1
$\begingroup$

$\lbrace Y_1, Y_2, \boldsymbol{X} \rbrace$ are jointly normally distributed (it is not essential to assume normality, I think). Let $\Sigma_{X}$ be the variance-covariance matrix of $\boldsymbol{X}.$ Let $\Sigma_{1X}$ and $\Sigma_{2X}$ be covariance vectors of $Y_1$ and $Y_2$ with $\boldsymbol{X},$ respectively. Let $\sigma_1$ and $\sigma_2$ be the standard deviations of $Y_1$ and $Y_2.$ Suppose I have a sample drawn from this multivariate distribution: $\lbrace y_{1i}, y_{2i}, \boldsymbol{x}_i \rbrace, i=1, \ldots, N$. I can estimate $S_1 = (1/\sigma_1^2) \Sigma_{1X} \Sigma_{X}^{-1} \Sigma_{1X}$ as the $R^2$ from the linear regression of $y_1$ versus $\boldsymbol{x}$. Similarly I can estimate $S_2 = (1/\sigma_2^2) \Sigma_{2X} \Sigma_{X}^{-1} \Sigma_{2X}$ as the $R^2$ from the linear regression of $y_2$ versus $\boldsymbol{x}$.

Now here is my question: How can I estimate the quantity $T = (1/\sigma_1 \sigma_2)\Sigma_{2X} \Sigma_{X}^{-1} \Sigma_{1X}$ without directly computing the covariance matrices? Is there a good interpretation of $T?$

I tried combining the $R^2$ estimates from the 3 regression models ($y_1 \sim \boldsymbol{x}$, $y_2 \sim \boldsymbol{x}$, and $y_2-y_1 \sim \boldsymbol{x}$) to estimate $T$, but it is not working.

$\endgroup$

2 Answers 2

0
$\begingroup$

I was able to get an answer as follows: $$T = \left[ S_1 \sigma_1^2 + S_2 \sigma_2^2 - S_{12}(\sigma_1^2 + \sigma_2^2 - 2\rho \sigma_1 \sigma_2) \right] /(2\sigma_1 \sigma_2),$$ where $\rho$ is the correlation between $Y_1$ and $Y_2.$ I would still appreciate other/better ways of estimating this, as well as how to interpret this quantity $T.$

$\endgroup$
0
$\begingroup$

I found an even simpler answer: $$\hat{T} = \mbox{Cov}(\boldsymbol{X \hat{\beta}}_1,\boldsymbol{X \hat{\beta}}_2)/(\hat{\sigma}_1 \hat{\sigma}_2).$$

This also provide a clean interpretation of $T.$

Furthermore: $$\hat{S}_1 = \mbox{var}(\boldsymbol{X \hat{\beta}}_1)/\hat{\sigma}_1^2; \, \mbox{ and } \, \hat{S}_2 = \mbox{var}(\boldsymbol{X \hat{\beta}}_2)/\hat{\sigma}_2^2.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.