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A standardized Gaussian distribution on $\mathbb{R}$ can be defined by giving explicitly its density: $$ \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$

or its characteristic function.

As recalled in this question it is also the only distribution for which the sample mean and variance are independent.

What are other surprising alternative characterization of Gaussian measures that you know ? I will accept the most surprising answer :)

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12 Answers

up vote 21 down vote accepted

My personal most surprising is the one about the sample mean and variance, but here is another (maybe) surprising characterization: if $X$ and $Y$ are IID with finite variance with $X+Y$ and $X-Y$ independent, then $X$ and $Y$ are normal.

Intuitively, we can usually identify when variables are not independent with a scatterplot. So imagine a scatterplot of $(X,Y)$ pairs that looks independent. Now rotate by 45 degrees and look again: if it still looks independent, then the $X$ and $Y$ coordinates individually must be normal (this is all speaking loosely, of course).

To see why the intuitive bit works, take a look at

$$ \left[ \begin{array}{cc} \cos45^{\circ} & -\sin45^{\circ} \newline \sin45^{\circ} & \cos45^{\circ} \end{array} \right] \left[ \begin{array}{c} x \newline y \end{array} \right]= \frac{1}{\sqrt{2}} \left[ \begin{array}{c} x-y \newline x+y \end{array} \right] $$

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Jay - this is basically a re-statement of the mean and variance being independent. $X+Y$ is a re-scaled mean and $X-Y$ is a rescaled standard deviation. – probabilityislogic Mar 15 '11 at 7:24
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@probabilityislogic - I like the intuition of what you said, but I don't think it is exactly a restatement because $X - Y$ isn't exactly a rescaling of the SD: the SD forgets the sign. So independence of mean and SD follows from independence of $X+Y$, $X-Y$ (when $n = 2$), but not the other way around. That may have been what you meant by "basically". Anyway, it's good stuff. – G. Jay Kerns Mar 15 '11 at 16:20
is there connection to Box-Muller transformation ? – Qbik Feb 13 at 22:12
up vote 16 down vote

The continuous distribution with fixed variance which maximizes differential entropy is the Gaussian distribution.

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up vote 11 down vote

Gaussian distributions are the only sum-stable distributions with finite variance.

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5  
That they are sum stable and that they are the unique ones with finite variance are both forced on us by the CLT. The interesting part of this assertion is that there exist other sum-stable distributions! – whuber♦ Nov 9 '10 at 23:12
@whuber: indeed! this characterization is a bit contorted, and the other sum-stable distributions are perhaps more curious. – shabbychef Nov 10 '10 at 16:58
@whuber actually, I do not see how the CLT implies this fact. It only seems to tell us that asymptotically, the sum of normals is normal, not that any finite sum is normally distributed. Or do you have to somehow use Slutsky's theorem as well? – shabbychef Nov 11 '10 at 4:20
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Adopting the usual standardization, a sum of two normals is the sum of one normal distribution X_0 plus the limiting distribution of a series X_1, X_2, ..., whence the sum is the limiting distribution of X_0, X_1, ..., which by the Lindeberg-Levy CLT is normal. – whuber♦ Nov 11 '10 at 13:28
up vote 10 down vote

There's an entire book written about this: "Characterizations of the normal probability law", A. M. Mathai & G. Perderzoli. A brief review in JASA (Dec. 1978) mentions the following:

"Let $X_1, \ldots, X_n$ be independent random variables. Then $\sum_{i=1}^n{a_i x_i}$ and $\sum_{i=1}^n{b_i x_i}$ are independent, where $a_i b_i \ne 0$, if and only if $X_i$ [are] normally distributed."

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there must be a condition such $<a,b>=0$ missing ? for example if n=2 $a_i=b_i=1$ $X_1+X_2$ and $X_1+X_2$ are not independant. – robin girard Dec 17 '10 at 8:38
@robin good catch. I've been puzzling over the implicit quantifiers, too. Unfortunately, all I have access to is that (titillating) quotation from the review, not the book. It would be fun to find it in a library and browse through it... – whuber♦ Dec 17 '10 at 23:03
This feels like a generalization of G. Jay Kerns' (currently #1) answer. – vqv Dec 19 '10 at 21:53
I think you may be looking for the Lukacs & King (1954) paper. See this answer at math.SE with a link to the aforementioned paper. – cardinal Mar 14 '12 at 4:00
Where this proposition says "where $a_i b_i\ne 0$", does it mean for EVERY set of scalars where $a_i b_i \ne 0$"? I hate seeing "where" used in place of "for every" or "for some". "Where" should be used to explain one's notation, as in "where $c$ is the speed of light and $g$ is the gross domestic product", etc. – Michael Hardy Feb 9 at 1:23
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up vote 9 down vote

Stein’s Lemma provides a very useful characterization. $Z$ is standard Gaussian iff $$E f’(Z) = E Z f(Z)$$ for all absolutely continuous functions $f$ with $E|f’(Z)| < \infty$.

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up vote 6 down vote

Let $\eta$ and $\xi$ be two independent random variables with a common symmetric distribution such that

$$ P\left ( \left |\frac{\xi+\eta}{\sqrt{2}}\right | \geq t \right )\leq P(|\xi|\geq t).$$

Then these random variables are gaussian. (Obviously, if the $\xi$ and $\eta$ are centered gaussian, it is true.)

This is the Bobkov-Houdre Theorem

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up vote 6 down vote

This is not a characterization but a conjecture, which dates back from 1917 and is due to Cantelli:

If $f$ is a positive function on $\mathbb{R}$ and $X$ and $Y$ are $N(0,1)$ independent random variables such that $X+f(X)Y$ is normal, then $f$ is a constant almost everywhere.

Mentioned by Gérard Letac here.

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it is good that you mention it! I can't figure out the intuition, do you? – robin girard Feb 15 '11 at 17:51
@robin This is what makes this conjecture so special: a completely elementary statement, some obvious approaches which fail miserably (characteristic functions), and one is left with nothing to grasp... By the way, should one bet on the conjecture being true or false? Even that is not obvious (to me). – Did Feb 15 '11 at 18:13
If Gérard Letac has not managed to prove it, it could stay an open conjecture for quite a while...! – Xi'an Jan 12 '12 at 20:31
@Xi'an: I fully agree, of course. (Didn't know you were roaming in these quarters of the web... Good news that you are.) – Did Jan 12 '12 at 21:14
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@Xi'an Here is a preprint by Victor Kleptsyn and Aline Kurtzmann with a counterexample to the Cantelli conjecture. The construction uses a new tool, which the authors call the Brownian mass transport, and yields a discontinuous function $f$. The authors state that they believe that Cantelli conjecture holds if one asks that $f$ is continuous (theirs is a mixture of two continuous functions). – Did Apr 26 '12 at 8:11
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up vote 3 down vote

Theorem [Herschel-Maxwell]: Let $Z \in \mathbb{R}^n$ be a random vector for which (i) projections into orthogonal subspaces are independent and (ii) the distribution of $Z$ depends only on the length $||Z||$. Then $Z$ is normally distributed.

Cited by George Cobb in Teaching statistics: Some important tensions (Chilean J. Statistics Vol. 2, No. 1, April 2011) at p. 54.

Cobb uses this characterization as a starting point for deriving the $\chi^2$, $t$, and $F$ distributions, without using Calculus (or much probability theory).

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up vote 3 down vote

A more particular characterisation of the normal distribution among the class of infinitely divisible distributions is presented in Steutel and Van Harn (2004).

A non-degenerate infinitely divisible random variable $X$ has a normal distribution if and only if it satisfies $$-\limsup_{x\rightarrow\infty}\dfrac{\log{\mathbb P}(\vert X\vert>x)}{x\log(x)}=\infty.$$

This result characterises the normal distribution in terms of its tail behaviour.

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A short proof of the stated limit goes as follows: If $X$ is standard normal, then $x \mathbb P(X>x)/\varphi(x)\to 1$ as $x\to \infty$, so $\log\mathbb P(X>x)−\log \varphi(x) + \log x \to 0$. But $2 \log \varphi(x) \sim −x^2$ and so the result follows. A rough sketch for the case of the Poisson seems to indicate that the given limit is $\lambda$, but I didn't check that too closely. – cardinal Jun 6 '12 at 19:41
up vote 2 down vote

Suppose one is estimating a location parameter using i.i.d. data $\{x_1,...,x_n\}$. If $\bar{x}$ is the maximum likelihood estimator, then the sampling distribution is Gaussian. According to Jaynes's Probability Theory: The Logic of Science pp. 202-4, this was how Gauss originally derived it.

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up vote 1 down vote

Its characteristic function has the same form as its pdf. I am not sure of another distribution which does that.

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See [this answer] of mine for ways of constructing random variables whose characteristic functions are the same as their pdfs. – Dilip Sarwate Jul 31 '12 at 1:23
up vote -2 down vote

The expectation plus minus the standard deviation are the saddle points of the function.

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7  
This is a property of the Normal distribution, to be sure, but it does not characterize it, because plenty of other distributions also have this property. – whuber♦ Nov 10 '10 at 21:18

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