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.
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
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 ~ , ~ be Gaussian random variables (not necessarily independent).
Prove that , are independent Gaussian random variables
Look at the bivariate normal density.
In THAT case if , 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 and are indep, then this is correct.
You can prove this via MGFs.
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?
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.)