Extracting a conditional density from a formula involving 2 stochastic integrals

Question

I have a problem coming from a financial maths application, that involves trying to extract the conditional density of a variable expressed as an integral over a Brownian motion, conditioned on another integral over the same Brownian motion. It feels like there should be a simple result hiding in here but I can't find a way to get at it!

Consider two variables that can be written in terms of stochastic integrals over the same Brownian motion

$$ A(t)=\int_{t_{0}}^{t} f(t^{\prime})\mathrm{d}W(t^{\prime}) $$ and $$ x(t)=\int_{t_{0}}^{t} \mathrm{d}W(t^{\prime}) $$ I am looking for a formula for the conditional density $p(A(t)=A^{\prime}|x(t)=x^{\prime})$.

The function $f(t)$ is non-anticipating and can be considered as smooth as necessary (or indeed monotonic). Think exponential.

The only way I could think to proceed is to consider the discrete case and write the joint probability as $$ \int \int \ldots \int\int p(\phi_{1})p(\phi_{2})\ldots p(\phi_{N-1})p(\phi_{N}) \mathrm{d}\phi_{1}\mathrm{d}\phi_{2}\ldots\mathrm{d}\phi_{N-1}\mathrm{d}\phi_{N} $$ subject to the constraints $$ A^{\prime}=\sum_{i}f_{i}\phi_{i} $$ and $$ x^{\prime}=\sum_{i}\phi_{i} $$ where the $\phi_{i}$'s are independent normals of variance $\mathrm{d}t$

To fit the constraints, solve for $\phi_{N}$, $\phi_{N-1}$ $$ \phi_{N-1}=\frac{\sum_{i=1}^{N-2} (f_{N}-f_{i})\phi_{i} - (A^{\prime}-x^{\prime}f_{N}) }{f_{N}-f_{N-1}} $$ $$ \phi_{N}=\frac{\sum_{i=1}^{N-2} (f_{N-1}-f_{i})\phi_{i} + (A^{\prime}-x^{\prime}f_{N-1}) }{f_{N}-f_{N-1}} $$ Given that the joint probability is $$ \simeq \int\ldots\int \exp \left( -\sum_{i=1}^{N}\phi_{i}^{2} \right) \mathrm{d}\phi_{1}\ldots \mathrm{d}\phi_{N} $$ My hope was to substitute the expressions for $\phi_{N-1},\phi_{N}$ into this and hope to work it back to something that will resolve into clean sums of $\phi$'s that will look like integrals. However it doesn't look that promising.

Is there a know result that might resemble this?

Answer 1

First note that $x(t)$ is nothing but $W_t-W_{t_0}$. If $f$ is deterministic, then the process $A(t)=\int_{t_0}^t f(s)dW_s$ is Gaussian. It can be show, see for example this answer here, that the vector $$\bigg(W_t-W_{t_0},\int_{t_0}^t f(s)dW_s\bigg)$$ is Gaussian for each $t$. Hence the process $(x(t),A(t))_{t\geq t_0}$ is Gaussian and we can characterize Gaussian processes by computing their means $\mu(t)$ as well as their covariance functions. The mean of this vector should be obvious, you can calculate the variance of the components using Itô's isometry and the covariances can also be easily computed (e.g. by using the covariation of the two processes $x$ and $A$).

Hence you will get a gaussian joint density $\phi_{\mu(t),\Sigma(t)}$ of the vectors $(x(t),A(t))$ with which you can calculate the conditional densities, see e.g. here for some basic rules.