# expression for transition density of multivariate geometric brownian motion

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 $$\sigma$$.

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.