Express $\rho_{X+U,Y}$ , in terms of, $\rho_{X,Y}$ ,$\rho_{U,Y}$ ,$\sigma{X}$ and $\sigma{U}$ where $X$ and $U$ are independent

Question

I need to express the correlation of $(X+U)$ and $Y$, in terms of the correlation between $X$ and $Y$ , the correlation between $U$ and $Y$ and the standard deviations of $X$ and $U$.

I know that $\text{Cov}(X,U) = 0$ implying $\rho_{X,U} = 0$, and know that: $\text{Cov}(X+U,Y) = E((X+U) - \mu_{X+U}) * (Y) - \mu_{Y}))$, but got stuck trying to simplify

Answer 1

We can compute \begin{align*} \text{Var}(X+U) & = \mathbb{E}[(X+U)^2] - [\mathbb{E}[X+U]]^2 \\ & = \mathbb{E}[X^2]+\mathbb{E}[U^2]+2\mathbb{E}[XU]-\mathbb{E}[X^2]-\mathbb{E}[U^2]-2\mathbb{E}[X]\mathbb{E}[U] \\ & = \mathbb{E}[X^2]-[\mathbb{E}[X]]^2 + \mathbb{E}[U^2]-[\mathbb{E}[U]]^2 \\ & = \text{Var}(X)+\text{Var}(U). \end{align*} Therefore $\sigma_{X+U}=\sqrt{\sigma_X^2+\sigma_U^2}$.

We also know that, since covariance is linear in each slot, \begin{align*} \text{Cov}(X+U,Y)=\text{Cov}(X,Y)+\text{Cov}(U,Y).\end{align*} I would say this is sufficiently simplified, but in order to achieve the goal stated in the title of the question, we want to rewrite $$\text{Cov}(X,Y)+\text{Cov}(U,Y)=\rho_{X,Y}\sigma_X\sigma_Y + \rho_{U,Y}\sigma_U\sigma_Y.$$

Now we divide. \begin{align*} \rho_{X+U,Y} & = \frac{\text{Cov}(X+U,Y)}{\sigma_{X+U}\sigma_Y} = \frac{\rho_{X,Y}\sigma_X\sigma_Y + \rho_{U,Y}\sigma_U\sigma_Y}{\sqrt{\sigma_X^2+\sigma_U^2}\sigma_Y} \\ & = \frac{\sigma_X}{\sqrt{\sigma_X^2+\sigma_U^2}}\cdot \rho_{X,Y} + \frac{\sigma_U}{\sqrt{\sigma_X^2+\sigma_U^2}}\cdot \rho_{U,Y} \end{align*}

Intuitively, this is a weighted sum of the individual correlations $\rho_{X,Y}, \rho_{U,Y}$, where the weights' squares sum to $1$ and the weights are according to the square roots of their proportions of the variance.

This would remain true with more variables. Let $X=\sum_{i=1}^n X_i$, $X_i$ uncorrelated. Then $\text{Var}(X)=\sum_{i=1}^n \text{Var}(X_i)=\sum_{i=1}^n \sigma_{X_i}^2$ and $\text{Cov}(X,Y)=\sum_{i=1}^n \text{Cov}(X_i,Y) = \sum_{i=1}^n \rho_{X_i}\sigma_{X_i}\sigma_Y$. Then $$\rho_{X,Y} = \frac{\sigma_Y\sum_{i=1}^n \sigma_{X_i}\rho_{X_i,Y}}{\sigma_Y\sqrt{\sum_{i=1}^n \sigma_{X_i}^2}} = \sum_{i=1}^n \frac{\sigma_{X_i}}{\sqrt{\sum_{j=1}^n \sigma_{X_j}^2}}\cdot \rho_{X_i,Y}.$$

The weight $\sigma_{X_i}/\sqrt{\sum_{j=1}^n \sigma_{X_j}^2}$ is the square root of the proportion of the variance of $X=\sum_{j=1}^n X_j$ which is contributed by $X_i$.