# Brownian Motions and Martingales [closed]

State whether each of the statement is true or false.

• If $Z$ is a standard normal random variable then $X_t = Z\sqrt t$ is a Brownian motion.
• If $W_t$ and $W_t'$ are two independent Brownian motions and $\rho$ is a fixed constant such that $|\rho|<1$, then $X_t = \rho W_t+ W'_t \sqrt{1-\rho%2}$ is a brownian motion.
• If $W_t$ is a Brownian motion then $X_t = \exp\left(\alpha W_t - \alpha^2t/2\right)$ is a martingale, where $\alpha$ is a fixed constant.

Any help is appreciated! As a recall, the conditions for a brownian motion are :

1. $W_0=0$
2. The sample path of $W$ is continuous and nowhere differentiable.

3. For $s\leqslant t$, $W_t-W_s$ is independent of all $W_u$ for $u\leqslant s$ and $W_t-W_s \sim N(0, t-s)$.

## closed as off-topic by Did, user223391, NCh, Claude Leibovici, Chris CusterApr 8 '18 at 7:15

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• Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – GNUSupporter 8964民主女神 地下教會 Apr 7 '18 at 22:24
• Perhaps a typo in your definition for Brownion motion $u \le s$ instead of $u \le t$ – Youem Apr 7 '18 at 23:07

## 1 Answer

The first one is clearly false because $(\sqrt t - \sqrt s) Z$ is not independent from $\sqrt s Z$. The second is also false because $E[X_t^2] = \rho^2 t + (1-\rho) t \neq t$. For the last it is correct and to prove it use the characteristic function of normal distribution variable.

• thanks for you answer. Could you please clarify a little on the first one? Why are they not independent? – TedMosby Apr 8 '18 at 20:23
• It is the same variable – Youem Apr 8 '18 at 20:24
• okay yes I see it. How about the second one, I'm not sure why your argument works? – TedMosby Apr 8 '18 at 20:38
• I am just computing the expectation of the square of $X_t$ – Youem Apr 8 '18 at 20:40
• In that case it will be a brownian motion – Youem Apr 8 '18 at 20:44