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I'd like to solve for the pdf of position $$P(x,t) = \Big\langle \delta\Big(x-\int_0^t dt_1 \int_0^{t_1}dt_2 \xi(t_2)\Big)\Big\rangle $$ for the second order Brownian motion given by a Langevin-type equation $$ \ddot{x}(t) = \xi(t),$$ where $\xi(t)$ is a Gaussian white noise with correlation function $ \langle \xi(t)\xi(s)\rangle = 2D\delta(t-s).$

This does not seem as if it'd be a complex problem, but I have not been able to find references on it. Any advice would be appreciated, including mathematical advice or references. In particular I'd like to understand how to derive the PDE governing $P(x,t)$. I suppose I'd be able to solve it once I had it.

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Regarding the joint distribution of position and velocity:

From my previous answer we know the variance $E[(\int_0^tW(u)\,du)^2]=t^3/3\,$ of the position. It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $$ E[\textstyle(\int_0^tW(u)\,du) W(t)]=t^2/2\,. $$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, the joint PDF of position and velocity is \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{\sqrt{3}}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{\sqrt{3}}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray} This function satisfies the Kolmogorov PDE $$ \partial_t P=-v\,\partial_x P+\frac{1}{2}\partial_{vv}P $$ for the diffusion process $$ \left(\begin{array}{c}dx\\dv\end{array}\right)=\left(\begin{array}{c}v\\0\end{array}\right)\,dt+\left(\begin{array}{cc}0&0\\1&0\end{array}\right)\left(\begin{array}{c}dW\\dW_v\end{array}\right) $$ where $W_v$ is a dummy BM that is not driving anything.

The Book of Karatzas and Shreve (Brownian Motion and Stochastic Calculus) is very useful for such problems.

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I would suggest using the Fourier transform: $$ \delta(x)=\int\frac{dk}{2\pi}e^{ikx} $$ therefore $$ P(x,t) = \left\langle \int\frac{dk}{2\pi}e^{ik\left[x-\int_0^tdt_1\int_0^{t_1}dt_2\xi(t_2)\right]}\right\rangle= \int\frac{dk}{2\pi}e^{ikx}\left\langle e^{-ik\int_0^tdt_1\int_0^{t_1}dt_2\xi(t_2)}\right\rangle $$ (The order of integration and averaging can be interchanged.)

Now $$ z= \int_0^tdt_1\int_0^{t_1}dt_2\xi(t_2) $$ is a Gaussian random variable, for which we can use the identity $$ \langle e^{\alpha z}\rangle = e^{\frac{\alpha^2\sigma_z^2}{2}} $$ where $\sigma_z$ can be easily calculated. The final Fourier transform is will be a Gaussian integral.

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    $\begingroup$ Thanks @Roger ! This looks great. I was playing with this but could not figure it out with the additional integral in $z$. Recognizing that the entire integral is a gaussian random variable is the key. $\endgroup$ Jun 21, 2021 at 17:03
  • $\begingroup$ I am curious though about the joint distribution of position and velocity. In this case from an analogous method should I start from $P(x,v,t) = \langle \delta(x-\int_0^t dt_1 \int_0^{t_1} dt_2 \xi(t_2)) \delta(v-\int_0^t dt_1 \xi(t_1))\rangle$ and take fourier transforms over both $x$ and $v$? Or am I misunderstanding how to write the joint distribution as an expectation of delta functions? $\endgroup$ Jun 21, 2021 at 17:05
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    $\begingroup$ @kevinkayaks indeed, an integral is a sum, whereas a sum of gaussian random variables is a gaussian random variable. $\endgroup$ Jun 21, 2021 at 17:06
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By definition of white noise, the integral (velocity) $$ W(t)=\int_0^t\xi(u)\,du $$ is a standard Brownian motion. The double integral in your $P(x,t)$ formula is then (position) $$ \int_0^tW(u)\,du\,. $$ It is known that this integral is normaly distributed. It has the variance $t^3/3$:

\begin{eqnarray*} \textstyle E\Big[(\int_0^tW(u)\,du)^2\Big]&=&\textstyle E\Big[(\int_0^tW(u)\,du)(\int_0^tW(v)\,dv)\Big]=E\Big[\int_0^t\int_0^tW(u)W(v)\,du\,dv\Big]\\ &=&\textstyle\int_0^t\int_0^t\min(u,v)\,du\,dv =\textstyle\int_0^t\Big(\int_0^vu\,du+\int_v^tv\,du\Big)\,dv\\ &=&\textstyle\int_0^t\frac{v^2}{2}+vt-v^2\,dv =\textstyle\frac{t^3}{6}+\frac{t^3}{2}-\frac{t^3}{3}=\frac{t^3}{3}\,. \end{eqnarray*} This gives us the density of position as the density of a normal distribution with variance $t^3/3\,$: $$ P(x,t)=\frac{\sqrt{3}}{\sqrt{2\pi t^3}}\exp\Big(-\frac{3x^2}{2t^3}\Big)\,. $$ For an arbitrary diffusion coefficient $D$ the formulas are easily modified.

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  • $\begingroup$ I am not sure it's a D dimensional problem.. probably D is just the diffusion coefficient. $\endgroup$
    – Quillo
    Jun 20, 2021 at 11:43
  • $\begingroup$ In that case, even simpler. We can drop the index $i$ and have a variance of $D^2t^3/3$ of the position $D\int_0^tW(s)\,ds\,.$ Same calculation. $\endgroup$
    – Kurt G.
    Jun 20, 2021 at 17:02
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    $\begingroup$ Yes, 1D Kurt - $D$ is just a diffusivity. Regardless the variance is not challenging, and the velocity distribution satisifies a diffusion equation $ \partial_t P(v,t) = \frac{\partial^2}{\partial v ^2} P(v,t)$, but what equations do the position distribution or the joint distribution of position and velocity satisfy? $\endgroup$ Jun 20, 2021 at 18:37
  • $\begingroup$ $E[W(u)W(v)] = \delta(u-v)$ by definition, does it not ? This gives $E(x^2) = t^3/2$, not $t^3/3$. $\endgroup$ Aug 17, 2021 at 21:54
  • $\begingroup$ Per your earlier answer here physics.stackexchange.com/questions/653279/… you simply forgot to eat up one of the integrals between writing $E(W(u)W(v))$ and $\min(u,v)$ if I am correct. $\endgroup$ Aug 17, 2021 at 22:03

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