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 ?

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$S,K$ 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.

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The theorem that Gortaur is citing is: two gaussian random variables are independent iff their covariance is zero. This is a special property of gaussian random variables that is not true in general. –  Tim van Beek Jun 12 '11 at 8:55
@Tim van Beek: Is what you say correct? Suppose $X$ has a standard normal distribution, and $Y = X$ when $|X| \ge k$ and $Y = -X$ when $|X| \lt k$, then $Y$ also has a standard normal distribution and there is a $k$ slightly less than 1.54 where the covariance is zero, but $X$ and $Y$ are not at all independent (one determines the other). This doesn't affect Gortaur's proof, which uses the general point that non-zero covariance implies lack of independence. –  Henry Jun 12 '11 at 9:29
I should have been more precise, the two random variables need to be jointly normally distributed, see en.wikipedia.org/wiki/… –  Tim van Beek Jun 12 '11 at 9:53
I am curious why downvote? –  Ilya Jun 12 '11 at 17:42

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 ${\rm E}[e^{iu_1 ( - Y) + iu_2 Y} ] \ne {\rm E}[e^{iu_1 ( - Y)} ]{\rm E}[e^{iu_2 Y} ]$.)

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Of course, Gortaur's approach is more appropriate for the question at hand, where the covariance exists. –  Shai Covo Jun 12 '11 at 10:37