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:

$X$ ~ $N(0,1)$
$Y$ ~ $N(0,1)$

What is the distribution of $X + Y$ ?

My Working
$X+Y$ ~ N($\mu_1 + \mu_2$, $\sqrt{\sigma_1^2 + \sigma_2^2})$
$X+Y$ ~ N($0 + 0$,$\sqrt{1^2+1^2}$
$X+Y$ ~ N($0$, $\sqrt{2}$)

Does this look correct?

-
Your working is correct assuming that you are using the expression $N(a, b)$ to mean a normal random variable with mean $a$ and standard deviation $b$. I have seen this usage several times in this forum and so I assume that it is becoming common in statistical circles. The notation $N(a,b)$ is also used for a normal random variable with mean $a$ and variance $b$, and in this notation, $$\text{independent} X_i \sim N(\mu_i, \sigma_i^2) \Rightarrow X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$$ cf. answer by Tal Galili –  Dilip Sarwate Nov 1 '11 at 11:54
Thanks Dilip, yes, in my University course, a Normal Distribution is modeled as N(Mean, Stdev) instead of N(Mean, Variance). I suppose different people use different notations –  Arvin Nov 1 '11 at 12:11
Yes, Arvin, you are correct in your supposition about notation. For instance, Wolfram Alpha agrees with your notation, not with Dilip's or @Tal's. (Others, especially in a Bayesian context, even parameterize Normals by their precision, as in $N(\mu, 1/\sigma^2)$.) –  whuber Nov 1 '11 at 13:43
We should probably merge this with some previous questions. I'll try to find some relevant links. –  cardinal Nov 1 '11 at 14:52
In addition to @whuber's remarks on notation, there are also the natural parameters of the normal, which probably look quite unnatural to most, though a very good reason exists for calling them as such. –  cardinal Nov 1 '11 at 14:54

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 $\dfrac{1}{b^2}$.

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!

-

Almost,

The variance should be written and not the s.d

See here:

http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

-
I noticed that actually but in the maths course I'm doing, a Normal distribution is defined as N(Mean, Stdev). –  Arvin Nov 1 '11 at 11:47
@Arvin You should understand that the normal distribution has a unique definition $$\frac{1}{\sigma\sqrt{2\pi}}\exp(-(x-\mu)^2/(2\sigma^2))$$ (where $\mu$ is the mean and $\sigma$ the standard deviation) that everyone agrees on, but how it is denoted (whether as $N(\mu,\sigma^2)$ or $N(\mu,\sigma)$ or $N(\mu,1/\sigma^2)$ (cf. comment by whuber) or Gaussian$(\mu, \sigma^2)$, etc) is different depending on the user. –  Dilip Sarwate Nov 1 '11 at 15:00
Hi Dilip, I had never encountered such variations on how to denote the distribution - thanks for sharing. –  Tal Galili Dec 2 '11 at 11:36