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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?

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It would be great if someone could tell me how the Itô integral in question is solved as soon as possible, because that I really need for the exam in 2 days!!! Many thanks in advance! –  Good guy Mike Dec 10 '12 at 23:07
    
I tried adding in the PDE so that I have: $\frac{\partial F}{\partial t}(t,x) + \mu(t,x)\frac{\partial F}{\partial x}(t,x) + \frac {1}{2}\sigma^2(t,x)\frac{\partial^2 F}{\partial t^2}(t,x) + rF(t,x) = 0$ So applying the Itô integral (which I understood now) $\frac{\partial F}{\partial t} + \mathcal{A}F(t,x) = r F(t,x)$ (instead of 0) So now I get, after taking expectation $E_{t,x}[\Phi(X_T)] = F(t, x) + E_{t,x}[\int^T_t rF(s,X_s)ds] + (\dots) = {Fubini's} = F(t, x) + \int^T_t rE_{t,x}[F(s,X_s)]ds + (\dots)$ From here I'm not certain on how to continue. –  Good guy Mike Dec 11 '12 at 12:47
    
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. –  Good guy Mike Dec 11 '12 at 12:52

1 Answer 1

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.

Put $Z_t = e^{-\int_t^T r\,ds}F$. Then

$dZ_t = -re^{-\int_t^T r\,ds}Fdt + e^{-\int_t^T r\,ds}dF$

Thus

$Z_T = Z_t + \int_t^T e^{-\int_s^T r\,du}(F_t + F_x\mu + F_{xx}\frac{1}{2}\sigma -rF)\,ds + \int_t^T e^{-\int_s^T r\,du}F_x\sigma\,dW_s$.

The time integral vanishes. Then taking expectations yields, since the expectation of the stocahstic integral is zero, that:

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

Concluding:

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

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