0
$\begingroup$

Assume that we have 3 3D points to make this questions simpler.

$$ pt_1 = \begin{bmatrix} x_1 & y_1 & z_1 \end{bmatrix} \\ pt_2 = \begin{bmatrix} x_2 & y_2 & z_2 \end{bmatrix} \\ pt_3 = \begin{bmatrix} x_3 & y_3 & z_3 \end{bmatrix} $$

and we also have the variance-covariance matrix for each point

$$ c_1 = \begin{bmatrix} s_{x_1} & s_{x_1y_1} & s_{x_1z_1} \\ s_{x_1y_1} & s_{y_1} & s_{y_1z_1} \\ s_{x_1z_1} & s_{y_1z_1} & s_{z_1} \end{bmatrix}, c_2 = \begin{bmatrix} s_{x_2} & s_{x_2y_2} & s_{x_2z_2} \\ s_{x_2y_2} & s_{y_2} & s_{y_2z_2} \\ s_{x_2z_2} & s_{y_2z_2} & s_{z_2} \end{bmatrix} c_3 = \begin{bmatrix} s_{x_3} & s_{x_3y_3} & s_{x_3z_3} \\ s_{x_3y_3} & s_{y_3} & s_{y_3z_3} \\ s_{x_3z_3} & s_{y_3z_3} & s_{z_3} \end{bmatrix} $$

I can easily calculate the average 3d point comprised of the 3 points above. How would I calculate the variance-covariance matrix of the calculated average point ?

$\endgroup$
3
  • 1
    $\begingroup$ Hint: covariance is a linear operation. $\endgroup$
    – amoeba
    May 23, 2016 at 19:12
  • $\begingroup$ I read that the variance parts are pretty much the quadrature over 3, in this case. So there is no difference for the correlations then I guess. $\endgroup$
    – xerion
    May 23, 2016 at 19:22
  • $\begingroup$ Just to clarify my previous comment: assuming that all between-point covariances are zero, you have simply $c=(c_1+c_2+c_3)/9$. There is no need to go through any computations at all. $\endgroup$
    – amoeba
    May 23, 2016 at 20:26

1 Answer 1

0
$\begingroup$

Let $\overline{pt}=\frac{pt_1+pt_2+pt_3}{3}$. Then the covariance matrix of $\overline{pt}$ should be:

$C_\overline{pt} = \left[\matrix{Cov(\bar{x},\bar{x}) & Cov(\bar{x},\bar{y}) & Cov(\bar{x},\bar{z}) \\ Cov(\bar{x},\bar{y}) & Cov(\bar{y},\bar{y}) & Cov(\bar{y},\bar{z})\\ Cov(\bar{x},\bar{z}) & Cov(\bar{y},\bar{z}) & Cov(\bar{z},\bar{z})\\}\right]$

Note that $Cov(A,A)=Var(A)$ for any random variable $A$. This will simplify all of the elements on the diagonal.

In addition, consider $Cov(\bar{x},\bar{y})$.

$\begin{eqnarray*} Cov(\bar{x},\bar{y}) &=& Cov\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right) \\ &=& \frac{1}{3}\times\frac{1}{3}\times Cov\left(x_1+x_2+x_3,y_1+y_2+y_3\right) \\ &=& \frac{1}{9}\left[Cov(x_1,y_1)+Cov(x_2,y_1)+Cov(x_3,y_1)+Cov(x_1,y_2)+Cov(x_2,y_2)+Cov(x_3,y_2)+Cov(x_1,y_3)+Cov(x_2,y_3)+Cov(x_3,y_3)\right] \\ &=& \frac{1}{9}\left[s_{x_1y_1}+s_{x_2y_1}+s_{x_3y_1}+s_{x_1y_2}+s_{x_2y_2}+s_{x_3y_2}+s_{x_1y_3}+s_{x_2y_3}+s_{x_3y_3}\right] \\ &=& \frac{1}{9}\sum_{i=1}^3\sum_{j=1}^3s_{x_iy_j} \end{eqnarray*}$

This, of course, generalizes to $Cov(\bar{y},\bar{z})$ and $Cov(\bar{x},\bar{z})$.

Combining the above results, the covariance matrix $C_\overline{pt}$ simplifies to the following:

$C_\overline{pt}=\left[\matrix{s_\bar{x} & \frac{1}{9}\sum_{i=1}^3\sum_{j=1}^3s_{x_iy_j} & \frac{1}{9}\sum_{i=1}^3\sum_{j=1}^3s_{x_iz_j} \\ \frac{1}{9}\sum_{i=1}^3\sum_{j=1}^3s_{x_iy_j} & s_\bar{y} & \frac{1}{9}\sum_{i=1}^3\sum_{j=1}^3s_{y_iz_j}\\ \frac{1}{9}\sum_{i=1}^3\sum_{j=1}^3s_{x_iz_j} & \frac{1}{9}\sum_{i=1}^3\sum_{j=1}^3s_{y_iz_j} & s_\bar{z}\\}\right]$

$\endgroup$
4
  • $\begingroup$ I guess this is the full blown out formula when there is also correlation between the 3 points themselves. In my case this $$ \frac{1}{9}\left[s_{x_1y_1}+s_{x_2y_1}+s_{x_3y_1}+s_{x_1y_2}+s_{x_2y_2}+s_{x_3y_2}+s_{x_1y_3}+s_{x_2y_3}+s_{x_3y_3}\right] $$ will be simplified to $$ \frac{1}{9}\left[s_{x_1y_1}+s_{x_2y_2}+s_{x_3y_3}\right] $$ $\endgroup$
    – xerion
    May 23, 2016 at 19:37
  • $\begingroup$ If you can assume independence between $pt_1$, $pt_2$, and $pt_3$, then yes. $\endgroup$
    – Matt Brems
    May 23, 2016 at 19:38
  • $\begingroup$ And just confirming that indeed the diagonal terms are $$ \frac{\sqrt{(s_{x1}^2+s_{x2}^2+s_{x3}^2)}}{3} $$ $\endgroup$
    – xerion
    May 23, 2016 at 19:42
  • $\begingroup$ Assuming $pt_1$, $pt_2$, and $pt_3$ are independent of one another, $Var(\overline{x})=Var(\frac{x_1+x_2+x_3}{3}=\frac{1}{9}\left[Var(x_1)+Var(x_2)+Var(x_3)\right]=\frac{1}{9}\left[s_{x_1}+s_{x_2}+s_{x_3}\right]$. If they are not independent of one another, then there will be additional covariance terms on the back end. ($2s_{x_1x_2}$,$2s_{x_1x_3}$,$2s_{x_2x_3}$) $\endgroup$
    – Matt Brems
    May 23, 2016 at 19:49

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.