Correlation between 2 linear combinations of the same variables / variables in the linear span of 2 vectors

I have 2 normally distributed variables $$X_1$$ and $$X_2$$.

$$Y_1 = aX_1 + bX_2$$

$$Y_2 = cX_1 + dX_2$$

There is no $$\epsilon$$.

ie.

$$Y_1 = aX_1 + bX_2 + (\epsilon=0)$$

$$Y_2 = cX_1 + dX_2 + (\epsilon=0)$$

I wish to know if $$Y_1$$ and $$Y_2$$ are correlated.

I may use the formula to find the covariance of $$Y_1$$ and $$Y_2$$:

$$cov(Y_1,Y_2)=$$

$$cov(aX_1 + bX_2,cX_1 + dX_2)$$ = ac $$Var(X_1)$$ +ad $$cov(X_1,X_2)$$ +bc $$cov(X_1,X_2)$$ + bd $$Var(X_2)$$

There does seem some covariance.

Now I am unable to see if Y_1 and Y_2 are correlated intuitively.

Example 1: Suppose $$X_1$$ and $$X_2$$ are independent. Then the above covariance will only have the variance terms on the RHS (since the covariance terms will be equal to zero) .

Example 2: Suppose further that ac = -bd and $$Var(X_1)$$ = $$Var(X_2)$$, then the RHS = 0.

Can I make a comment on the correlation or covariance of $$Y_1$$ and $$Y_2$$ which is precise?

$$Y_1$$ and $$Y_2$$ belong to the linear span of $$X_1$$ and $$X_2$$. Does that shed light on the correlation between them ? May I know how close to 1 will the correlation between them can be?

Another example for intuition: $$X_1$$ and $$X_2$$ may be equal to a common variable $$X_3$$ + white noise. So when I compute $$Y_1$$ = $$1 X_1$$ + $$1 X_2$$ and $$Y_2$$ = $$1 X_1 - 1 X_2$$ the $$X_3$$ cancels out and I am left with only the noise in the $$Y_2$$.

• You should express this mapping as a matrix vector product: $\mathbf{y} = \mathbf{A}\mathbf{x}$, where $\mathbf{A}$ contains the coefficients $a,b,c,d$. Then, we can use the formula $\mathbb{V}[\mathbf{A}\mathbf{x}] = \mathbf{A}\mathbb{V}[\mathbf{x}]\mathbf{A}^\top$. $Y_1$ and $Y_2$ are correlated iff the top right/bottom left elements of this matrix are nonzero. If $X_1$ is indep of $X_2$ with same variance, this is true iff $[a,b]^\top [c,d]=0$. PS: the linear span of $X_1,X_2$ is not a concept that makes sense since these are scalars. PPS: these conclusions hold w/o normality assumptions. Commented Aug 2, 2023 at 12:09
• @JohnMadden That should be an answer! not a comment. Commented Aug 2, 2023 at 13:21
• @JohnMadden In $y=Ax$, I can assume that $x$ is a column of 2 vectors $X_1$ and $X_2$ and that $y$ is a column of 2 vectors $Y_1$ and $Y_2$. Then $Y_1$ and $Y_2$ are 2 arbitrary but fixed vectors in the linear span of $X_1$ and $X_2$, no? Commented Aug 3, 2023 at 5:21
• @user2338823 When you say "a column of two vectors $X_1$ and $X_2$", it sounds like in your notation $X_1$ and $X_2$ are vectors, which isn't obvious from the question: it sounds like from the question like these are (scalar) gaussian variables. If you meant that you stack the scalars $X_1$ and $X_2$ to form the vector $\mathbf{x}$, we're still not there: for $\mathbf{y}=\mathbf{A}\mathbf{x}$ to be in the span of $\mathbf{x}$ would mean that $\mathbf{y} = \lambda \mathbf{x}$ for some constant $\lambda$, or in other words, that $\mathbf{x}$ is an eigenvector of $\mathbf{A}$. Commented Aug 3, 2023 at 12:03
• While @JonMaddens matrix approach is great, i have to disagree with him about one thing: The set of real valued random variables is closed under both addition ($X +Y$) and scalar multiplication ($rX, r\in \mathbb R$), making it a vector space over $\mathbb R$ and so it is perfectly valid to speak about the span of such random variables. Commented Aug 3, 2023 at 19:04