Given a Brownian motion $B_t$ on a filtered probability space, how can I prove that $W_t=B_t+\alpha t$ is still a Brownian motion, with $\alpha \in \mathbb{R}$? Is it always true? Do I need necessarly to use Girsanov Theorem?

  • 2
    $\begingroup$ $B_t$ and $W_t$ can not be Brownian motions under the same measure. $\endgroup$
    – Gordon
    Commented Mar 25, 2021 at 13:12
  • $\begingroup$ And under two different measures? $\endgroup$
    – RedLapm
    Commented Mar 25, 2021 at 13:53

1 Answer 1


I try to clarify. First, let us be under the real world probability measure $\mathbb{P}$. Say that the process $Y_t$ is a $\mathbb{P}$ standard Brownian motion. Then the following is true by definition (assume wlg. that s < t):

  • $(Y_t - Y_s) \sim N(0, t-s)$

Now, using your definition of $W_t$:

  • $\mathbb{E} \left[(W_t - W_s)\right] = \mathbb{E} \left[(B_t - B_s)\right] + \alpha t - \alpha s \neq 0$

therefore, $W_t$ is NOT a $\mathbb{P}$ standard Brownian motion.

However, the process $W_t$ can be seen as a SBM under a new prospective, in other words, under a new measure. Say we have a constant $\alpha \in \mathbb{R}$. Then, we make use of Novikov criterion, which states that, if $\mathbb{E} \left[exp (\frac{1}{2} \int_0^T \alpha^2 \ ds )\right] < \infty$ (which is clearly true for a constant), then we can define the stochastic exponential $Z_T = \mathcal{E}_T(- \alpha \cdot B)$ and we have $\mathbb{E} [ Z_T ] = 1$.

Now, we are ready to define the RN density $\frac{d \mathbb{Q}}{d \mathbb{P}} = Z_T$ which defines a new measure (our $\mathbb{Q})$. Now, Girsanov comes into play, saying that a new process defined as:

  • $W_t = B_t + \int_0^t \alpha ds = B_t + \alpha t$

is indeed a $\mathbb{Q}$ standard Brownian motion.


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