# Sum & difference of dependent normal variables?

• Apr 14th 2009, 07:48 PM
HD09
Sum & difference of dependent normal variables?
Given two normal random variables, which are not necessarily independent, with different means but the same variance, I need to show that:

1) The sum of these two variables is normal
2) The difference of these two variables is normal
3) The sum and the difference are independent

Can someone start me off/point me in the right direction/suggest a resource for this question? The "not necessarily independent" part has me stumped
• Apr 14th 2009, 08:37 PM
matheagle
What if I let Y=-X, they have the same variance and the sum is ZERO, which is not normal.
I'm confused.
(3) Cov(X+Y,X-Y)=V(X)-V(Y)=0 which would imply indep IF we have normality.
• Apr 14th 2009, 09:25 PM
HD09
Regarding (3) - I know independence impies zero correlation, but I thought the converse wasn't necessarily true?

For (1) & (2)... Maybe I've missed something in the wording of the question. This is what I have, word for word:

Let $\displaystyle Z_{1}$~$\displaystyle N(m_{1},\sigma^{2})$, $\displaystyle Z_{2}$~$\displaystyle N(m_{2},\sigma^{2})$ be Gaussian random variables (not necessarily independent).

Prove that $\displaystyle X_{1} = Z_{1}+Z_{2}$, $\displaystyle X_{2} = Z_{1}-Z_{2}$ are independent Gaussian random variables
• Apr 14th 2009, 10:04 PM
matheagle
Look at the bivariate normal density.
In THAT case if $\displaystyle \rho =0$, then the joint density factors, hence they are indep.
see Multivariate normal distribution - Wikipedia, the free encyclopedia

Again, what if X=Y, then the difference is not normal.
IF $\displaystyle Z_1$ and $\displaystyle Z_2$ are indep, then this is correct.
You can prove this via MGFs.
• Apr 14th 2009, 10:30 PM
HD09
Yes, I agree. If Z1 and Z2 were independent then X1 and X2 would be JOINTLY normal and in that case zero correlation would imply independence. That's fine.

BUT, I'm given specifically that Z1 and Z2 are NOT necessarily independent...

Which throws into question whether X1 and X2 are even seperately normally distributed. The only clue seems to be the equal variances thing...

Re: "zero is not normal" - could you not consider zero to be a normal random variable with zero mean and zero variance?
• Apr 14th 2009, 10:36 PM
matheagle
That is a discrete rv known as $\displaystyle P(X=0)=1$.

How can you write that as $\displaystyle {1\over \sqrt{2\pi}\sigma}e^{-(x-\mu)^2\over 2\sigma^2}$

You can't divide by zero.
• Apr 14th 2009, 10:50 PM
HD09
Well, the pdf wouldn't exist, but I've seen that convention adopted in a couple of textbooks... p247 of Bertsekas & Tsitsiklis for example

So you're saying the question makes no sense?
• Apr 14th 2009, 10:58 PM
matheagle
A normal rv is a continuous rv defined on (-infinity,infinity).
P(X=0)=1 is certainly not defined on the entire real line.
• Apr 14th 2009, 11:14 PM
HD09
I'm not saying it is.

Quote:

Originally Posted by HD09
but I've seen that convention adopted in a couple of textbooks... p247 of Bertsekas & Tsitsiklis for example

So you're saying the question makes no sense?

• Apr 16th 2009, 02:36 PM
awkward
Quote:

Originally Posted by HD09
Given two normal random variables, which are not necessarily independent, with different means but the same variance, I need to show that:

1) The sum of these two variables is normal
2) The difference of these two variables is normal
3) The sum and the difference are independent

Can someone start me off/point me in the right direction/suggest a resource for this question? The "not necessarily independent" part has me stumped

I think you must assume that the two normal variables have a joint bivariate normal distribution. Let's say the variables are X and Y. Define U = X + Y and V = X - Y. Then apply the change of variables theorem to find the joint distribution of U and V. Once that is done the rest of the problem should be easy.

(Whether this works in the case where the correlation is -1 is a matter of definition, and I don't know whether the commonly accepted definition of the normal distribution is broad enough to encompass this case.)