Adding Drift to Brownian Motion and Changing Probability Measures
Summary
The document asks whether adding a constant linear drift to Brownian motion leaves the process Brownian and whether Girsanov’s theorem is needed. Under the original probability measure, the shifted process has increments with mean equal to the drift multiplied by the elapsed time. Since standard Brownian increments have zero mean, the shifted process is not standard Brownian motion under that same measure.
The answer describes using Girsanov’s theorem to define a new probability measure under which the drift-adjusted process is standard Brownian motion. For constant drift over a finite horizon, it invokes Novikov’s criterion to establish that the stochastic exponential used as the Radon–Nikodym density has expectation one. This is a measure-dependent result: the process’s Brownian property changes with the probability measure, and the explanation assumes the stated constant drift and finite horizon.
Key ideas
- Adding constant drift changes the increment means under the original probability measure.
- The shifted process is not standard Brownian motion under the original measure.
- Girsanov’s theorem can identify a new measure under which the shifted process is Brownian.
- For constant drift on a finite horizon, the answer uses Novikov’s criterion to justify the measure change.
Tags
Full text
# How to prove that the following is still a Brownian motion
# How to prove that the following is still a Brownian motion
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?
## Answer by Yoda And Friends (score 1, accepted)
https://quant.stackexchange.com/a/61988
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.