Here is the problem:

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I am interested in problem b.

So, my idea is the following: We know, that if $X = AZ + a$, where $ A$ is $m*m$ and $a$ is constant, than $X \sim N(A\mu + a,A\Sigma A^T)$

I want to use this fact and we have two equalities: $$ A\mu + a = 0$$ $$A\Sigma A^T = I$$

And my question is: am I on the right way to solution(i need to find A and a)? If no, give me a hint please. If yes, how can i derive A from the second equality?



When $m = 1$ the proper transformation is

$$ \frac{Z - \mu}{\sigma} = \sigma^{-1} Z - \sigma^{-1} \mu. $$

Since $\Sigma = \sigma^2$ in this case, what does that tell you about $A$ and $a$? Can you now generalize this for $m \geq 1$?

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  • $\begingroup$ Write $A$ in terms of $\Sigma$ using the fact that $\Sigma$ is just a scalar equal to $\sigma^2$, this should make this part of the answer clear. Then use the fact that $\text{E}(AZ) = A \mu$. $\endgroup$ – dsaxton Nov 4 '16 at 15:22
  • $\begingroup$ So, my answer is wrong? $\endgroup$ – Daniel Yefimov Nov 4 '16 at 15:37
  • $\begingroup$ I didn't really follow your last comment, but it's not true that $A = 1$. Just try to apply what I said above. $\endgroup$ – dsaxton Nov 4 '16 at 16:10
  • $\begingroup$ A! I got it. $A = 1/{\sigma}$ and $a = -{\mu}/{\sigma}$ It was SO obvious, but sometimes such things is't visible. Thought, it would harder problem. $\endgroup$ – Daniel Yefimov Nov 4 '16 at 17:06
  • $\begingroup$ Right, now how would you state this in terms of $\Sigma$ and $\mu$, and can you verify the matrix calculations? $\endgroup$ – dsaxton Nov 4 '16 at 17:11

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