Mark Joshi, The concepts and practice of mathematical finance chapter 6 exercise 6 [duplicate]

Question

Suppose a stock allows a geometric Brownian motion in a Black-Scholes world. Develop an expression for the price of an option that pays $S^2 - K$ if $S^2 > K$ and zero otherwise. What PDE will this option satisfy?

We have $$dS_t = \mu S_tdt + \sigma S_t dW_t$$ and $F_T(t) = e^{r(T-t)}S_t$. I am lost where to go, unless we are in the risk-neutral measure world I can go on to show that

$$dF_T(t) = \sigma F_T(t)dW_t$$

But I cannot follow Joshi's solution to this, any suggestions are appreciated. The suggested post for pricing the square or nothing option does not seem the same as what I am asking for here.

Answer 1

I don't have Joshi's book with me. But I guess you can use Feynman-Kac, right?

It says if X(t) follows the stochastic differential equation:

$dX(u) = \beta(u, X(u))du + \gamma(u, X(u))dW(u)$

and $g(t,x) = E^{t,x}[e^{-r(T-t)}h(X(T))]$, then $g(t,x)$ follows:

$g_t(t,x) + \beta(t, x) g_x(t, x) + \frac{1}{2}\gamma^2(t,x)g_{xx}(t,x) = rf(t,x)$

One way is that you can solve the above PDE with the terminal condition $g(T,x) = (x^2-K)^+$.

Alternatively, I can rewrite $g(t,x)$ as a function of $x^2$:

$g(t,x) = E^{t,S^2}[e^{-r(T-t)}(S^2-K)^+] = E^{t,x}[e^{-r(T-t)}(x-K)^+]$ where $x = S^2$. Then I only need to figure out what's the SDE for $S^2$, which is:

$d(S^2) = 2SdS + dS^2 = 2S^2(\mu + \frac{1}{2}\sigma^2)dt+\sigma S^2dW_t$

With $x = S^2$, the above SDE becomes:

$dx = 2x(\mu + \frac{1}{2}\sigma^2)dt+\sigma xdW_t\Rightarrow \frac{dx}{x} = 2(\mu + \frac{1}{2}\sigma^2)dt+\sigma dW_t$

Then the PDE becomes:

$g_t(t,x) + 2(\mu + \frac{1}{2}\sigma^2)x g_x(t, x) + \frac{1}{2}\sigma^2 x^2 g_{xx}(t,x) = rf(t,x)$

So it's the same PDE as Black-Scholes but with different $\mu$ and $\sigma$. Does it make sense?