Why a Scaled Single Normal Variable Does Not Form Brownian Motion

Summary

The document tests whether the process formed by multiplying one standard normal random variable by the square root of time is Brownian motion. Although each value has a normal distribution with mean zero and variance equal to time, those marginal distributions are not enough to establish Brownian motion. The increments must also have the required time-dependent distributions and be independent of the process’s history.

For two times, the increment is the difference between their square roots multiplied by the same random variable. Its variance is therefore the square of that difference, rather than the elapsed time. This fails the required normal increment distribution, so the process is not Brownian motion. The argument does not need to resolve the separate independence condition once this distribution requirement already fails. The example highlights why checking individual-time distributions alone is insufficient when identifying a stochastic process.

Key ideas

• Brownian motion requires more than continuous paths and normal marginal distributions.
• Its increments must have variance equal to the elapsed time and be independent of the prior history.
• The proposed process uses the same normal variable at every time point.
• Its increment variance does not equal the time between observations, so the process is not Brownian motion.

Tags

# Do we have a Brownian motion


# Do we have a Brownian motion












Background Information:

The process $W = (W_t:t\geq 0)$ is a $\mathbb{P}$-Brownian motion if and only if

i) $W_t$ is continuous, and $W_0 = 0$

ii) the value of $W_t$ is distributed, under $\mathbb{P}$, as a normal random variable $N(0,t)$,

iii) the increment $W_{s+t} - W_{s}$ is distributed as a normal $N(0,t)$, under $\mathbb{P}$, and is independent of $\mathcal{F}_s$, the history of what the process did up to time $s$.

Question:

If $Z$ is a normal $N(0,1)$, then the process $X_t = \sqrt{t}Z$ is continuous and is marginally distributed as a normal $N(0,t)$. Is $X$ a Brownian motion?

From the definition Brownian motion above it seems that we directly satisfy the 2 conditions. Although I believe we need to show the third condition to indeed conclude that $X$ is of Brownian motion. I am just not sure how to provide a formal solution to this question.

## Answer by Gordon (score 5, accepted)

https://quant.stackexchange.com/a/31123

Aside from the independence requirement for the increments, that is, the independence of $X_{s+t}-X_s$ and $\mathcal{F}_s$, you can check whether the increment $X_{s+t}-X_s$ has the distribution of $N(0, t)$. In fact, note that \begin{align*} X_{s+t}-X_s &= (\sqrt{s+t}-\sqrt{s}) Z\\ &\sim N\left(0,\, (\sqrt{s+t}-\sqrt{s})^2\right), \end{align*} which obviously does not have the distribution of $N(0, t)$. That is, the process $\{X_t, \, t\ge 0\}$ is not a Brownian motion.

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