I have an object that is accelerated with acceleration $A(t)$ for a period $T$. I am interested in how far it travels, i.e. $$ s = \int^T \int^t A(\tau)\ d\tau \ dt $$
Now, $A(\tau)$ is a gaussian process with gaussian noise, i.e. it has a constant mean $a$ and variance $\sigma^2$. In other words, the intended acceleration $a$ is corrupted by uncorrelated white noise.
How might I go about solving this for the expected distance travelled $\mathbb{E}[s]$, and the variance of distance travelled $\mathbb{E}[(s-\bar{s})^2]$?
Using the examples on this site, I have done everything except for variance of the distance travelled. $$ v(\tau) = \int_0^\tau A(t) \ dt = \int_0^\tau \left(a+\sigma n(t)\right)dt $$ where each $n(t)$ is an independent normal random variable. So, at time $\tau$, $$ \mathbb{E}[v(\tau)] = \int_0^\tau a\ dt=a\tau $$
$$ Var[v(\tau)] = \mathbb{E}[(v(\tau)-\bar{v}(\tau))^2] = \sigma^2\mathbb{E}\left[ \int_0^\tau n^2(\tau)d\tau \right] $$ then because noise is independent of time, $$ = \sigma^2 \int_0^\tau\mathbb{E}\left[n^2(t) \right]= \sigma^2\tau $$ Then for distance, $$ \mathbb{E}[s] = \mathbb{E}\left[\int_0^T v(\tau)d\tau\right] = \mathbb{E}\left[ \int_0^T a\tau \ d\tau\right] = \frac{1}{2}aT^2 $$
but now, how can I obtain $Var[s]$ ? I got this far: $$ \mathbb{E}[(s-\bar{s})^2] = \mathbb{E}\left[ \left(\int_0^T v(t) \right)^2 \right] - \bar{s}^2 $$ $$ = \mathbb{E}\left[ \left( \int_0^T\int_0^\tau (a+\sigma n(t))\ dt \ d\tau \right)^2 \right]-\bar{s}^2 $$ Then $\bar{s}^2$ cancels giving $$ = \mathbb{E}\left[ 2a\sigma \int_0^T\int_0^\tau n(t) \ dt\ d\tau \right] + \mathbb{E}\left[ \sigma^2\left(\int_0^T\int_0^\tau n(t)dt\ d\tau\right)^2 \right] $$ and the left term has zero expectation, giving $$ =\sigma^2\mathbb{E}\left[ \left( \int_0^T\int_0^{\tau_1} n(t) \ dt\ d\tau_1 \right)\cdot\left( \int_0^T\int_0^{\tau_2} n(t) \ dt\ d\tau_2 \right) \right] $$ It is clear that the variance of position should increase with $T$ and $\sigma^2$. But now I'm stuck!
In general, I'm trying to do this for an arbitrary simple acceleration function $a(t)$ with multiplicative noise $\sigma^2(t)=\sigma_0^2a^2(t)$. But I can't even get the constant case to work... Any pointers for either the constant or general case are appreciated!
Actually, you've already claculated the variance of $v(\tau)$ wrong. Correctly, you have \begin{align} \big(v(\tau)-\bar v(\tau)\big)^2 &= \Big(\int_0^\tau a(t)dt - a\tau\Big)^2 = \Big(\sigma \int_0^\tau n(t)dt\Big)^2 = \\ &= \sigma^2 \int_0^\tau \int_0^\tau n(t_1)n(t_2) \,dt_1dt_2 \neq \sigma^2\int_0^\tau n(t)^2 dt \end{align} Therefore you have $$ \mathbb E\big[\big(v(\tau)-\bar v(\tau)\big)^2\big] = \sigma^2 \int_0^\tau \int_0^\tau \mathbb E[ n(t_1)n(t_2)] \,dt_1dt_2 $$ You say that $n(t)$ are independent of each other, so you'll get $$ \mathbb E\big[\big(v(\tau)-\bar v(\tau)\big)^2\big] = \sigma^2 \int_0^\tau \int_0^\tau \big(\mathbb E[ n(t_1)]\mathbb E[n(t_2)] + \delta_{t_1,t_2}\big) \,dt_1dt_2 = 0 $$ where $$ \delta_{t_1,t_2} = \left\{\begin{array}{ll}1&\text{for }t_1=t_2\\0&\text{for }t_1\neq t_2\end{array}\right.$$ To undertand why you get $0$ variance, notice that a finite change in acceleration for an infinitezimal amount of time would only change the overall velocity by an infinitezimal amount. If you want these infinitezimal changes to compound to anything, you need the change to persist for some time, and for that you need a corelation between $n(t_1)$ and $n(t_2)$. Alternatively, you may allow for $a(t)$ to have infinite variance, so that $\mathbb E[(a(t_1)-a)(a(t_2)-a)] = \sigma^2\delta(t_1-t_2)$, where $\delta(x)$ is the Dirac delta. Then you're able to get finite variance for $v(\tau)$ even with no corelations.
Later, with the distance, you'll have a similar situation. You'll have $$ \mathbb E\big[\big(s-\bar s\big)^2\big] = \sigma^2 \int_0^T d\tau_1 \int_0^T d\tau_2\int_0^{\tau_1} dt_1 \int_0^{\tau_2}dt_2 \,\mathbb E[ n(t_1)n(t_2)] $$ and if $n(t)$ are independent, then the integral is equal to $0$.