# joint distribution of random vector

I want to find the joint distribution of the random vector $(W_t, \int_0^t W_s \; \mathrm ds)$

where $W_t$ is Brownian motion. I know $W_t \sim N(0,t)$, but I don't know how to calculate the distribution of the integral

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In this question math.stackexchange.com/questions/27841/… it was shown that the second coordinate of your vector is a Gaussian Process, so this two Gaussian Process have multivariated normal distribution. See the link for more details. –  Leandro Oct 22 '11 at 7:22
@Leandro, the second coordinate of your vector is a Gaussian Process, so this two Gaussian Process have multivariated normal distribution... No, the distributions of the coordinate processes are not enough. –  Did Dec 16 '11 at 23:17

For every $t\geqslant0$, let $Y_t=\displaystyle\int_0^tW_s\mathrm ds$ and $Z_t=(W_t,Y_t)$. One calls $(Y_t)_{t\geqslant0}$ the Langevin process and $(Z_t)_{t\geqslant0}$ the Kolmogorov process. The Langevin process is not a Markov process but the Kolmogorov process is a diffusion, with generator $\frac12\partial^2_{xx}+x\partial_y$. For every given $t$, the distributions of $W_t$ and $Y_t$ do not fully determine the distribution of $Z_t$ (despite what one might be led to believe by a formulation in your question).
When $Z_0=(0,0)$, the joint distribution of $Z_t$ is $$\mathrm d\mathrm P_{Z_t}(x,y)=\frac{\sqrt3}{\pi t^2}\exp\left(-\frac6{t^3}y^2+\frac6{t^2}xy-\frac2tx^2\right)\,\mathrm dx\,\mathrm dy.$$ In particular, $Z_t$ is centered gaussian${}^1$ with covariance matrix $$\mathrm E(Z_t^TZ_t)=\begin{pmatrix}t & \frac12t^2 \\ \frac12t^2 & \frac13t^3\end{pmatrix}.$$ To fully characterize the distribution of the Kolmogorov process, thanks to the semi-group property one only needs to know the distribution of $Z_t$ for every starting point $Z_0=(x_0,y_0)$. The formulas for a general $(x_0,y_0)$ are analogous to the one given above for $(0,0)$ and are written down in Emmanuel Jacob's recent PhD thesis (page 19), for example.
Note (1): Recall that the density probability of a 2D centered gaussian vector with positive covariance matrix $C$ is proportional to $\exp(-\frac12(x,y)\cdot C^{-1}\cdot(x,y)^T)$. In the case above, one must find the unique symmetric $2\times2$ matrix $C$ such that, for every $(x,y)$, $$(x,y)\cdot C^{-1}\cdot(x,y)^T=\frac{12}{t^3}y^2-\frac{12}{t^2}xy+\frac4tx^2,\quad\text{hence}\ C^{-1}=\begin{pmatrix}\frac4t & -\frac6{t^2} \\ -\frac6{t^2} & \frac{12}{t^3}\end{pmatrix},$$ and the inverse of $C^{-1}$ is $C=\mathrm E(Z_t^TZ_t)$.
+1. The process $Z_t$, I believe, appears in the Kloeden-Platen-Schurz method of numerical integration of scalar SDEs. Thanks for the thesis link. –  Sasha Oct 22 '11 at 9:47
How can you assume the covariance to be $\frac{1}{2}t^2$ since we don't even know the correlation here? –  user18079 Oct 23 '11 at 4:48
@Steven Ap: We know thata $X_t$ and $Y_t$ are correlated, because they are solutions to differential equations driven by the same white noise. Using notations of Didier's answer, $\mathrm{d} (X_t, Y_t) = (0, X_t) \mathrm{d} t + (1, 0) \mathrm{d} W_t$ with initial conditions $X_0 = 0$ and $Y_0 = 0$. –  Sasha Oct 23 '11 at 5:30