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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?

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    $\begingroup$ 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 $\endgroup$
    – Arvin
    Commented Nov 1, 2011 at 12:11
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    $\begingroup$ 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 $N(\mu, 1/\sigma^2)$.) $\endgroup$
    – whuber ♦
    Commented Nov 1, 2011 at 13:43
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    $\begingroup$ We should probably merge this with some previous questions. I'll try to find some relevant links. $\endgroup$
    – cardinal
    Commented Nov 1, 2011 at 14:52
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    $\begingroup$ @Dilip I thought you had caught me in a contradiction :-) (because I usually use the SD as a parameter), but not quite: in the multivariate case one doesn't usually represent the covariance matrices as squares. The moral is that if there's a chance of confusion, we should be clear about our parameterization. In the present case, the use of the squares and square roots in the formulas make the meaning obvious, so I don't think there was any need to spell it out. $\endgroup$
    – whuber ♦
    Commented Nov 1, 2011 at 14:53
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    $\begingroup$ 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. $\endgroup$
    – cardinal
    Commented Nov 1, 2011 at 14:54

1 Answer 1

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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 $b$.
    (Some people write $b$ as $\left(\sqrt b \right)^2$ to emphasize that the second parameter is the variance, not the standard deviation).

  • $X \sim N(a,b)$ means that the variance of $X$ is $\dfrac{1}{b}$.
    (In a comment on the Question, Moderator whuber says that $b$ is called the precision, especially in a Bayesian context, and is often written as $\dfrac{1}{\sigma^2}$ where $\sigma$ denotes the standard deviation).

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!

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    $\begingroup$ True, @dilip-sarwate . Not to speak of N($\mu,\tau$), where $\tau=\frac{1}{\sigma^2}$ is the precision. For my own sake, I prefer using lowercase $\phi$ to denote the normal distribution, so as to tie in with the symbol for its cumulative distribution, $\Phi$. $\endgroup$
    – Svein Olav Nyberg
    Commented Jun 15, 2016 at 20:20
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    $\begingroup$ Sorry this is 7 years late. But what if the expected value of two independent random variables are far apart, won't we have a double humped distribution in that case? $\endgroup$
    – q126y
    Commented Dec 12, 2018 at 14:09
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    $\begingroup$ @q126y The pdf of the sum of two random variables, whether independent or not, is not the sum of the pdfs of the random variables (and to forestall your followup query, it is not a weighted sum of the pdfs either); the pdf of the sum of two independent random variables is the convolution of their individual pdfs. So, it does not matter in the least whether the means are vastly different or nearly the same; the pdf of the sum of two independent normal random variables is a single-humped camel with mean and variance as described above, not the double-humped monstrosity that you envision. $\endgroup$
    – Dilip Sarwate
    Commented Dec 12, 2018 at 19:08
  • $\begingroup$ Whenever does b in N(a,b) mean the variance is 1/b.... lol $\endgroup$
    – John D
    Commented Jan 23, 2023 at 2:28
  • $\begingroup$ Who in the world uses the denotes the variance by 1/b...the simplest clear notation is to use sigma or sigma^2. $\endgroup$
    – Abhishek Divekar
    Commented Aug 30, 2023 at 18:12

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