# 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.

• For each fixed $t$, the random vector $(W(t),I(t))$ is jointly normal with mean zero and the covariance matrix $$\begin{pmatrix} t & t^2/2 \\ t^2/2 & t^3/3 \end{pmatrix}.$$ Is this the kind of information that you are looking for? – Sangchul Lee Mar 8 '17 at 0:19
• @SangchulLee Why jointly normal? – bcf Mar 8 '17 at 0:45
• Because each $(W(t),I(t))$ is a linear transform of the gaussian family $(W(s))_{0\leqslant s\leqslant t}$. – Did Mar 9 '17 at 9:06
• Re references, the keywords integrated Brownian motion should provide a wealth of information about the process $(W,I)$. – Did Mar 9 '17 at 9:08

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 $0$.
• I know this is an old question. But I wonder how we can compute $\mathbb{E}[I_t W_t]$ ? – MathMen Oct 13 '18 at 11:03
• @MathMen Apply Fubini's theorem: $$\mathbb{E}(I_t W_t) = \mathbb{E} \left( \int_0^t W_s W_t \, ds \right) = \int_0^t \mathbb{E}(W_s W_t) \, ds.$$ – saz Oct 13 '18 at 18:48
• @saz How to deduce that the vector $X_n$ is Gaussian for each $n\in\mathbb{N}, t>0?$ – Idonknow Feb 14 at 15:08
• @Idonknow Since Brownian motion is a Gaussian process, the vector $Y:=(W_{t_1},\ldots,W_{t_n})$ is Gaussian. Linear combinations of Gaussian random vectors are Gaussian, i.e. $A \cdot Y + b$ is Gaussian for any matrix $A$ and for any vector $b$. Pick suitable $A$ and $b$ which allows you to write X_n = AY+b\$. – saz Feb 14 at 18:05