[Math] Sufficient and Necessary condition for the sum of brownian motions to be a brownian motion

Question

The question has two parts :

Part a): Let $(B_1(t), B_2(t), B_3(t))$ be standard a brownian motion in $R^3$. Write down necassary and sufficient condition for $\sum\limits_{i=1}^3 a_iB_i(t)$ to be a one dimensional browian motion.

Here is my attempt :
We have to make sure that the sum is gaussian with mean zero and variance $t$ (since the starting brownian motion is standard).
Sum of gaussians is gaussian, also sum of 3 mean zero random variables has mean zero.
For variance, $Var(\sum\limits_{i=1}^3 a_iB_i(t)) = a_1^2Var(B_1(t)) +a_2^2Var(B_2(t))+a_3^2Var(B_3(t))$ (since they are independent) we conclude that
$\sum\limits_{i=1}^3 a_i = 1$ is sufficient and necassary.

What is the flaw in this reasoning? I was told that the above answer is not correct, however, I am unable to see why.

Part b) Suppose that $X(t)$ is a stochastic process with $X(0) = 0$ and, continuous sample paths, and X(t) is normal mean $0$ and variance $t$.
Prove or disprove : $X(t)$ is a brownian motion.

Here is my attempt :
I think it just follows by Levy characterization right? or is there more to it ?

I appreciate help/hints/suggestions.

Answer 1

1. Why do you conclude $$\sum_{i=1}^3 a_i = 1?$$ Your calculation of the variance actually shows that $$\sum_{i=1}^3 a_i^2 = 1$$ is a necessary condition.
2. Lévy's characterization requires that the processes $(X_t,\mathcal{F}_t)_{t \geq 0}$ and $(X_t^2 -t, \mathcal{F}_t)_{t \geq 0}$ are martingales. This is not assumed here. Think about the following process: Let $Y \sim N(0,1)$ be a centered Gaussian random variable with variance $1$ and define $$X_t := \sqrt{t} Y, \qquad t \geq 0.$$ Does $(X_t)_{t \geq 0}$ satisfy the given assumptions? Is it a Brownian motion?