# How to calculate a cross-product like R^2?

$$\lbrace Y_1, Y_2, \boldsymbol{X} \rbrace$$ 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.

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