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QUESTION

If $Z$ is a normal $N(0,1)$, then is the process $X_t=X\sqrt{t}$
continuous? What is it's distribution? Is $X_t$ a Brownian motion?


ANSWER

$Z\in N(0,1)$ is a continuous random variable. $t$ is continuous
time. Thus $Z\sqrt{t}$ is a continuous random variable. It's
distribution follows from the standard tranformation between
normal distributions as per handout.

If $Z\in N(0,1)$ then $$X_t=\underbrace{\sqrt(t)}_{\textrm{new
standard deviation}}Z+\underbrace{0}_{\textrm{new mean}}$$ a
continuous random variable $X_t\in N(0,t)$. Is it Brownian? Check
conditions in notes (p.24)

\begin{description}

\item[(i)]
$X_t$ is continuous and $0=X_0$

\item[(ii)]
$X_t\in N(0,t)$

\item[(iii)]
$X_{s+t}-X_s\in N(0,t)$ and independent of time$&lt;s$:
$X_{s+t}-X_s=\left(\sqrt{s+t}-\sqrt{s}\right)Z\in
N(0,\sqrt{s+t}-\sqrt{s})$ by above. Therefore (iii) is not
satisfied so it is not Brownian.

\end{description}



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