Black and Scholes pricing

Question

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?

Answer 1

At least two ways to price this:

1. Use Carr-Madan
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:

3. 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.

EDIT3: Details to answers:

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

Ad 1. Carr-Madan:

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).

Ad 2. Power numeraire:

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 $1$.

Ad 3. Put-call symmetry

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.