Variance of Gaussian linear combination I have two independent gaussian distributions to combine and I have a doubt.
Let's say we have $X \sim N(\mu_x,\sigma^2_x)$ and $Y \sim N(\mu_y,\sigma^2_y)$. I want to mix the two variables with a 0.5 weight each, so I obtain a new variable $Z=0.5*X+0.5*Y$ with mean $\mu_z=0.5*\mu_x+0.5*\mu_y$.
Is it right to say that $\sigma^2_z=0.5^2*\sigma^2_x+0.5^2*\sigma^2_y$?
Because I clearly obtain less variance than intuitively expected when combining distributions.
 A: This answer should be viewed as being a compilation of comments which resulted in the OP's confusion being resolved, and is not intended as being the most general or rigorous statement on the subject. Its goal is to aid interpretation.
Jeremias K's answer shows the formula for the variance of a weighted sum of random variables.  In particular, taking the special case of all weights being $1/n$ for a weighted sum of n independent random variables, the variance of the weighted sum equals $1/n^2$ times the sum of the variances of the individual random variables.  If the random variables are also identically distributed, then the variance of the weighted sum equals $1/n$ times the variance of an individual random variable.
As you take the average of more and more i.i.d. (Gaussian) random variables, the variance of that average will decrease, as seen from the preceding formulation, which is a result which should be quite familiar.  Why does the variance of that average decrease vs. an individual random variable? Well, taking the case of 2 independent random variables, an extreme value in one variable tends not to be matched with a similar direction extreme value in the other variable, hence making the average less extreme than the individual random variables, hence the decreased variance. The magnitude of this effect increases as the number of random variables being averaged increases.
A: If $X$ and $Y$ are independent, $Cov(X,Y) = 0$. Since also, $Var(aX+bY) = a^2Var(X) + b^2Var(Y) + 2abCov(X,Y)$, it is in fact right to calculate the variance as you did.
I am not sure I can give an answer to the part of your question about intuition. I cannot really follow your intuition as to why the variance from the combination should be larger. Can you elaborate?
