sums of normal random variables need not be normal

A common misconception among students of probability theory is the belief that the sum of two normally distributed ( random variablesMathworldPlanetmath is itself normally distributed. By constructing a counterexample, we show this to be false.

It is however well known that the sum of normally distributed variables X,Y will be normal under either of the following situations.

  • ā€¢

    X,Y are independentPlanetmathPlanetmath.

  • ā€¢

    X,Y are joint normal (

The statement that the sum of two independent normal random variables is itself normal is a very useful and often used property. Furthermore, when working with normal variables which are not independent, it is common to suppose that they are in fact joint normal. This can lead to the belief that this property holds always.

Another common fallacy, which our example shows to be false, is that normal random variables are independent if and only if their covarianceMathworldPlanetmath, defined by


is zero. While it is certainly true that independent variables have zero covariance, the converseMathworldPlanetmath statement does not hold. Again, it is often supposed that they are joint normal, in which case a zero covariance will indeed imply independence.

We construct a pair of random variables X,Y satisfying the following.

  1. 1.

    X and Y each have the standard normal distribution.

  2. 2.

    The covariance, Covā”(X,Y), is zero.

  3. 3.

    The sum X+Y is not normally distributed, and X and Y are not independent.

We start with a pair of independent random variables X,Ļµ where X has the standard normal distribution and Ļµ takes the values 1,-1, each with a probability of 1/2. Then set,

Y={Ļµā¢X,ifĀ ā¢|X|ā‰¤1,-Ļµā¢X,ifĀ ā¢|X|>1.

If S is any measurable subset of the real numbers, the symmetry of the normal distribution implies that ā„™(XāˆˆS) is equal to ā„™(-XāˆˆS). Then, by the independence of X and Ļµ, the distributionPlanetmathPlanetmath of Y conditionalMathworldPlanetmathPlanetmath on Ļµ=1 is given by,


It can similarly be shown that ā„™(YāˆˆSāˆ£Ļµ=-1) is equal to ā„™(XāˆˆS). So, Y has the same distribution as X and is normal with mean zero and varianceMathworldPlanetmath one.

Using the fact that Ļµ has zero mean and is independent of X, it is easily shown that the covariance of X and Y is zero.


As X and Y have zero covariance and each have variance equal to 1, the sum X+Y will have variance equal to 2. Also, the sum satisfies

X+Y={2ā¢Ļµā¢X,ifĀ ā¢|X|ā‰¤1,0,ifĀ ā¢|X|>1.

In particular, this shows that ā„™(|X+Y|>2)=0. However, normal random variables with nonzero variance always have a positive probability of being greater than any given real number. So, X+Y is not normally distributed.

This also shows that, despite having zero covariance, X and Y are not independent. If they were, then the fact that sums of independent normals are normal would imply that X+Y is normal, contradicting what we have just demonstrated.

Title sums of normal random variables need not be normal
Canonical name SumsOfNormalRandomVariablesNeedNotBeNormal
Date of creation 2013-03-22 18:43:44
Last modified on 2013-03-22 18:43:44
Owner gel (22282)
Last modified by gel (22282)
Numerical id 5
Author gel (22282)
Entry type Example
Classification msc 62E15
Classification msc 60E05