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 $s$.


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.

  • 4
    $\begingroup$ Can you please check whether $X_{s+t}-X_s$ has the distribution of $N(0, t)$? $\endgroup$
    – Gordon
    Nov 22, 2016 at 15:20

1 Answer 1


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.

  • 2
    $\begingroup$ Who could have guessed! :) $\endgroup$
    – Quantuple
    Nov 22, 2016 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.