4
$\begingroup$

I have the following problem right now: solve $$F_t(t,x) + rxF_x(t,x) + \frac{\sigma^2}{2}F_{xx}(t,x) = rF(t,x), \\ F(T,x) = (x - K)^2.$$

How do I solve this?

There exists a theorem to solve this, which represents $F(t,x)$ as the discounted expected value of $(X - k)^2$ where $dX = rXds + \sigma X dW$ and with $X_t = x$.

However, my issues are mainly centered around how messy my calculations get. If I try to solve the above by first calculating $X_s$ since it satisfies a GBM, and then calculating its expectation after first squaring it in the way I need to $(X - k)^2$, the calculations get incredibly messy, and all I get is a whole lot of $\exp$'s to a bunch of powers that vaguely resemble one another but it still doesn't simplify easily.

Have I made a mistake, or is the answer really that messy?

$\endgroup$
4
$\begingroup$

Martingale Approach

As you noted, you need to solve

\begin{eqnarray} F(0) & = & e^{-r T} \mathbb{E} \left[ \left( X_T - K \right)^2 \right]\\ & = & e^{-r T} \left( \mathbb{E} \left[ X_T^2 \right] - 2 K \mathbb{E} \left[ X_T \right] + K^2 \right) \end{eqnarray}

Let $Y_t = X_t^2$. Then, by applying the Itō formula, we get

\begin{eqnarray} \mathrm{d}Y_t & = & 2 X_t \mathrm{d}X_t + \mathrm{d} \langle X \rangle_t\\ & = & \left( 2 r + \sigma^2 \right) Y_t \mathrm{d}t + 2 \sigma Y_t \mathrm{d}W_t. \end{eqnarray}

It follows that

\begin{eqnarray} \mathbb{E} \left[ X_T \right] & = & X_0 e^{r T},\\ \mathbb{E} \left[ X_T^2 \right] & = & X_0^2 e^{\left( 2 r + \sigma^2 \right) T}. \end{eqnarray}

Consequently,

\begin{equation} F(0) = e^{-r T} \left( X_0^2 e^{\left( 2 r + \sigma^2 \right) T} - 2 K X_0 e^{r T} + K^2 \right). \end{equation}

PDE Approach

Alternatively, you can apply a change of variables to the PDE. Define

\begin{equation} \tau = (T - t), \quad \xi = r - \frac{1}{2} \sigma^2, \quad y = \ln(x) + \xi \tau, \quad F(t, x) = e^{-r \tau} G(\tau, y). \end{equation}

After carefully applying the chain rule, we obtain the PDE

\begin{equation} \frac{\partial G}{\partial \tau} = \frac{1}{2} \sigma^2 \frac{\partial^2 G}{\partial y^2} \end{equation}

with initial condition

\begin{equation} G(0, y) = \left( e^y - K \right)^2. \end{equation}

The fundamental solution is the heat kernel given by

\begin{equation} \phi(\tau, y) = \frac{1}{\sqrt{2 \pi \sigma^2 \tau}} \exp \left\{ -\frac{y^2}{2 \sigma^2 \tau} \right\}. \end{equation}

We obtain $G(\tau, y)$ through the convolution

\begin{eqnarray} G(\tau, y) & = & \int_{-\infty}^\infty G(0, z) \phi(\tau, y - z) \mathrm{d}z\\ & = & \int_{-\infty}^\infty \left( e^{2 z} - 2 K e^z + K^2 \right) \phi(\tau, y - z) \mathrm{d}z. \end{eqnarray}

Next note that for some $\alpha \in \mathbb{R}$,

\begin{eqnarray} \int_{-\infty}^\infty e^{\alpha z} \phi(\tau, y - z) \mathrm{d}z & = & e^{\alpha y + \alpha^2 \sigma^2 \tau / 2} \end{eqnarray}

Thus

\begin{equation} G(\tau, y) = e^{2 y + 2 \sigma^2 \tau} - 2 K e^{y + \sigma^2 \tau / 2} + K^2 \end{equation}

and substituting back yields

\begin{equation} F(0, x) = e^{-r T} \left( x^2 e^{\left( 2 r + \sigma^2 \right) T} - 2 K x e^{r T} + K^2 \right) \end{equation}

as before.

$\endgroup$
  • $\begingroup$ Does this satisfy the condition of integrability that the used result takes as given? $\endgroup$ – Rama Jan 22 '17 at 20:11
  • $\begingroup$ Do you mean if the payoff function satisfies the integrability condition such that the Feynman-Kac theorem can be applied? Yes it does. $\endgroup$ – LocalVolatility Jan 22 '17 at 21:02

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.