Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The sum of two Gaussian variables is another Gaussian.
It seems natural, but I could not find a proof using Google.

What's a short way to prove this?

Edit: Provided the two variables are independent.

share|cite|improve this question
What search terms were you using? I found this as the top result on Google. – Larry Wang Jul 21 '10 at 1:31
Please make sure to use well-defined terms and more appropriate tags. – user126 Jul 21 '10 at 1:33
@Nicholas: You need to mention that the two variables are independent, otherwise it's not true (if X is N(0,1), then so is Y=-X, but their sum X+Y=0 is not normally distributed). – Simon Nickerson Jul 21 '10 at 16:50
@Simon +1 indeed thanks! I edit the question. – Nicolas Raoul Jul 28 '10 at 0:53
up vote 6 down vote accepted

I prepared the following as an answer to a question which happened to close just as I was putting the finishing touches on my work. I posted it as a different (self-answered) question but following suggestions from Srivatsan Narayanan and Mike Spivey, I am putting it here and deleting my so-called question.

If $X$ and $Y$ are independent standard Gaussian random variables, what is the cumulative distribution function of $\alpha X + \beta Y$?

Let $Z = \alpha X + \beta Y$. We assume without loss of generality that $\alpha$ and $\beta$ are positive real numbers since if, say, $\alpha < 0$, then we can replace $X$ by $-X$ and $\alpha$ by $\vert\alpha\vert$. Then, the cumulative probability distribution function of $Z$ is $$ F_Z(z) = P\{Z \leq z\} = P\{\alpha X + \beta Y \leq z\} = \int\int_{\alpha x + \beta y \leq z} \phi(x)\phi(y) dx dy $$ where $\phi(\cdot)$ is the unit Gaussian density function. But, since the integrand $(2\pi)^{-1}\exp(-(x^2 + y^2)/2)$ has circular symmetry, the value of the integral depends only on the distance of the origin from the line $\alpha x + \beta y = z$. Indeed, by a rotation of coordinates, we can write the integral as $$ F_Z(z) = \int_{x=-\infty}^d \int_{y=-\infty}^{\infty}\phi(x)\phi(y) dx dy = \Phi(d) $$ where $\Phi(\cdot)$ is the standard Gaussian cumulative distribution function. But, $$d = \frac{z}{\sqrt{\alpha^2 + \beta^2}}$$ and thus the cumulative distribution function of $Z$ is that of a zero-mean Gaussian random variable with variance $\alpha^2 + \beta^2$.

share|cite|improve this answer
I like this proof very much, because it explicitly uses the rotational symmetry, and therefore makes it clear why the Gaussian has this property but other distributions do not. – Michael Lugo Sep 19 '11 at 23:34

I don't know how I missed that one, indeed:
Thanks Kaestur Hakarl!

share|cite|improve this answer
Heh, at least you have your solution now. :) – Noldorin Jul 21 '10 at 16:54

I posted the following in response to a question that got closed as a duplicate of this one:

It looks from your comment as if the meaning of your question is different from what I thought at first. My first answer assumed you knew that the sum of independent normals is itself normal.

You have $$ \exp\left(-\frac12 \left(\frac{x}{\alpha}\right)^2 \right) \exp\left(-\frac12 \left(\frac{z-x}{\beta}\right)^2 \right) = \exp\left(-\frac12 \left( \frac{\beta^2x^2 + \alpha^2(z-x)^2}{\alpha^2\beta^2} \right) \right). $$ Then the numerator is $$ \begin{align} & (\alpha^2+\beta^2)x^2 - 2\alpha^2 xz + \alpha^2 z^2 \\ \\ & = (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz\right) + \alpha^2 z^2 \\ \\ & = (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz + \frac{\alpha^4}{(\alpha^2+\beta^2)^2}z^2\right) + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2 \\ \\ & = (\alpha^2+\beta^2)\left(x - \frac{\alpha^2}{\alpha^2+\beta^2}z\right)^2 + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2, \end{align} $$ and then remember that you still have the $-1/2$ and the $\alpha^2\beta^2$ in the denominator, all inside the "exp" function.

(What was done above is completing the square.)

The factor of $\exp\left(\text{a function of }z\right)$ does not depend on $x$ and so is a "constant" that can be pulled out of the integral.

The remaining integral does not depend on "$z$" for a reason we will see below, and thus becomes part of the normalizing constant.

If $f$ is any probability density function, then $$ \int_{-\infty}^\infty f(x - \text{something}) \; dx $$ does not depend on "something", because one may write $u=x-\text{something}$ and then $du=dx$, and the bounds of integration are still $-\infty$ and $+\infty$, so the integral is equal to $1$.

Now look at $$ \alpha^2z^2 - \frac{\alpha^4}{\alpha^2+\beta^2} z^2 = \frac{z^2}{\frac{1}{\beta^2} + \frac{1}{\alpha^2}}. $$

This was to be divided by $\alpha^2\beta^2$, yielding $$ \frac{z^2}{\alpha^2+\beta^2}=\left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2. $$ So the density is $$ (\text{constant})\cdot \exp\left( -\frac12 \left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2 \right) . $$ Where the standard deviation belongs we now have $\sqrt{\alpha^2+\beta^2}$.

share|cite|improve this answer
Did you read this? – Did Sep 19 '11 at 5:54
I may be the author of that. It depends on whether, and how much, others may have edited that in recent years. – Michael Hardy Sep 19 '11 at 17:56
From the wikipedia history page: "cumulative density" is an idiotic self-contradictory phrase... Nice! – The Chaz 2.0 Sep 19 '11 at 19:31
@The I disagree with that sentiment. It's not like all terminology in math makes a lot of sense. And I somehow always remember it cumulative density, but in my defense I just thought of it as some "cumulative of the density". :-) – Srivatsan Sep 19 '11 at 22:54
@Sri : I just thought it quotable. You'll have to consult the author (who is notiffied already)! – The Chaz 2.0 Sep 19 '11 at 23:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.