1. Difference of Random Variables

In a solution I am trying to understand, we are given:
$\displaystyle A \sim N(\mu, 0.04)$
$\displaystyle B \sim N(\mu, 0.04)$

And then the next step is written as:
$\displaystyle (\bar{A}-\bar{B}) \sim N(0, 0.08)$

Can someone please explain to me how this step is made? I can see why the mean would be zero, but why are the variances summed? Thank you.

2. Originally Posted by doopokko
In a solution I am trying to understand, we are given:
$\displaystyle A \sim N(\mu, 0.04)$
$\displaystyle B \sim N(\mu, 0.04)$

And then the next step is written as:
$\displaystyle (\bar{A}-\bar{B}) \sim N(0, 0.08)$

Can someone please explain to me how this step is made? I can see why the mean would be zero, but why are the variances summed? Thank you.
I assume A and B are independent.

So you want a difference of two normal random variables to have a variance of zero. Which would mean no uncertainty at all in the value of A - B .......

It can be proved that if $\displaystyle U = a_1 X_1 + a_2 X_2$ then:

$\displaystyle E(U) = a_1 E(X_1) + a_2 E(X_2)$

$\displaystyle Var(U) = a_1^2 Var(X_1) + a_2^2 Var(X_2) + 2 a_1 a_2 Cov(X_1, X_2)$.

If $\displaystyle X_1$ and $\displaystyle X_2$ are independent then $\displaystyle Cov(X_1, X_2) = 0$.

Property 2 here: http://en.wikipedia.org/wiki/Normal_distribution

is of related interest.