Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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

1 Answer 1

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 $C=\mathrm E(Z_t^TZ_t)$.

share|improve this answer
    
+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
    
Jacob studies the properties of the Langevin process when reflected at some inelastic boundary, see also papers by Jean Bertoin and slightly older ones by Aimé Lachal. –  Did Oct 22 '11 at 10:13
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.