2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ $h(S_T)=S_T^3 - S_T^2$ is not equivalent to $(S_T-1)^+.$ Take for example $S_T = 2.$ Then $h(2)=2^3 - 2^2 = 4 \neq 1 = (2 - 1)^+.$ $\endgroup$ – UBM Jan 11 at 15:17
  • $\begingroup$ @UBM thank you! $\endgroup$ – Finance Student Jan 11 at 16:52
6
$\begingroup$

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:

  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.

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ @ilovevalatility , are you saying as per your method 2, my what I did is correct? $\endgroup$ – Finance Student Jan 11 at 15:43
  • $\begingroup$ 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 $r$. $\endgroup$ – ilovevolatility Jan 11 at 15:55
  • $\begingroup$ @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. $\endgroup$ – Gordon Feb 11 at 16:52
  • $\begingroup$ @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. $\endgroup$ – Idonknow Jun 10 at 14:28
  • $\begingroup$ @Gordon finally partially answered your question. $\endgroup$ – ilovevolatility Jun 10 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.