# Is squared Brownian Motion a gaussian process?

I am working at the following SP, given by $(X_t)_{t\geq0} = \alpha W_t^2+\beta t$ where $W_t$ is Brownian motion and $\alpha,\beta$ real. I managed to calculate mean and covariance function and now I try to find out if this process is gaussian. So I started right away with $P(\alpha W_{t_1}^2+\beta t_1\leq x_1,\dots,\alpha W_{t_p}^2+\beta t_p\leq x_p)=\\P(-\sqrt{\frac{x_1-\beta t_1}{\alpha}}\leq W_{t_1}\leq\sqrt{\frac{x_1-\beta t_1}{\alpha}},\dots,-\sqrt{\frac{x_p-\beta t_p}{\alpha}}\leq W_{t_p}\leq\sqrt{\frac{x_p-\beta t_p}{\alpha}})$ I know that $(W_{t_1},\dots,W_{t_p})$ is gaussian distributed. My idea was now to denote by $F$ the distribution function of $(W_{t_1},\dots,W_{t_p})$ to get $P(\alpha W_{t_1}^2+\beta t_1\leq x_1,\dots,\alpha W_{t_p}^2+\beta t_p\leq x_p)=\\F(\sqrt{\frac{x_1-\beta t_1}{\alpha}},\dots,\sqrt{\frac{x_p-\beta t_p}{\alpha}})-F(-\sqrt{\frac{x_1-\beta t_1}{\alpha}},\dots,-\sqrt{\frac{x_p-\beta t_p}{\alpha}})$. I was thinking about differentiating this to get the density function but I am confused about how e.g. $\frac{\partial}{\partial x_j}F(x_1,\dots,x_p)$ is what? Maybe someone can help me or give me an easier way, best, Alex

Hint Assume without loss of generality that $\alpha \geq 0$. Then, $$X_t = \alpha W_t^2 + \beta t \geq \beta t.$$ In particular, there exists $x \in \mathbb{R}$ such that $\mathbb{P}(X_t<x)=0$. Can such a process be Gaussian?
• No, because a guassian distribution has always probability weight in intervals $[-\infty,x)$, right? – AlexConfused Jun 29 '14 at 9:31
• @AlexConfused Exactly. If $X \sim N(\mu,\sigma^2)$, then $\mathbb{P}(X \in (a,b))>0$ for any $-\infty \leq a < b \leq \infty$. (Here, we need that $\sigma^2>0$. The statement is not true for a degenerate Gaussian random variable, i.e. if $\sigma^2=0$.) – saz Jun 29 '14 at 9:36