Just need to check the answer for the following 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?

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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 Nov 1 '11 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 Nov 1 '11 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 Nov 1 '11 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 Nov 1 '11 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 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 $\dfrac{1}{b}$.

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!

  • $\begingroup$ 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 Jun 15 '16 at 20:20
  • $\begingroup$ 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 Dec 12 '18 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 Dec 12 '18 at 19:08
  • $\begingroup$ @DilipSarwate Thanks. If you have some spare time, can you please take a look at this math.stackexchange.com/questions/3033702/… $\endgroup$ – q126y Dec 13 '18 at 6:14


The variance should be written and not the s.d

See here:


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    $\begingroup$ I noticed that actually but in the maths course I'm doing, a Normal distribution is defined as N(Mean, Stdev). $\endgroup$ – Arvin Nov 1 '11 at 11:47
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    $\begingroup$ @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. $\endgroup$ – Dilip Sarwate Nov 1 '11 at 15:00
  • $\begingroup$ Hi Dilip, I had never encountered such variations on how to denote the distribution - thanks for sharing. $\endgroup$ – Tal Galili Dec 2 '11 at 11:36

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