Skip to content
All library documents

Girsanov Measure Changes and the Drifted Brownian Motion Law

Article Quant Q&A · Author: user2781712
Read the original

Summary

The document asks whether the probability law of a Brownian motion with constant drift is the same as a measure under which that drifted process becomes standard Brownian motion. It correctly distinguishes the process’s original law from the changed measure: under the original law, its mean is drift times time, while under the new measure it has zero mean.

It identifies Girsanov’s theorem as the tool for changing measure, but does not derive the Radon–Nikodym derivative of the drifted process’s law relative to Wiener measure. The only response reiterates that a measure change is necessary to make the process Brownian; it does not give a formula or a detailed explanation. The note is therefore useful as a conceptual prompt about probability measures and drift, but incomplete as a computational guide.

Key ideas

  • A Brownian motion with constant drift has a different law from standard Brownian motion under its original measure.
  • Girsanov’s theorem changes the probability measure so a drifted process can become standard Brownian motion.
  • The process’s expectation depends on which measure is being used.
  • The document raises the likelihood-ratio question but provides no derivation of the Radon–Nikodym derivative.

Tags

Statistics
Full text
# How to compute the Radon-Nikodym derivative?


# How to compute the Radon-Nikodym derivative?












Suppose $B(t)$ is a standard Brownian motion, and $B_{1}(t)$ is given by $dB_{1}(t)=\mu dt+dB(t)$. Suppose $P$ is the Wiener measure induced by $B(t)$ on the $C[0,\infty)$, and $P_{1}$ is the Law induced by $B_{1}(t)$ on $C[0,\infty)$. Here we follow the definitions of law is referred to

https://math.stackexchange.com/questions/90268/how-is-the-law-of-a-stochastic-process-defined/557519#557519

According to Girsanov theorem ( for example, P155. Thereom 8.6.3 in Fifth Edtion, Stochastic Differential Equations: An introduction with Application), there exists a Law $Q$ such that $B_{1}(t)$ is a standard Brownian motion under $Q$.

Is $Q$ equal to $P_{1}$ ?

I thought they are not equal to each other. The reason is that for fixed time $t$ the expectation of $B_{1}(t)$ under $Q$ is 0, and under $P_{1}$ its expectation should be $\mu t$. If my derivation is wrong, please point out where is my mistake.

If I am correct, a new question is how to compute $\frac{d P_{1}}{dP}$?

Recall that $\frac{d Q}{dP}$ is given by Girsanov theorem. Any references are very appreciated.

## Answer by david (score 2)

https://quant.stackexchange.com/a/9393

No, that's the point of Girsanov's theorem. If $Q$ is equal to $P_1$, then nothing has changed. In order to make $B_1(t)$ a standard BM we need to transition to a new Law. Namely, $Q$.

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.