Xrfgtjtk

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

$F(T,X_T) = F(t,X_t) + int^T_t big(frac{partial F}{partial t}(s,X_s) + mathcal{A}F(s, X_s)big)ds + int^T_tsigma(s, X_s)frac{partial F}{partial x}(s,X_s)dB_s$
(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?

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

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

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:

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