$W(t)$ standard Wiener process, is the process $X(t)=|W(t)|$ Guassian? Find $E(X(t))$ I have that a random vector(of order $n$) is guassian if it's density function is equal to
$f(x) = Ce^{-Q(x)}$ where $Q(x)$ is positively definite form $x=(x_1,x_2,...,x_n)$
So first I need to find the density function of this random process, with time variable $t$. I know that:
$W(t): N(0,t)$- normal distribution with parameters $0$ and $t$, meaning that the density function here is $$f_{W(t)}(x)= \frac{1}{\sqrt{2\pi t}}e^{\frac{-x^{2}}{2t}}$$
Then I also know that:
(density function of $X(t)$
$$f_{X(t)}(x)=\frac{d}{dx}F_{X(t)}(x)=\frac{d}{dx}Pr\{X(t)<x\}=\frac{d}{dx}Pr\{-x<W(t)<x\}=\frac{d}{dx}\int_{-x}^{x} \frac{1}{\sqrt{2\pi t}}e^{\frac{-x^{2}}{2t}} =2 \frac{d}{dx}\int_{0}^{x} \frac{1}{\sqrt{2\pi t}}e^{\frac{-x^{2}}{2t}}dx= \frac{2}{\sqrt{2\pi t}}e^{\frac{-x^{2}}{2t}}=Ce^{-Q(x)}$$
meaning that it is in fact Guassian.
Finding the expected value would be just two times the expected value of the normal distribution $N(0,t)$ which is $0$.
Am I correct in this?
 A: By definition of a standard Wiener process, $W(t) = \big( W_1(t), \ldots, W_n(t) \big)$, we know that $W_j$ is independent of $W_k$, for all $1 \leq j,\,k \leq n$, and $j \neq k$.
For each $1 \leq k \leq n$
$$ W_k(t) = N(0,t) = \sqrt{t} \, N(0,1)$$ 
where the equality is in distribution, and $N(\mu, \, \sigma^2)$ denote the Normal distribution with mean $\mu$ and variance $\sigma^2$. Consequently we have (where again, equality is in distribution)
\begin{align*}
| W(t) | \, \colon & = \left( \sum_{k=1}^n W_k(t)^2 \right)^{1/2} \\
& = \left( \sum_{k=1}^n t\, N(0,1)^2  \right)^{1/2} \\
& = t^{1/2} \chi^2(n),
\end{align*}
where $\chi^2(n)$ denotes the Chi-squared distribution with $n$ degrees of freedom.
Then, using properties of the $\chi^2$-distribution, we have
$$ \mathbb{E} \big[ | W(t)| \big] = t^{1/2} \, \mathbb{E} \big[ \chi^2(n) \big] = n t^{1/2}.$$ 

Proof that $\mathbb{E} \big[ \chi^2(n) \big] = n t^{1/2}$
Including this in case you have not seen it before, and because it is a neat trick.
Let $f_n(x)$ denote the probability distribution function for a $\chi^2(n)$ distribution
$$f_n(x) = 2^{-n/2} \Gamma \left( \textstyle\frac{n}{2} \right)^{-1} x^{n/2 - 1} e^{-x/2} = C_n \,x^{n/2 - 1} e^{-x/2},$$
where $C_n = 2^{-n/2} \Gamma \left( \textstyle\frac{n}{2} \right)^{-1} $, and $\Gamma$ denotes the Gamma function. Then
\begin{align*}
\mathbb{E}\left[ \chi^2(n) \right] & =
\int_0^\infty x f_n(x) d x \\
& = C_n \int_0^\infty x x^{n/2 - 1} e^{-x/2} d x \\
& = C_n \int_0^\infty x^{(n+2)/2 - 1} e^{-x/2} d x.
\end{align*}
Looking at the form of the integrand in the last line, we see that it is the non-normalized density of a $\chi^2(n+2)$ distribution; i.e.
$$\int_0^\infty x^{(n+2)/2 - 1} e^{-x/2} d x  = C_{n+2}^{-1}$$
Hence
\begin{align*}
\mathbb{E}\left[ \chi^2(n) \right] & =
\frac{C_n}{C_{n+2}} \\
& =
\frac{2^{(n+2)/2} \Gamma \left( \frac{n+2}{2} \right)}{2^{n/2} \Gamma \left( \frac{n}{2} \right)} \\
& =
2\frac{\Gamma \left( \frac{n}{2} +1 \right)}{ \Gamma \left( \frac{n}{2} \right)} \\
& = 2 \frac{n}{2} \\
& = n,
\end{align*}
where the Gamma functions cancel due to the property $\Gamma(z+1) = z \Gamma(z)$, which can be proven using integration by parts.
