Finite Dimensional Distributions of Stochastic Process

Question

If $X(t) = \int_0^t \! B(s)ds$ where $B(s)$ for $s > 0$ is a Brownian motion process.

Part a) what are the finite dimensional distributions of $X(t)$? (not an explicit formula, you don't need to evaluate integrals)

Part b) Compute the quadratic variation of $X(t)$.

Part c) is $X(t)$ a markov process?

My attempt:

Part a) I don't know how to do it without writing down the integral expression. Is it just Gaussian? will saying that be enough? I don't know what I am supposed to say for this part.

Part b) $X(t) = \int_0^t \! B(s)ds$. So $<X(t)> = \int_0^t \!1 <B(s)>ds = \int_0^t \! sds = 1/2t^2$ Is this correct ? (since quadratic variation of $B(s)$ is just $s$.

Part c) for this part is it enough to say $X(t)$ is a markov process because it is a function of another markov process ($B(s)$) and that function (integration) is invertible? I think I had seen this as a lemma somewhere.

I appreciate help/hints/suggestions.

Answer 1

1. Yes, they are Gaussian. Have a look at this question.
2. No, this is not correct; the identity $$\langle \int B_s \, ds \rangle = \int \langle B_s \rangle \, ds$$ does not hold. How did you define the quadratic variation? Hint: The process $X(t)$ is of bounded variation.
3. No, the process is not a Markov process, but one can show that the process $(X_t,B_t)_{t \geq 0}$ is a Markov process. To see that $(X_t)_{t \geq 0}$ is not a Markov process, calculate expressions of the form $$\mathbb{E}(f(X_t) \mid \mathcal{F}_s)$$ for $s \leq t$ and bounded, continuous functions $f$.

---

Lemma: $\langle X \rangle_t = 0$.

Proof: Note that

$$|X_t-X_s| = \left| \int_s^t B_r \, dr \right| \leq |t-s| \sup_{r \in [s,t]} |B_r|$$

for any $0 \leq s \leq t$. Consequently,

$$\begin{align*} \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^2 &\leq \sup_{i=0,\ldots,n} \sup_{r \in [t_i,t_{i+1}]} |B_r|^2 \cdot \sum_{i=0}^n |t_{i+1}-t_i|^2 \\ &\leq |T_n| \cdot \sup_{r \in [0,t]} |B_r|^2 \cdot \underbrace{\sum_{i=0}^n |t_{i+1}-t_i|}_{t}\end{align*}$$

for any partition $T_n$ of the interval $[0,t]$. Letting $|T_n| \to 0$, we find

$$\langle X \rangle_t = \lim_{|T_n| \to 0} \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^2 = 0.$$

Remark: The proof actually shows $$\langle X \rangle_t =0$$ for any process of the form $$X_t(\omega) = \int_0^t f(s,\omega) \, ds$$ where $f(\cdot,\omega)$ is a continous function for each fixed $\omega \in \Omega$.