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I am given $3$ normal random variables $X_1$,$X_2$,$X_3$ which are linear combinations of $Z_1$,$Z_2$,$Z_3$. $Z_1$,$Z_2$,$Z_3$ are mutually independent standard normal variables.

I am given a correlation matrix

$$X=\begin{bmatrix} 1 & 0.3 & 0.4 \\0.3 & 1 & 0.5 \\0.4 & 0.5 & 1 \end{bmatrix}$$

I have to find $M$ which is $3 \times 3$ such that

$$\begin{bmatrix} X_1 \\X_2 \\X_3 \\ \end{bmatrix} =M \cdot\begin{bmatrix} Z_1 \\Z_2 \\Z_3 \\ \end{bmatrix}$$

Any hints on how to approach this, based on my research , I am getting a stong correlation of google hits on multivariate normal distribution and cholesky method, but I am still not able to relate how cholesky method will help if it is M would be relevant.

octave-3.2.4.exe:10> Q3

Q3 =

1.00000 0.30000 0.40000

0.30000 1.00000 0.50000

0.40000 0.50000 1.00000

octave-3.2.4.exe:11> chol(Q3)

ans =

1.00000 0.30000 0.40000

0.00000 0.95394 0.39835

0.00000 0.00000 0.82542

Thanks for your help!!

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Unrelated to the Math in the question: Please go through the Edits I have made to your post. It might be of help the next time you post here. – user21436 Mar 18 '12 at 4:16

The correlation matrix of $Z_1,Z_2,Z_3$ is the identity matrix. You get the correlation matrix of $X_1,X_2,X_3$ from this by multiplying by $M$ from the left and $M^\top$ from the right. Thus the correlation matrix of $X_1,X_2,X_3$ is $MM^\top$. Cholesky decomposition gives you a matrix $M$ such that $MM^\top$ is a given matrix. Note that this is only one solution, the one in which $M$ is lower triangular; you can right-multiply this $M$ by any orthogonal matrix to get another solution.

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Thanks Joriki!!!! – newbietolinalg Mar 18 '12 at 16:56

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