Looking for the distribution of the difference of two Gaussians in a weird relationship

Question

I have a covariate $B$ (let's say age) and two different responses $T_1$ and $T_2$. The bivariate distributions of $B,T_1$ as well as $B,T_2$ are bivariate normal and known:

$$ \begin{pmatrix}B\\T_1\end{pmatrix} \sim N \left[ \begin{pmatrix}\mu_B\\\mu_1\end{pmatrix} , \begin{pmatrix}\ \sigma_B^2 & \sigma_{1B} \\ \sigma_{1B} & \sigma_1^2 \end{pmatrix} \right] $$

and

$$ \begin{pmatrix}B\\T_2\end{pmatrix} \sim N \left[ \begin{pmatrix}\mu_B\\\mu_2\end{pmatrix} , \begin{pmatrix}\ \sigma_B^2 & \sigma_{2B} \\ \sigma_{2B} & \sigma_2^2 \end{pmatrix} \right] $$

My goal now is to find the distribution of $\begin{pmatrix}B\\T_1-T_2\end{pmatrix}$.

Here is an image to visualize. The distribution of the black ($T_1$) and red ($T_2$) cloud in the top figure are given, and I am looking for the distribution of the bottom cloud.

I have everything except for the covariance between $T_1$ and $T_2$:

• I know my distribution is a bivariate normal again
• The mean vector of that distribution is simple: $(\mu_B, \mu_1-\mu_2)^T$
• From the covariance matrix, the top left element is $\sigma_B^2$, and the two off-diagonal elements are $\sigma_{1B}-\sigma_{2B}$.
• The variance of $T_1-T_2$ is a problem. I know: $$\text{Var}(T_1-T_2) = \sigma_1^2 + \sigma_2^2 - 2 \cdot \text{Cov}(T_1,T_2)$$

But I have no idea how to compute $\text{Cov}(T_1-T_2)$. I also know:

$$ \text{Cov}(T_1,T_2) = \text{E}(T_1T_2) - \mu_1 \mu_2$$

So equivalently, I am looking for $\text{E}(T_1T_2)$ .

Does anyone have an idea how to get one of these two values?

Edit: Just to clarify: I am not looking for the sample covariance, but for the 'real' covariance. The ovservations in the image are just for visualization

Answer 1

$cov(T_1,T_2)$ is not uniquely determined by the information you have given.

You have not specified that $B,T_1, T_2$ are 3-dimensional Normal, but I will assume you have intended this. Therefore, you should think of a 3 by 3 covariance matrix for this 3-dimensional Normal, whose only constraints are that the covariance entries you have provided in the 2 by 2 matrices hold, and that the 3 by 3 matrix is (symmetric) positive semidefinite.

As an example, let all 3 component variances = 1, and $cov(B,T_1) = cov(B,T_2) = 0.1$. In this example, as I determined by applying semidefinite optimization, $cov(T_1,T_2)$ can be anywhere in the range [-0.98,1], and the 3 by 3 matrix will indeed be positive semidefinite.

Of course, if you change the data in my above example to other values, the range of possible $cov(T_1,T_2)$ will change accordingly.