# Black and Scholes pricing

I want to price B&S with $$S_t$$ stock price that has payoff, $$h(S_T)=(S_T^3-S_T^2)^+$$. Would it be wrong if I solved as $$(S_T^3-S_T^2)^+\implies (S_T^3\geq S_T^2) \implies (S_T\geq 1) \implies (S_T-1)^+$$ and use regular B&S formula with $$K=1$$? I am assuming $$S_t$$ positive since exponential. However, I seem to get different answer if I used discounted method of payoff for call price, i.e. $$c_t=e^{-rt}h(S_T)$$ and put price by call-put parity. Which is correct?

• $h(S_T)=S_T^3 - S_T^2$ is not equivalent to $(S_T-1)^+.$ Take for example $S_T = 2.$ Then $h(2)=2^3 - 2^2 = 4 \neq 1 = (2 - 1)^+.$ – user39119 Jan 11 at 15:17
• @UBM thank you! – Finance Student Jan 11 at 16:52

At least two ways to price this:

2. Use $$S^2$$ as a (power) numeraire, in which case you can price the payoff $$(S_T - 1)_+$$ under the power numeraire measure.

EDIT:

1. Put-call symmetry.

Maybe I can get another -1 for my answer. Is the purpose of answering questions here to do homework for someone else or to stimulate further study and generate discussion?

EDIT2: I just saw this question has been bumped up by the community, and I notice that I was in a rather foul mood when I wrote my answer above (apologies). Please bear with me, I will write a more extended answer shortly.

Let $$dS_t = \sigma S_t dW_t$$ Generalization to deterministic carry is straightforward.

Please see here for the Carr-Madan formula. Since $$S_t$$ is strictly positive, we can write $$f(x) = (x^3-x^2)_+ = x^2 (x-1)_+$$ We therefore need to calculate $$f'(x)$$ and $$f''(x)$$. So, $$f'(x) = (3x^2 - 2x) \theta(x-1)$$ where $$\theta(\cdot)$$ is the Heaviside / step function. And $$f''(x) = (6x -2) \theta(x-1) + (3x^2 -2x)\delta(x-1)$$ with $$\delta(\cdot)$$ the Dirac delta function. Plug all this into the Carr-Madan formula and you obtain the replicating portfolio and hence the price for the claim. However, there are some discontinuities, so it's a bit messy (i.e. in theory it works, in practice there will be slippage).

Let $$N_t = S_t^2$$ which is a strictly positive process. We will use $$N_t$$ as numeraire. Its process is $$dN_t = \sigma^2 N_t dt + 2\sigma N_t dW_t$$ Hence, the process $$S_t / N_t = 1/S_t$$ satisfies $$d(S_t/N_t) = d(1/S_t) = (\sigma^2 / S_t) dt - (\sigma / S_t) dW_t$$ To find a change of measure that makes $$S_t/N_t$$ a martingale is the same as finding the change of measure that turns $$1/S_t$$ into a martingale, which is $$dW = \widetilde{dW} + \sigma dt$$ Under this new measure, $$dS = \sigma^2 S dt + \sigma S \widetilde{dW}$$ Hence, the price of the claim is $$E_t(S_T^3 - S_T^2) = S_t^2 \widetilde{E_t} (S_T - 1)_+$$ where the expectation on the right hand side is now under the power measure. But $$\widetilde{E_t} (S_T - 1)_+$$ is just the Black-Scholes formula with a drift equal to $$\sigma^2$$ and strike equal to $$1$$.

I cannot for the life of me recall or understand now what I meant :-D As my foul moods are highly correlated to brain-farts (although I am not sure what the exact causal relationship is), it is highly likely it was a brain-fart. However, to atone somewhat for my sins here is another method:

3'. Local-time method

Using the Ito-Tanaka formula, we can write $$E_t (S_T^3-S_T^2)_+ = (S_t^3-S_t^2)_+ + \frac{\sigma^2}{2} E_t \left( \int_t^T f''(S_u) S_u^2 du \right)$$ with $$f''(S_u)$$ as in $$f''(x)$$ above. I am not sure this can be simplified further - need to think about it more.

In any case, the power-numeraire method (2) appears to be the simplest among the three methods.

• @ilovevalatility , are you saying as per your method 2, my what I did is correct? – Finance Student Jan 11 at 15:43
• Not entirely. You will need to write down the dynamics of $S$ under the power measure. In other words you will need to find a Girsanov transform such that $S/S^2 = 1/S$ is a martingale. This will give a different drift for $S$ compared to the risk neutral measure. Once you have this adjusted drift you can use BS formula, but with the adjusted drift instead of $r$. – ilovevolatility Jan 11 at 15:55
• @ilovevolatility: Can you elaborate more on the Carr-Madan and Put-Call symmetry methods. In particular, I do not see how Put-Call symmetry will help as it only changes the payoff to another similar payoff. – Gordon Feb 11 at 16:52
• @ilovevolatility I am interested to see an answer of using the Girsanov theorem as my background towards applying it is not firm yet, particularly the Radon-Nikodym derivative part. I know how to use it to price European call option though. – Idonknow Jun 10 at 14:28
• @Gordon finally partially answered your question. – ilovevolatility Jun 10 at 16:46