2
$\begingroup$

Consider the following boundary value problem in the domain $[0,T]$ x $R$ for an unknown function F.

$\frac{\partial F}{\partial t}(t,x) + \mu(t,x)\frac{\partial F}{\partial x}(t,x) + \frac {1}{2}\sigma^2(t,x)\frac{\partial^2 F}{\partial t^2}(t,x) = 0$

$F(T,x) = \Phi(x)$

$\Phi, \mu, \sigma$ are assumed to be known functions.

Derive a stochastic representation formula for this problem. Make sure it is clear at which points the functions should be evaluated.

So this is how I think you do this, but I need some help understanding the steps.

We first assume that it actually exists such stochastic representation that is the solution to the SDE

$dX_s = \mu(t,X_s)ds + \sigma(t,X_s)dB_s$

$X_t = x$

And the infinitesimal generator $\mathcal{A}$ of X is

$\mu(t,x)\frac{\partial F}{\partial x}(t,x) + \frac {1}{2}\sigma^2(t,x)\frac{\partial^2 F}{\partial t^2}(t,x)$

so we can rewrite the the PDE as

$\frac{\partial F}{\partial t} + \mathcal{A}F(t,x) = 0$ (or should it be a minus sign)

$F(T,x) = \Phi(x)$ $(\star)$

And now we apply the Itô formula on $F(s,X_s)$ and this step I don't understand (if someone could explain it I would be very pleased, I know the Itô formula but on this problem I don't get it), but if I'm correct it we get

$F(T,X_T) = F(t,X_t) + \int^T_t \big(\frac{\partial F}{\partial t}(s,X_s) + \mathcal{A}F(s, X_s)\big)ds + \int^T_t\sigma(s, X_s)\frac{\partial F}{\partial x}(s,X_s)dB_s$ (where the ds-integral is $0$ and $F(T,X_t) = \Phi(X_T)$ by assumption)

So now we take expectations on both sides and we get:

$E_{t,x}[\Phi(X_T)] = F(t, x) + E_{t,x}[\int^T_t \sigma(s,X_s)\frac{\partial F}{\partial x}(s,X_s)dB_s]$

Where the integral is $0$ if $\sigma(s,X_s)\frac{\partial F}{\partial x}(s,X_s)$ is sufficiently nice.

So we then have our stochastic representation of $F(t,x) = E_{t,x}[\Phi(X_T)]$

So, how do you apply the Itô formula on $(\star)$? Also I'm a bit confused if it should be a minus sign at $(\star)$ aswell, I think it should? Is this the Kolmogorovs backward equation? And if you instead have a initial condition in the PDE you get the foward equation? I think we should also be able to this if you add some nice function to the PDE, say $r(x)$. How would that change the derivation of this stochastic representation?

  • 0
    My idea was to put $E_{t,x}[F(s,X_s)] = m(s)$ so we have that, after differentiating, $m'(t) = r m(t)$ so that means that $m(t) = e^{rt}$. But then the solution should be that $F(t,x) = e^{-rs}E_{x,t}[\Phi(X_T)]$ and that is not what I get from this.2012-12-11

1 Answers 1

2

I use the notation $\frac{\partial}{\partial t}F = F_t$.

$dF = F_t\,dt + F_x\,dX_t + F_{xx}\frac{1}{2}\sigma\,dt$

Assume that your original PDE is equal to $rF$ instead of 0 (set $r$ to 0 to get the original PDE).

Put $Z(s) = e^{-\int_t^s r\,du}F(s, X(s))$. Then

$dZ(s) = -re^{-\int_t^s r\,du}Fds + e^{-\int_t^s r\,du}dF = e^{-\int_t^s r\,du}(dF - rFds)$

So, $Z(T) - Z(t) = \int_t^T dZ(s)$ and

$Z(T) = Z(t) + \int_t^T e^{-\int_t^s r\,du}(F_t + F_x\mu + F_{xx}\frac{1}{2}\sigma -rF)\,ds + \int_t^T e^{-\int_t^s r\,du}F_x\sigma\,dW(s)$.

The time integral vanishes. Then, since the expectation of the stochastic integral is zero, the expected value of $Z(T)$ is:

$E_t[Z(T)] = E_t[Z(t)] = e^{-\int_t^t r\,ds}F(t, X(t)) = F(t, x)$.

On the other hand, $Z(T)$ is defined as $e^{-\int_t^T r\,ds}F(T, X(T)) = e^{-\int_t^T r\,ds}\Phi(X(T))$ so we conclude:

$E_t[e^{-\int_t^T r\,ds}\Phi(X(T))] = F(t, x)$.