# How to prove that the following is still a Brownian motion [closed]

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?

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

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.