# Properties of the time integral of Wiener process

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?

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.

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*}

• Are you sure? For example, suppose $t>s>0$. Then $\int_0^t \int_0^s \min(u,v) dv du = (1/6) (3 s^2 t -s^3)$, whereas this answer gives $(1/2) s^2 t$ Apr 27, 2019 at 19:04