# Thread: Sum of normal distributions

1. ## Sum of normal distributions

Does anyone know what distribution the sum of two normal distributions will have if they are not independent, but have co-variance a?

If it is needed, they can have the same mean, but I would rather not have this condition, if possible.

I have been looking for this, but I can only find it for independent RVs.

2. Originally Posted by Oli
Does anyone know what distribution the sum of two normal distributions will have if they are not independent, but have co-variance a?

If it is needed, they can have the same mean, but I would rather not have this condition, if possible.

I have been looking for this, but I can only find it for independent RVs.
The sum will be normally distributed. The mean will be the sum of the means,
and to find the variance:

V=E([(x1+x2)-(mu1+mu2)]^2)=E( [(x1-mu1)+((x2-mu2)]^2)

Now expand this and simplify.

RonL

3. Do you know why they will be normally distributed?

I can see that is how to find the mean and variance if they are.

Thankyou very much.

4. The question is related to question 4c) of this:
http://www.maths.ox.ac.uk/filemanager/active?fid=4675

This is the solution sheet, but in his proof of 4c) he appears to have said that two non-independent normal distribtions sum to give a normal distribution with the variance shown (1). (If you are right about two correlated normal distributions, then this is fine).

But he has then said by (2) that cov(X,Y)=0 implies independence. I did not think this was true. Any idea why he would say such a thing?

5. Originally Posted by Oli
[snip]
But he has then said by (2) that cov(X,Y)=0 implies independence.[snip]
This statement is completely wrong. Independence => Cov = 0, but Cov = 0 does NOT => independence.

6. Originally Posted by Oli
Does anyone know what distribution the sum of two normal distributions will have if they are not independent, but have co-variance a?

If it is needed, they can have the same mean, but I would rather not have this condition, if possible.

I have been looking for this, but I can only find it for independent RVs.
Oli,

You have to be careful how you phrase your question.

If X and Y have a joint normal distribution then their sum has a normal distribution. However, if X and Y each have a normal distribution then it is not necessarily the case that they have a joint normal distribution, and then the result does not follow.

See the Wikipedia entry on the multivariate normal distribution,
Multivariate normal distribution - Wikipedia, the free encyclopedia
especially the section titled "A Counterexample".

The same web page explains why the sum is normally distributed if X and Y have a joint normal distribution; see the section titled "Affine Transformation".

It is also true that if X and Y have a joint normal distribution and are uncorrelated then they are independent. The Wikipedia entry says so but does not give a proof. However, a proof is fairly simple. If you look at the joint pdf when the correlation is zero, you will see that it factors into the product of a function of X times a function of Y, which shows that X and Y are independent.

Be careful: If X and Y do not have a joint normal distribution, then being uncorrelated does not imply their independence.

7. Think that sounds exactly right, thanks awkward.

8. Originally Posted by mr fantastic
This statement is completely wrong. Independence => Cov = 0, but Cov = 0 does NOT => independence.
But it does for bivariate normals (because the covariance matrix is diagonal
and so joint distribution comes apart as the product of two normal densities
in the two variables)

RonL