Rotational Invariance of X and Y on Rk: Proving Independence & Distribution

[dymin3]

suppose that X and Y are independent and each rotationally invariant on R^k

a) Let P denote any orthogonal projection with dim P = k₁
determine the distribution of the correlation coefficient r= X'PY/(|PX||PY|)

I think r is a special case of ∑(Xi-barX)(Yi-barY) = X'PX where P = I-n^(-1)11'
but what should I do after?
b) Let X = X1 with Xi on R^ki where i =1,2 and prove
X2
- Each Xi is rotationally invariant in its own right

I tried to prove Maxwell-Hershell [Maxwell-Hershell: x1,...,xn iid N(0,σ²) iff x is rotationally invariant and x1,..,xn are independent] but i was not sure...

- If Vi is in R[from K1 to K] with V'V = I, then V'X = X1

This one, I was not sure how to start T-T

 

[mark726]

a) To determine the distribution of the correlation coefficient r, we can use the formula r= X'PY/(|PX||PY|). Since X and Y are independent, we know that the covariance between X and Y is zero. Therefore, the numerator of the correlation coefficient will be zero. Additionally, since X and Y are rotationally invariant, their covariance matrices will be diagonal matrices with equal variances along the diagonal. This means that |PX| and |PY| will be equal. Therefore, the correlation coefficient r will have a distribution of zero.

b) To prove that each Xi is rotationally invariant in its own right, we can use the definition of rotational invariance. This means that the distribution of X must remain unchanged under any rotation. Since X is composed of X1 and X2, we can look at each component separately. For X1, we can use the Maxwell-Hershell theorem, which states that if x1,...,xn are independently and identically distributed as N(0,σ2), then x is rotationally invariant. Since Xi is a subset of X, it will also be rotationally invariant.

For X2, we can use a similar argument. Since X2 is a subset of X, it will also be rotationally invariant. Additionally, if Vi is in R[from K1 to K] with V'V = I, then V'X = X1. This means that rotating X2 by V will not change its distribution. Therefore, X2 is also rotationally invariant.

For the second part of the question, we can start by using the definition of rotational invariance again. Since V'V = I, this means that V is an orthogonal matrix. This means that rotating X by V will not change its distribution. Therefore, V'X = X1 is also rotationally invariant.