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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:

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

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?

whuber
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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](http://www.wolframalpha.com/input/?i=normal+distribution%280%2C+2%29) 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](http://stats.stackexchange.com/questions/17710/weighted-mean-of-two-2d-gaussian-random-variables/17718#17718) 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](http://en.wikipedia.org/wiki/Exponential_family#Normal_distribution:_Unknown_mean_and_unknown_variance), 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

2 Answers2

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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 $b$.

  • $X \sim N(a,b)$ 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!

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
  • 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
  • @DilipSarwate Thanks. If you have some spare time, can you please take a look at this https://math.stackexchange.com/questions/3033702/combining-chi-square-charecterisitc – q126y Dec 13 '18 at 06:14
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Almost,

The variance should be written and not the s.d

See here:

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

Tal Galili
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    I noticed that actually but in the maths course I'm doing, a Normal distribution is defined as N(Mean, Stdev). – Arvin Nov 01 '11 at 11:47
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    @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 01 '11 at 15:00
  • Hi Dilip, I had never encountered such variations on how to denote the distribution - thanks for sharing. – Tal Galili Dec 02 '11 at 11:36