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I am having some troubles with a question regarding a joint distribution with several normal distributions.

The question is put up like this:

$X_1 , X_2$ are two normal distributions with the distribution $N(0,I_1) $. $Z_1 = \frac{X_1 +X_2}{2}, Z_2 = \frac{X_1 -X_2}{2}, Z_1 = \frac{X_2 -X_1}{2}$.

Find the joint distribution of $(Z_1, Z_2, Z_3)$.

What I did so far was to try find the distributions of $Z_1, Z_2, Z_3$, where I used the rules for sums of normal distributions. This gives me the result:

$Z_1 \sim N(\frac{1}{2}\cdot 0 + \frac{1}{2}\cdot 0, \frac{1}{2}^2\cdot I_1 +\frac{1}{2}^2\cdot I_1 ) = N(0, \frac{1}{2}I_1)$ $Z_2 \sim N(\frac{1}{2}\cdot 0 + \frac{-1}{2}\cdot 0, \frac{1}{2}^2\cdot I_1 +\frac{-1}{2}^2\cdot I_1 ) = N(0, \frac{1}{2}I_1)$ $Z_2 \sim N(\frac{1}{2}\cdot 0 + \frac{-1}{2}\cdot 0, \frac{1}{2}^2\cdot I_1 +\frac{-1}{2}^2\cdot I_1 ) = N(0, \frac{1}{2}I_1)$

I then try using the rules for multivariate normal distributions to find the joint distribution of $Z_1, Z_2, Z_3 $.

This gives: $ Z \sim N(\mu , \sum) $

where

$\mu = [E[Z_1],E[Z_2], E[Z_3]]$

$\sum = [Cov[Z_i,Z_j]] \; i = 1,2,3 \; j = 1,2,3$

This however just gives me a vector with zeroes and a matrix with zeroes with my calculations. What am I doing wrong in my assumptions?

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  • $\begingroup$ "a matrix with zeroes with my calculations" Some off-diagonal entries are $0$, yes, but every diagonal entry is $\frac12$ and a pair of off-diagonal entries is not $0$ either. $\endgroup$ – Did Sep 27 '16 at 6:13
  • $\begingroup$ But isn't for example $ Cov[Z_1,Z_1] = E[(Z_1 - \mu_1)(Z_1 - \mu_1)] = 0$? Since the expected value for $Z_1=0$, right? $\endgroup$ – Pannkakan Sep 27 '16 at 7:36
  • $\begingroup$ No, not right, $E((Z_1-\mu_1)^2)$ is the variance of $Z_1$, not $0$. $\endgroup$ – Did Sep 27 '16 at 11:22

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