Pricing a Call on the Squared Stock Price with Feynman–Kac
Summary
The document explains how to formulate the price and pricing PDE for a European payoff that is positive when the square of a stock price exceeds a strike. It presents two routes: apply Feynman–Kac directly to the stock price with terminal payoff equal to the positive part of its square minus the strike, or transform the state variable to the squared stock price and derive its stochastic differential equation using Itô’s lemma.
Under the transformed variable, the answer gives drift and diffusion terms and substitutes them into a Black–Scholes-style PDE with a terminal payoff defined on the squared price. The discussion is a forum response rather than a complete derivation or numerical valuation. Its displayed equation uses the drift parameter and includes a possibly inconsistent right-hand-side notation; in risk-neutral pricing, the valuation measure and discounting convention must be specified carefully before using such a PDE.
Key ideas
- Feynman–Kac links the discounted expected terminal payoff to a pricing PDE.
- The terminal condition is the positive part of the squared stock price minus the strike.
- Itô’s lemma gives a stochastic differential equation for the squared stock price.
- Changing to the squared price as the state variable produces a Black–Scholes-style PDE with transformed coefficients.
- The pricing measure and discounting convention must be clear when applying the formula.
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Full text
# Mark Joshi, The concepts and practice of mathematical finance chapter 6 exercise 6
# Mark Joshi, The concepts and practice of mathematical finance chapter 6 exercise 6
> 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 by Lipton (score 4, accepted)
https://quant.stackexchange.com/a/37795
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?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.