Let $B(t)$ be $n$-dimensional Brownian motion.
I want to show that $E|B(t)-B(s)|^4 = n(n+2)|t-s|^2$ so that Kolmogorov's continuity theorem can be used to show that Brownian motion has a continuous version.
I tried the following:
$$E|B(t)-B(s)|^4 = E\left[ \sqrt{\sum_{i=1}^n (B_i(t)-B_i(s))^2}^4 \right]=E\left[ \left(\sum_{i=1}^n (B_i(t)-B_i(s))^2\right)^2 \right]$$
$$= \sum_{i,j=1}^n E\left[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2\right]$$
Since the individual components of Brownian motion are independent, if $i \neq j$ we get
$$E[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2] = (t-s)^2$$
while if $i=j$ we get
$$E[(B_i(t)-B_i(s))^4]= 3(t-s)^2$$
Adding together we see the RHS is then equal to
$$3n (t-s)^2 + (n^2-n)(t-s)^2$$
and, since $3n + n^2 - n = n(n+2)$, the result follows by Kolmogorov's Continuity Theorem that a continuous version of Brownian motion exists.
Is this correct?