Suppose $\beta_1 \sim N(m_1,s_1^2)$, $\beta_2 \sim N(m_2,s_2^2)$ and $cov(\beta_1,\beta_2) = s_{12}$. Now generate draws of these random coefficients from draws of two independent standard normal distributed variables $z_1$ and $z_2$. Give the $2\times 2$ matrix $L$ that satisfies: $$ \begin{bmatrix}\beta_1\\\beta_2\end{bmatrix} \sim \begin{bmatrix}m_1\\m_2\end{bmatrix} + L \begin{bmatrix}z_1\\z_2\end{bmatrix}$$

Hint: make a guess and verify that L is Cholesky decomposition of $$\Sigma = \begin{bmatrix} s_1^2 & s_{12} \\ s_{12} & s_2^2\end{bmatrix}$$ by computing $LL'$.

What I tried is to substitute $$L = \begin{bmatrix} a & b \\ c & d\end{bmatrix}$$ carry out the matrix multiplication and addition to get: $$\begin{bmatrix}\beta_1\\\beta_2\end{bmatrix} \sim \begin{bmatrix} m_1 + az_1 + bz_2\\ m_2 + cz_1 + dz_2 \end{bmatrix}$$

My understanding is that this means that $\beta_1$ is distributed as $m_1 + az_1 + bz_2$ and $\beta_2$ as $m_2 + cz_1 + dz_2$. Is this correct?

My next step is to find the expected values, variances and covariance of the RHS and set them equal to the given moments for $\beta_1$ and $\beta_2$. This gives me the following system of non-linear equations: \begin{align} a^2 + b^2 & = s_1^2 \\ c^2 + d^2 &= s_2^2\\ ac+bd &= s_{12} \end{align}

But how to continue from here? Set for example $a=1$ or $a=0$ and try to solve the system? Or am I thinking in the wrong direction? Thanks.

  • $\begingroup$ $L$ is lower triangular, allowing you to solve for $a$; that lets you solve for $c$, and then for $d$ $\endgroup$
    – Glen_b
    Feb 6, 2020 at 1:58

1 Answer 1


I think you used the Cholesky decomposition where $L$ is a lower triangular matrix, meaning $b=0$. I hope this helps.


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