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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.

multivariate gaussian pic

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

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    $\begingroup$ For what reason can you not calculate $COV(T_1,T_2)$? Is this output from a model? You will need to estimate the covariance for this problem. $\endgroup$ – Zachary Blumenfeld Feb 12 '16 at 20:33
  • $\begingroup$ Based on your graphs, It looks like you can estimate the covariance using the sample estimate $\endgroup$ – Zachary Blumenfeld Feb 12 '16 at 20:36
  • $\begingroup$ I obtained the samples in the bottom graph through a very inconvenient detour, with kernel density estimates and stuff. I could estimate that sample's covariance, but I would prefer to have the 'true' value, given the known parameters of the other two distributions. $\endgroup$ – Alexander Engelhardt Feb 12 '16 at 20:39
  • $\begingroup$ I don't know what "kernel density estimation and stuff" is (maybe you can provide more context here). I do know that, in general, you cannot estimate/know the $var(T_1-T_2)$ without estimating/knowing the $cov(T_1,T_2)$. $\endgroup$ – Zachary Blumenfeld Feb 12 '16 at 20:45
  • $\begingroup$ Exactly. That's why I am looking for $cov(T_1,T_2)$ here. The 'kernel density estimation' part is, as far as I see it, irrelevant to my question. I just used it to generate the plot. $\endgroup$ – Alexander Engelhardt Feb 12 '16 at 20:50
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$cov(T_1,T_2)$ 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.

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