Independence of sums of gaussian random variables Say, I have independent gaussian random variables $t1, t2, t3, t4, t5$ and I have two new random variables $S = t1 + t2 - t3$ and $K = t3 + t4$.
Are $S$ and $K$ independent or is there any theorem about independece of random variables formed by sum of independent gaussians ? 
 A: $S,K$ are both normal random variables, hence their independence depends only on the covariance. $cov(S,K) = -Var(t_3)\neq 0$, so they are not independent.
A: In fact, the distribution of the $t_i$ plays no significant role here, and, moreover, existence of the covariance is not necessary.
Let $S=X-Y$ and $K=Y+Z$, where $X$, $Y$, and $Z$ are independent random variables generalizing the role of $t1+t2$, $t3$, and $t4$, respectively.
Note that, by independence of $X$, $Y$, and $Z$, for any $u_1,u_2 \in \mathbb{R}$ it holds
$$
{\rm E}[e^{iu_1 S + iu_2 K} ] = {\rm E}[e^{iu_1 X + iu_1 ( - Y) + iu{}_2Y + iu_2 Z} ] = {\rm E}[e^{iu_1 X} ]{\rm E}[e^{iu_1 ( - Y) + iu{}_2Y} ]{\rm E}[e^{iu_2 Z} ]
$$
and
$$
{\rm E}[e^{iu_1 S} ] {\rm E}[e^{iu_2 K} ] = {\rm E}[e^{iu_1 X} ]{\rm E}[e^{iu_1 (-Y)} ]{\rm E}[e^{iu_2 Y} ]{\rm E}[e^{iu_2 Z} ].
$$
The following basic theorem then shows that $S$ an $K$ are generally not independent.
Theorem. Random variables $\xi_1$ and $\xi_2$ are independent if and only 
$$
{\rm E}[e^{iu_1 \xi _1  + iu_2 \xi _2 } ] = {\rm E}[e^{iu_1 \xi _1 } ]{\rm E}[e^{iu_2 \xi _2 } ]
$$
for all $u_1,u_2 \in \mathbb{R}$.
(In particular, note that if $-Y$ and $Y$ are not independent, then there exist $u_1,u_2 \in \mathbb{R}$ such that
${\rm E}[e^{iu_1 ( - Y) + iu_2 Y} ] \ne {\rm E}[e^{iu_1 ( - Y)} ]{\rm E}[e^{iu_2 Y} ]$.)
