Is this stochastic process a martingale?

Question

I have the following process:

$X_t=tB_t-\int^{t}_{0}B_s \ ds$ where $B_t $ is a Brownian motion.

Is this a Gauß-process and/or a martingale?

Can someone help me with this? And how can I calculate the $\int^{t}_{0}B_s \ ds$ part?

Answer 1

Hints: (Martingale) Fix $s \leq t$.

1. Write $$\mathbb{E}(t B_t \mid \mathcal{F}_s) = \mathbb{E}(t (B_t-B_s) \mid \mathcal{F}_s) + \mathbb{E}(t B_s \mid \mathcal{F}_s).$$ For the first term, use that $(B_t)_{t \geq 0}$ has independent increments, i.e. $B_t-B_s$ and $\mathcal{F}_s$ are independent, and $B_t-B_s \sim N(0,t-s)$. What about the second term?
2. Use the identity $$\int_0^t B_r \, dr = \int_0^s B_r \, dr + \int_s^t B_r \, dr = \int_0^s B_r \, dr + \int_s^t (B_r-B_s) \, dr+ (t-s)B_s$$ and a similar reasoning as in step 1 to calculate $$\mathbb{E} \left( \int_0^t B_r \, dr \mid \mathcal{F}_s \right).$$
3. Conclude.

Hints: (Gaussian)

1. Set $t_i := \frac{i}{n} t$. Since $(B_t)_{t \geq 0}$ is Gaussian, we know that $(B_{t_1},\ldots,B_{t_n})$ is (jointly) Gaussian. Deduce from this fact that $$X_t^n := t B_{t_n} - \sum_{i=1}^n B_{t_i} (t_{i+1}-t_i)$$ is Gaussian.
2. Recall that $X_t^n \to X_t$ almost surely as $n \to \infty$.
3. Conclude.

Remark: This argumentation does not use Itô's formula. However, Itô's formula provides us with an alternative solution: It follows easily from Itô's formula that

$$X_t = \int_0^t s\, dB_s.$$

Since stochastic integrals are martingales (... at least if the integrand is "nice") and integrals of the form

$$\int_0^t f(s) \, dB_s$$

are Gaussian for any determinstic function $f$ whenever the integral exists, we find that $(X_t)_{t \geq 0}$ is Gaussian and a martingale.

Answer 2

Apply Itô formula : $$d(tB_t)=B_tdt+tdB_t$$ So that $$X_t=\int_0^t sdB_s$$ which is a Wiener integral (Gaussian) thus a martingale.