How Standard Brownian Motion Determines a Drifted Process Distribution

Summary

The document explains the relationship between standard Brownian motion and a Brownian process with drift and scale. Standard Brownian motion starts at zero, and its increments over an interval of length t are normally distributed with mean zero and variance t. Taking the starting time as zero gives the distribution of its value at time t.

A process with drift μ and volatility σ is constructed by scaling standard Brownian motion by σ and adding the deterministic term μt. Its increments over an interval have mean μ times the interval length and variance σ² times the interval length; starting from zero, its value at time t has the corresponding normal distribution. The explanation is conceptual and gives these distributional consequences, but it does not discuss proofs of the underlying Brownian motion assumptions or processes with nonzero initial values.

Key ideas

• Standard Brownian motion starts at zero and has normally distributed increments with variance equal to elapsed time.
• A Brownian motion’s value at time t has a normal distribution because it is its increment from time zero.
• Adding linear drift and scaling standard Brownian motion changes the increment mean and variance.
• For a process starting at zero, its value at time t is normally distributed with mean μt and variance σ²t.

Tags

# standard brownian vs brownian motion


# standard brownian vs brownian motion












We say Xt with paramters (µ,σ) is brownian process if (Xt-s - X t) ~N (µs,σ2 s) AMONG other conditons .

Here we don't speak about any particular distribution for X t. We only say it is a brownian motion and its increments are normally distributed.

But when it comes to standard brownian motion ( Wt) , why do we say it has a normal distribution i.e Wt ~ N(0,t).

Does that mean I can say any brownian motion process Xt with parameters is µ,σ is also normally distributed N(µt,σ2 s)?

I am new to this topic and if the question does not have a logic, please enlighten with your inputs

edit 1: it would be helpful if the explanation is more intuitive than mathematical

## Answer by Mehdi (score 1)

https://quant.stackexchange.com/a/41439

A standard Brownian motion $ \{W_t, t \in \mathbb{R} \}$ starts from 0 which means $W_{0} = 0$ with probability one, add to that $W_t - W_{s} \sim N(0,t-s)$. Replace s with 0 and the you get $W_t =W_t - W_{0} \sim N(0,t)$

From this standard Brownian motion you can construct another one with a "drift" $\mu$ : $X_t = \mu t + \sigma W_t$ and you'll get $X_t - X_{s} \sim N(\mu(t-s), \sigma^{2}(t-s))$ and then particularly:

$X_{t-s} - X_{t} \sim N(\mu s, \sigma^{2}s)$

$X_{t} \sim N(\mu t, \sigma^{2}t) $

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.