# Randomly stopped dynamics of $x(t)$: How can I find $\text{var}\{ x(t) \}$?

Consider the simple dynamical equation

$$\dot{x}(t) = u H(t-\tau),$$ where the timescale $$\tau$$ is an exponentially distributed random variable $$\tau \sim \omega \exp\{\omega \tau\}$$ and $$H(t) = 1-\theta(t)$$. That is, $$H(t) = \begin{cases} 1 & \text{if} & t<0 \\ 0 & \text{if} & t> 0\end{cases}.$$ This describes a particle which moves with velocity $$u$$ for a random time $$\tau$$ at which point its motion is cut off.

I can integrate for its position conditional on the random variable $$\tau$$ (given $$x(0)=0$$): $$x(t) = u \int_0^t dt' H(t-\tau),$$ which gives the mean value $$\langle x(t) \rangle = u \int_0^t dt' \langle H(t'-\tau) \rangle$$ I believe I can evaluate the ensemble average over $$H(t'-\tau)$$ using the exceedance probability of $$\tau$$. Since $$H(t-\tau)=1$$ when $$t<\tau$$, $$\langle H(t-\tau) \rangle = \text{Prob}(\tau > t') = \int_{t'}^\infty \omega \exp\{-\omega \tau\} = \exp\{-\omega t' \},$$ giving mean position $$\langle x(t) \rangle = \frac{u}{\omega}(1-\exp\{-\omega t\}).$$ When the mean cut-off time $$\langle \tau \rangle = 1/\omega \rightarrow \infty$$, I obtain the expected result in absence of the cut-off: $$\langle x(t) \rangle \approx ut + O(\omega),$$ so I believe my result is correct.

Now for my questions. First, have I done anything wrong? Second, given I'm on the right track, how can I compute the second moment of the position? That is, I'm seeking $$\langle x(t)^2 \rangle$$ which I believe should be given by:

$$\langle x(t)^2 \rangle = u^2 \int_0^t dt_1 \int_0^t dt_2 \langle H(t_1-\tau) \rangle \langle H(t_2-\tau) \rangle \\ = u^2 \int_0^t dt_1 \int_0^t dt_2 \text{Prob}(\tau>t_1)\text{Prob}(\tau>t_2) \\ = \Big(u \int dt \text{Prob}(\tau > t) \Big)^2 \\ = \langle x(t) \rangle^2$$ but this is unexpected: how is the second moment just the square of the mean? This implies a zero variance... I expected a transition in the second moment when $$t$$ surpasses $$1/\omega$$, since this is the point when most particles in the ensemble will have been trapped.

Any guidance is appreciated !

Almost there. The two $$\text{Prob}$$'s are correlated: $$u^2 \int_0^t dt_1 \int_0^t dt_2 \langle H(t_1-\tau) H(t_2-\tau) \rangle = u^2 \int_0^t dt_1 \int_0^t dt_2 \text{Prob}(\tau>\max(t_1, t_2)) = \dots = \frac{2u^2}{\omega^2}\left(1-(1+\omega t)\exp\{-\omega t\}\right)$$

Which you can verify with a quick Python script if so inclined:

import numpy as np
omega = 2.5
t = 0.75
u = 4.
N = 1000000

tau = np.random.exponential(1./omega, N)
x = u * ((tau > t) * t + (tau < t) * tau)

print (np.mean(x), u/omega * (1 - np.exp(-omega*t)))
print (np.mean(x**2), 2 * u**2/omega**2 * (1 - (1+omega*t) * np.exp(-omega*t)))

• Excellent @alarge ! Thank you. This makes perfect sense. – kevinkayaks Jun 25 at 21:53
• I am left with some confusion though: In this case the variance $\text{var}\{x(t)\} = \langle x(t)^2\rangle - \langle x(t) \rangle^2$ doesn't seem to make sense. It seems to become negative? Am I missing something? – kevinkayaks Jun 25 at 23:45
• @kevinkayaks Can you type what you got for the expression? Just substituting I have $\frac{2u^2}{\omega^2}\exp\{-\omega t\}(\sinh \omega t - \omega t)$ and of course $\sinh x \geq x$ for non-negative $x$. – alarge Jun 26 at 0:57
• Sorry I didn't remove this in time! .. I had missed the factor of $2$ out front in $\langle x(t)^2 \rangle$, so the variance was negative. Thanks a bunch! This helps. – kevinkayaks Jun 26 at 0:59