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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?

BeBlunt
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Arvin
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  • 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 01 '11 at 11:54
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    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 01 '11 at 12:11
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    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 01 '11 at 13:43
  • It is also possible that variances are denoted by $\sigma$ (with appropriate subscripts) rather than $\sigma^2$ with appropriate subscripts, as in this answer by @whuber. So notation does indeed vary a lot. – Dilip Sarwate Nov 01 '11 at 14:48
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    We should probably merge this with some previous questions. I'll try to find some relevant links. – cardinal Nov 01 '11 at 14:52
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    @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. – whuber Nov 01 '11 at 14:53
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    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 01 '11 at 14:54

1 Answers1

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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!

Dilip Sarwate
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    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, $\Phi$. – Svein Olav Nyberg Jun 15 '16 at 20:20
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    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? – q126y Dec 12 '18 at 14:09
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    @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. – Dilip Sarwate Dec 12 '18 at 19:08
  • Whenever does b in N(a,b) mean the variance is 1/b.... lol – John D Jan 23 '23 at 02:28
  • Who in the world uses the denotes the variance by 1/b...the simplest clear notation is to use sigma or sigma^2. – Abhishek Divekar Aug 30 '23 at 18:12
  • @AbhishekDivekar Please read the comments on the main question to discover those in the world who are familiar with the interpretation of the second parameter in $N(a,b)$ as $1/\sigma^2$, the reciprocal of the variance.. The suspects include Moderators whuber and cardinal. As Shakespeare wrote "There are more things in heaven and earth than are dreamt of in thy philosophy" – Dilip Sarwate Aug 30 '23 at 21:54
  • @DilipSarwate Agreed! My comment was a bit tongue-in-cheek. There's no objective right answer when it comes to notation. Certain notations feel more "natural" depending on the reader's background and supporting work of the field, not to mention personal preference. – Abhishek Divekar Aug 31 '23 at 08:42