Question

My Stochastic calculus professor always used to say "When in doubt use Ito"

So let $f(t,x) = t x $ and compute $\partial_t f(t,x) = x$, $\partial_x f(t,x) = t $ and $ \partial_{xx} f(t,x) = 0$

Now the Ito lemma says for $f$ twice differentiable with respect to $x$ and once differentiable with respect to $t$ then the following formula holds for any Ito process:

$f(t, X_t) = f(0,X_0) + \int_{0}^{t}\partial_s f(s,X_s) ds + \int_{0}^{t}\partial_x f(s,X_s) dX_s$ $+ \int_{0}^{t}\partial_{xx} f(s,X_s) d \left< X,X \right>_s $

So applying the above fact to the function $f(t,x) = tx$ gives:

$t X_t = 0 + \int_{0}^{t}X_s ds + \int_{0}^{t} s dX_s + 0$ or to give you a starting answer $\int_{0}^{t}X_s ds = t X_t - \int_{0}^{t} s dX_s$

Edit 3: The following only holds now if $X_t$ is a Gaussian process which is not true in general... So in vague words we have that The (Riemann) integral of an ito process is equal to the difference of two Gaussian processes which should again be Gaussian .... (if all this logic is correct it should then suffice to characterize the processes covariance structure in order to have a complete understanding of the law of the process.)

For example one can compute the variance if $X_t$ is standard Brownian motion:

$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^2 \mathbb{E}[(X_t)^2] -2t \mathbb{E}[X_t \int_{0}^{t} s dX_s] + \mathbb{E}[(\int_{0}^{t}sdX_s)^2] $

By the Ito isometry we have $\mathbb{E}[(\int_{0}^{t}sdX_s)^2] = \int_{0}^{t}s^2ds = t^{3}/3$.

To compute $\mathbb{E}[X_t \int_{0}^{t} s dX_s]$ notice first that the Ito integral of a deterministic function is always a Gaussian process. EDIT: Shavi has given that $\mathbb{E}[X_t \int_{0}^{t} s dX_s] = t^2/2$

$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^3 - t^3 + \frac{t^{3}}{3} = \frac{t^3}{3}$

Computing the covariance $\mathbb{E}[\int_{0}^{t}X_s ds \int_{0}^{u}X_s ds]$ involves dealing with terms $\mathbb{E}[\int_{0}^{t}s dX_s \int_{0}^{u}s dX_s]$ and $\mathbb{E}[X_t X_u]$ which are again probably well known in certain cases (the second term is obviously equal to $min(t,u)$ when $X$ is b.m.) but may be difficult to handle in your general case.

Edit 2: To give an approach to answer the question "Is it possible that the integrated processes have equivalent laws?"

Since $\int_{0}^{t}X_{s}^{(1)}ds$ and $\int_{0}^{t}X_{s}^{(2)}ds$ are Gaussian processes (we proved this using ito) it suffices to check if there covariance functions $g_{1}(t,u)=\mathbb{E}[\int_{0}^{t}X_{s}^{(1)}ds \int_{0}^{u}X_{s}^{(1)}ds]$ and $g_{2}(t,u) = \mathbb{E}[\int_{0}^{t}X_{s}^{(2)}ds \int_{0}^{u}X_{s}^{(2)}ds]$ are equal for all $t,u >0$ to show that the two processes have equivalent laws.

Now applying the result we got above from the ito calculation lets us start computing the covariance:

$\mathbb{E}[\int_{0}^{t}X_s ds \int_{0}^{u}X_s ds]$ = $\mathbb{E}[( t X_t - \int_{0}^{t} s dX_s)( u X_u - \int_{0}^{u} s dX_s)]$ $ = t u \mathbb{E}[ X_t X_u ] - t \mathbb{E}[X_t \int_{0}^{u} s dX_s ] -u \mathbb{E}[X_u \int_{0}^{t} s dX_s ]$ $+\mathbb{E}[\int_{0}^{t}s dX_s \int_{0}^{u}s dX_s]$

I refer to my above example on ways to deal with the terms in this expression given certain assumptions on $\mu$ and $\sigma$.Edit 3: Again this is just a way to start and obviously the calculations involving standard Brownian motion are trivial but the point is that the laws of $Y^{(1)}$ and $Y^{(2)}$ are equivalent (as opposed to equal) as soon as you show $g_1(t,u) = g_2(t,u)$ for all $t,u>0$.