Suppose we are in the following setting
\begin{gather}
dX_i(t)=X_i(t) \big( b_i(t)dt+ \sum \limits_{\nu=1}^{d} \sigma_{i \nu}(t) dW_{\nu}(t) \big) , \qquad X_i(0)=x_i, \ i=1, \ldots, n,
\end{gather}
with $d \geq n$, $d$-dimensionl standard Brownian motion $(W_1(\cdot), \ldots, W_d(\cdot))^{\intercal}$. Furthermore, the processes $b(\cdot)=(b_1(\cdot), \ldots, b_n(\cdot))^{\intercal}$ and $\sigma(\cdot)=(\sigma_{i \nu}(\cdot))_{1 \leq i \leq n, 1 \leq \nu \leq d}$ are progressively measurable with respect to the filtration generated by the brownian motion and fulfill the usual intergability conditions
Question: Does there exist a closed form espression for the transition density function of the process $X$? Does anyone know of a closed form expression for the simpler case where $b$ and $\sigma$ are constant? I'm interested in relatively straightforward derivations, i.e. not the one of the paper mentioned below (there is nothing wrong with it I just wonder if there is a simpler approach for the case of geometric brownian motion).
For example, in the one dimensional case, where $\sigma$ and $b \equiv 0$ are time independent we have the following transition density:
$$p(X(t),t;X(0),0)=\frac{1}{X(t)\sigma \sqrt{2\pi t}}\exp{\left(-\frac{1}{2}\left[\frac{\log(X_t)-\log(X_0)-\sigma^2 t/2}{\sigma \sqrt{t}}\right]^2\right)}$$
as seen here: Transition density of a Geometric Brownian-motion .
Explenation: I know how to derive the probability density function in the case where $n=1$, i.e., in the one dimensional case. The same approach does not work here because the components of $X$ are not independent and we can not simply write the density function as a product of the simpler density functions of the one dimensional case. There is some literature on the topic for general diffusion models, like here: https://www.princeton.edu/~yacine/multivarmle.pdf but the paper is very technical and several assumptions are made. This motivates my question about the existence of a simpler approach in the case of geometric brownian motion, or in the case where we make even the stronger restriction of considering time independent processes $b$ and $\sigma$.
Best Answer
First, a simplification. Let $Y_i(t):=\log(X_i(t))$ and $Y(t):=(Y_1(t),\ldots,Y_n(t))^T$. It is enough to obtain a transition function for $Y$. By using Ito's formula for function $\log x$, you can easily obtain the following. $$ dY_i(t)=b'_i(t)dt + \sum_j \sigma_{ij}(t)dW_j(t), $$ where $b'_i(t)=b_i(t)-\frac 12 \sum_j \sigma_{ij}(t)^2$. Since the right hand side does not depend on $Y$, you can take integral of both sides and compute $Y(t)$ explicitly as an Ito integral.
Now, assume that $b$ and $\sigma$ are constant. You can explicitly obtain $$ Y_i(t)= tb'_i + \sum_j \sigma_{ij}W_j(t). $$ Equivalently, $$ Y(t)=Y(0)+tb'+\sigma W(t). $$ Hence, $Y$ is a multivariate normal random variable with mean $Y(0)+tb'$ and covariance matrix $\sigma\sigma^T$. In the case $\sigma$ has rank $n$, this vector has a density which you can find here. If not, it does not have a density.
You can generalize this to the case where $b$ and $\sigma$ are deterministic but may depend on $t$. In this case, $Y(t)$ is normal with mean $Y(0)+\int_0^t b'(s)ds$ and covariance $\int_0^t \sigma(s)\sigma(s)^T ds$.
If $b$ and $\sigma$ are not deterministic, you cannot hope to find a general formula for transition probabilities. However, under some assumptions you can derive a differential equation for it, which is exactly what the Fokker-Planck equation does.