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I have measurements of 2 position vectors ($\mathbf p_1$ and $\mathbf p_2$):

  • Each with their own mean position vectors $(\overline x_1, \overline y_1, \overline z_1)^T$ and $(\overline x_2,\overline y_2,\overline z_2)^T$ respectively,
  • Each with their own $3 \times 3$ variance-covariance matrices ($\Sigma_1$ and $\Sigma_2 $) respectively.
  • $\mathbf p_1$ and $\mathbf p_2$ are independent.

How do I find the variance and covariance of ($\mathbf p_2 - \mathbf p_1$)? In other words, what is the variance and covariance of relative position vector $(\overline x_2 - \overline x_1, \overline y_2 - \overline y_1, \overline z_2 - \overline z_1)^T$?

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  • $\begingroup$ Welcome to Math.SE! Can you expand on what you already know yourself? For example, what is the definition of variance and covariance? $\endgroup$ Commented May 5, 2015 at 9:48
  • $\begingroup$ When I said variance, I mean $Var()$ and when I said covariance, I mean $Cov()$ $\endgroup$ Commented May 5, 2015 at 10:00
  • $\begingroup$ I am sorry, but that is extremely non-informative. What does $Var()$ mean? $\endgroup$ Commented May 5, 2015 at 10:21
  • $\begingroup$ Yes. You are right. It is not so informative. Sorry for that. Let me rephrase of what I meant. Variance of relative position vector means the diagonal terms of variance-covariance matrix, while covariance of relative position vector means off-diagonal terms of variance-covariance matrix. I hope this is acceptable. $\endgroup$ Commented May 5, 2015 at 12:28
  • $\begingroup$ For me this still does not mean anything, because I don't know what the variance-covariance matrix is. This might be general knowledge in your field, I cannot tell. $\endgroup$ Commented May 5, 2015 at 14:25

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At least if $\mathbf p_1$ and $\mathbf p_2$ are each multivariate (incl. bivariate) normal, then, as they are independent, you can just sum the covariance matrices to get the covariance matrix of the relative position vector:

$\Sigma_{2-1} = \Sigma_1 + \Sigma_2 $

for the derivation see e.g.: Multivariate Normal Difference Distribution

right now I am not unfortunately not sure about the general case (i.e. other distributions).

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    $\begingroup$ Do you means that I can obtain variance-covariance matrix of relative position vector as follows? $$\sigma^2_{x_{2-1}} = \sigma^2_{x_2} + \sigma^2_{x_1} $$ $$\sigma^2_{y_{2-1}} = \sigma^2_{y_2} + \sigma^2_{y_1} $$ $$\sigma^2_{z_{2-1}} = \sigma^2_{z_2} + \sigma^2_{z_1} $$ and $$Cov({x_{2-1}},{y_{2-1}}) = Cov({x_2},{y_2}) + Cov({x_1},{y_1})$$ $$Cov({x_{2-1}},{z_{2-1}}) = Cov({x_2},{z_2}) + Cov({x_1},{z_1})$$ $$Cov({y_{2-1}},{z_{2-1}}) = Cov({y_2},{z_2}) + Cov({y_1},{z_1})$$ $\endgroup$ Commented May 5, 2015 at 10:15
  • $\begingroup$ Yes, that's how I have understood it. If $p_1$ and $p_2$ are each multivariate normal, you can. For other distributions I'm just not sure at the moment. $\endgroup$ Commented May 5, 2015 at 10:26
  • $\begingroup$ think I missed the title - where it says bivariate, which is a special case of multivariate, so, yes, here you can just sum the covariance matrices. $\endgroup$ Commented May 5, 2015 at 13:24

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