Joint Distribution of Brownian Motion and its Time Integral

Let $W(t)$ be a standard Brownian motion and let $I(t) = \int_0^t W(s) \, \mathrm{d}s$ denote its time integral. I'm interested in the joint distribution of $(W(t), \,I(t))$, but I can't seem to find any references for this. Is it unknown? I can't seem to find even a covariance calculation.

Solution 1:

Since the Brownian motion $(W_t)_{t \geq 0}$ is a Gaussian process, the random vector $(W_{t_1},\ldots,W_{t_n})$ is Gaussian for any $t_1,\ldots,t_n \geq 0$, $n \in \mathbb{N}$. This implies that the vector

$$X_n := \bigg( W_t, \sum_{j=1}^n W_{t_j} (t_j-t_{j-1}) \bigg)$$

is Gaussian for each $n \in \mathbb{N}$, $t>0$ where $t_j:= \frac{j}{n} t$. As $X_n$ converges pointwise to $(W(t),\int_0^t W(s) \, ds)$, we find that the latter is Gaussian as a pointwise limit of Gaussian random variables. Since Gaussian random vectors are uniquely determined by their mean vector and covariance matrix, it just remains to calculate $\mathbb{E}(W_t^2)$, $\mathbb{E}(I_t W_t)$, $\mathbb{E}(I_t^2)$ and to note that the mean vector equals $0$.