The problem is: I have bivariate, normally distributed observations (x,y) to a model. But as I can't observe the original variables, I will observe them as (x+p,y+q) where p and q are some error terms which are assumed to be normal. Will the new data follow bivariate normal, or something else?
The new data will be bivariate normal, but: you can expect attenuated correlations and a decrease in power. This is because, since you have to evaluate A = X + p and B = Y + q, there are more sources of variation. E[A] = E[X + p] = E[X] + E[p], assuming that p is independent of X. Likewise, E[B] = E[Y + q] = E[Y] + E[q]. So your means for X and for Y will shift by mean(p) and mean(q), respectively. That's not so bad. But the increased variance comes about thusly: Var(A) = Var(X + p) = Var(X) + Var(p), again assuming p is independent of X, and Var(B) =Var(Y + q) = Var(Y) + Var(q), Var(A,B) = Var(A) + Var(B) + Cov(A,B) = Var(X) + Var(Y) + Cov(X,Y) + Var(p) + Var(q).
Well done Mary, but I think you might be doing BHARGAVARAMA's math stat homework!!
Mary and John, Thanks for your concern. I did this work already and this is not my home work :)- But my doubt is How we can conclude that (A,B) will be distributed as Bivariate normal even though A and B are normal variables.? There is a condition that to be checked that any linear combination of A and B is again a normal. This will be useful for me in my research doing on Measurement Error Models...
Once again thanks for your answers....
I was just joking around ..... I think you just need the theorem that any linear combination of multivariate normal random variables is again multivariate normal, There is not need to check conditions because this is a theorem which characterizes multivariate normal distributions.
See the link: http://www.statlect.com/normal_distribution_linear_combinations.htm
Hi john sir, I understood you were joking. We know that the linear combination of INDEPENDENT normal random variables is normal.But in the case of regression x and y need not to be independent. Thats where i am confusing...
Thanks in advance sir....
Bhargavarama, since (X,Y) as bivariate normal, then f(X,Y) is a function of g(X), h(Y), and their covariance. If the marginal distributions g(X) and h(Y) are normal, then you want to know if g(X+p) and h(Y+q) are normal, since the latter are marginal distributions of (A,B). I'll end this with every mathematician's favorite phrase "It can be shown..."
"We know that the linear combination of INDEPENDENT normal random variables is normal."
The independent case is just a special case of the broader theorem. Independence is not required. The theorem and proof are included in the link I posted where you should note that if independence were required the matrix V would be an identity matrix.
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