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Say I have two risk factors $X_1$ and $X_2$. Standard deviation for $X_1$ is $\sigma_1$ and $\sigma_2$ for $X_2$. Furthermore, $X_1$ has a mean of $\mu_1$ and $X_2$ has a mean of $\mu_2$. Correlation between $X_1$ and $X_2$ is $\rho$.

The system is as follows:

$$\begin{eqnarray} X_1 & = & \mu_1 + \lambda_{11} U_1 \\ X_2 & = & \mu_2 + \lambda_{21} U_1 + \lambda_{22} U_2 \end{eqnarray}$$

My book reads as follows:

$\lambda_{11}=\sigma_1$ (1)
$\lambda_{21}^2 + \lambda_{22}^2= \sigma_2^2$ (2)
$\lambda_{21} \lambda_{11}= \rho \sigma_1 \sigma_2$ (3)"

I don't understand how they work out lines (2) and (3).

Can anyone please help?

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up vote 3 down vote accepted

If $\Sigma$ is the variance/covariance matrix of random variables $U_1, U_2, \ldots U_n$, and $V = c + w_1 U_1 + \ldots + w_n U_n$, where $c$ is a constant, and we let $\mathbf{w}$ be the vector with the 'weights' $w_1, w_2, \ldots, w_n$, then the variance of $V$ is equal to $\mathbf{w}^{\top}\Sigma\mathbf{w}$. Moreover, if $T$ is another random variable described by weights vector $\mathbf{x}$, then the variance of $T$ is $\mathbf{x}^{\top}\Sigma\mathbf{x}$, and the covariance of $V$ and $T$ is equal to $\mathbf{x}^{\top}\Sigma\mathbf{w}$.

In your problem, $$ \Sigma = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$ and you are looking at weights vectors $[\lambda_{11}\,0]$ and $[\lambda_{21}\,\lambda_{22}]$, thus the variance of $X_1$ is $\lambda_{11}^2$, which is the first equation, and the variance of $X_2$ is $\lambda_{21}^2 + \lambda_{22}^2$, which is the second equation. Computing the covariance of $X_1$ and $X_2$ gives the third equation.

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You should offer more details [I assume U1 and U2 are N(0,1)] but I think you should read this: http://en.wikipedia.org/wiki/Cholesky_decomposition

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Thanks. I did forget to mention U1 and U2 were N(0,1). The link you provided is interesting. However I would appreciate a more specific reply especially how lines 2 and 3 are worked out from a mathematical point of view. Regards, Julien. – Julien May 31 '11 at 17:51

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