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?


2 Answers 2


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.

  • $\begingroup$ Can you explain why $\mathbb{E} (\int_s^t (B_r-B_s) \, dr+ (t-s)B_s|\mathcal{F}_s)=0$ ? $\endgroup$
    – kanbhold
    Mar 15, 2015 at 11:20
  • $\begingroup$ @kanbhold Where did I claim this...? The expression does not equal $0$. $\endgroup$
    – saz
    Mar 15, 2015 at 12:15
  • $\begingroup$ Ok, then I don't see why is that a martingale. $\endgroup$
    – kanbhold
    Mar 15, 2015 at 12:59
  • $\begingroup$ @kanbhold Without any further details (or your computations), there is nothing I can do about that. $\endgroup$
    – saz
    Mar 15, 2015 at 13:01
  • 1
    $\begingroup$ @Joel Where exactly are you stuck? I already gave a hint (=the identity) so what is stopping you from following this hint? $\endgroup$
    – saz
    Jan 15, 2020 at 14:47

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.


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