt is a discounted price process, then under RN measure, drift rate is 0 -- martingale.
3) if not a price process, then under RN measure, drift rate can be anything.

-- What if payout = max[0, (ST2) - K]?  This requires the CBS formula.
-- What if payout = max[0, (logST) - K]? Once you know the RN distribution of logST is normal, this is tractable.
-- what if payout = max[0, ST - K] but the stock pays continuous dividend rate q? Now the stock price process is not a tradeable.

No We don’t change the underlier to the tradeable bundle. We derive the RN dynamics of the non-tradeable price S as

dS = (r-q) S dt + σ S dW ... then apply CBS formula.

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So far all the “variables” are non-tradeable, so we can’t apply the MG formula

-- What if payout = STXT where both are no-dividend stock prices. Now this contract can be statically replicated. Therefore we take an even simpler approach. Price today is exactly S0X0