Properties of the time integral of Wiener process

Question

Let $W_t$ be a Wiener process and consider the time integral $$ X_T:= \int_0^T W_t dt $$

It is often mentionend in literature that $X_T$ is a Gaussian with mean 0 and variance $T^3/6$.

I am interested in learning more about the process $X_T$ for $T>0$. Except for the description of the individual random variable $X_T$ I have not found much.

But there must be more to it. It is if one stopped talking about a Wiener process after mentioning that $W_t$ is a Gaussian with mean 0 and variance $t$.

As an example, it would like to have sample paths discussed. I wonder, since as being defined by an integral over a continuous function it should be differentiable. On the other hand it feels wrong, since I have never encountered a non-degenerate stochastic process that is (a.s.) differentiable.

Do you know were this process is discussed in depth?

Answer 1

As an integral of a zero-mean Gaussian process, your $X_T$ is a zero-mean Gaussian process as well. Its covariance function can be calculated via

$$ c(s, t) = \int^s_0 \int^t_0 \min(u, v) \; du \; dv \; , $$

which yields

$$ c(s, t) = \frac{\min(s, t)^2}{6} \left( 3 \max(s, t) - \min(s, t) \right) . $$

In terms of sample functions, your expectation is correct. Being an integral of a random process with (almost surely) continuous sample paths, it indeed has (almost surely) differentiable sample paths.

It does not have independent increments, as discussed here.

Answer 2

The covariance computed above does not seem right. I can't see where the second term in parentheses comes from. I believe the right answer should be

\begin{align*} \operatorname { cov } \left( X(s) , X(t) \right) &= \mathbb{E} \left[ (X(s)-\mathbb{E}[X(s)]) \cdot (X(t)-\mathbb{E}[X(t)]) \right]\\ &= \mathbb{E} \left[ X(s) \cdot X(t) \right]\\ &= \mathbb{E} \left[ \int_{0}^{s} W(u) du \cdot \int_{0}^{t} W(u) du \right]\\ &=\int_{0}^{t} \int_{0}^{s} \mathbb{E}(W(u) W(v)) \mathrm{d} v \mathrm{d} u\\ &= \int_{0}^{t} \int_{0}^{s} \min \{u, v\} \mathrm{d} v \mathrm{d} u\\ &= \frac{1}{2}\min (s, t)^{2} \max (s, t) \end{align*}