What is the distribution of the sum of independent normal variables? Just need to check the answer for the following question:
Question
Suppose $X$ and $Y$ are two independent standard normal variables:
$X \sim \mathcal N (0,1)$
$Y \sim \mathcal N (0,1)$
What is the distribution of $X + Y$ ?
My Working
$X+Y \sim \mathcal N ( \mu_1 + \mu_2$, $\sqrt{\sigma_1^2 + \sigma_2^2}) $
$= X+Y \sim \mathcal N (0 + 0$,$\sqrt{1^2+1^2}) $
$= X+Y \sim \mathcal N ( 0, \sqrt{2}) $
Does this look correct?
 A: To sum up the long series of comments:
Yes, your working is correct.  More generally, if $X$ and $Y$ are independent
normal random variables with means $\mu_X$, $\mu_Y$ respectively
and variances $\sigma_X^2$ and $\sigma_Y^2$ respectively, then 
$aX+bY$ is a normal random variable with mean $a\mu_X+b\mu_Y$
and variance $a^2\sigma_X^2 + b^2\sigma_Y^2$.
The various comments by whuber, cardinal, myself, and the Answer 
by Tai Galili are all occasioned by the fact that there are at least
three different conventions for interpreting $X \sim N(a,b)$ as
a normal random variable.  Usually, $a$ is the mean $\mu_X$ 
but $b$ can have  different meanings.


*

*$X \sim N(a,b)$ means that the standard deviation of $X$ is $b$.
(This is the convention you are using).

*$X \sim N(a,b)$ means that the variance of $X$ is $b$.

*$X \sim N(a,b)$ means that the variance of $X$ is $\dfrac{1}{b}$.
Fortunately, $X \sim N(0,1)$ (which is what you asked about)
means that $X$ is a standard 
normal random variable in all three of the above conventions! 
