Prove that integral is a Gaussian random variable, compute its mean and variance

Question

I have to prove that $X_t=\int_0^t W_s ds$ is a Gaussian random variable. I need also to compute it's mean and variance.

My attempt:

Let $W_t$ be a simple adapted process $$W_t=\xi_0\mathbf{1}_0(t)+\sum_{i=0}^{n-1} \xi_i \mathbf{1}_{(t_i,t_{i+1}]}(t)$$ where $0=t_0<t_1<\dots<t_n=T$ and $\xi_i$ are measurable with respect to $F_{t_i}$ and $E[\xi_i^2]<\infty$

Then the integral of this process with respect to time is given by $$\int_0^T W_sds=\sum_{i=0}^{n-1} \xi_i (t_{i+1}-t_i)$$ If I can prove that this integral is Gaussian, I will prove that $X_t$ is Gaussian, since a limit of Gaussian processes is Gaussian. Any help?

Answer 1

Using the definition of a Riemann integral, we have that the Riemann sum for a partition $0=t_0<t_1<\dots<t_N=T$ is $$ S_N = \sum_{i=1}^N W(t_i)(t_i - t_{i-1}), $$ Now, we have that $W(t_i) = \sum_{k=1}^i[W(t_k) - W(t_{k-1})]$, so $$ S_N = \sum_{i=1}^N\sum_{k=1}^i[W(t_k) - W(t_{k-1})](t_i - t_{i-1}). $$ $$ S_N = \sum_{i=1}^N[W(t_i) - W(t_{i-1})](T - t_{i-1}). $$ Since the increments of the Wiener process are independent and Gaussian with variance equal to the time increment, we have that $$ E[S_N^2] = \sum_{i=1}^N[t_i - t_{i-1}](T - t_{i-1})^2. $$ The expression above is a Riemann sum too, so when the mesh of the partition vanishes and $N\to\infty$, we have that $$ E\left[\left(\int_0^T W(t)\,dt\right)^2\right] = \int_0^T (T-t)^2\,dt = \frac{T^3}{3} $$

Gaussian random variables form a linear subspace of $L_2$, since Gaussianity is closed with respect to addition and multiplication by scalar, i.e., if X and Y are Gaussian random variables then $$ Z = aX + bY $$ is also a Gaussian random variable. A subspace of a Hilbert space is also a Hilbert space, so it is complete. You can then use dominated convergence to show that $S_N$ converges in $L_2$ (mean square convergence), so that its limit is also Gaussian. Taking the expected value of $S_N$ it is also easy to see that the integral is zero-mean.

Answer 2

An alternate way of showing this result is to use the (stochastic)-integration by part formula :

$\int_0^tW_sds= W_t-W_0 - \int_0^tsdWs$ ( no quadratic covariation term has s has 0 quadratic variation)

The first term is normal and independent from the stochastic integral which is a Wiener integral. As Wiener integrals are normal random variables, we have written $\int_0^tW_sds$ as the sum of 2 independent normal variables showing that it is itself a normal random variable (BTW mean and variance are easy to get using Itô's isometry and martingale property of stochastic integrals).

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