Unconditional Simulation

It is a simple matter to produce an $N(0,1)$ random number, and by stacking k such numbers in a column vector you obtain a vector with independent standard normal components $\bW \sim N_ k(\mb {0},\bI )$. The meaning of the terms independence and randomness in the context of a deterministic algorithm required for the generation of these numbers is somewhat subtle; see Knuth (1973, Vol. 2, Chapter 3) for a discussion of these issues.

Rather than $\bW \sim N_ k(\mb {0},\bI )$, what is required is the generation of a vector $\bZ \sim N_ k(\mb {0},\bV )$—that is,

\[  \bZ = \left[ \begin{array}{c} Z_1\\ Z_2\\ \vdots \\ Z_ k \end{array} \right]  \]

with covariance matrix

\[  \bV = \left( \begin{array}{cccc} \sigma _{11} &  \sigma _{12} &  \cdots &  \sigma _{1k}\\ \sigma _{21} &  \sigma _{22} &  \cdots &  \sigma _{2k}\\ &  \ddots & & \\ \sigma _{k1} &  \sigma _{k2} &  \cdots &  \sigma _{kk} \end{array} \right)  \]


\[  \sigma _{ij} = \mbox{Cov}(Z_ i,Z_ j)  \]

If the covariance matrix is symmetric and positive definite, it has a Cholesky root $\bL $ such that $\bV $ can be factored as

\[  \bV =\bL \bL ’  \]

where $\bL $ is lower triangular. See Ralston and Rabinowitz (1978, Chapter 9, Section 3-3) for details. This vector $\bZ $ can be generated by the transformation $\bZ =\bL \bW $. Note that this is where the assumption of multivariate normality is crucial. If $\bW \sim N_ k(\mb {0},\bI _ k)$, then $\bZ =\bL \bW $ is also normal or Gaussian. The mean of Z is

\[  \mr {E}(\bZ ) = \bL (\mr {E}(\bW )) = 0  \]

and the variance is

\[  \mr {Var}(\bZ ) = \mr {Var}(\bL \bW ) = \mr {E}(\bL \bW \bW ’\bL ’) = \bL ~ \mr {E}(\bW \bW ’)~ \bL ’ = \bL \bL ’ = \bV  \]

Finally, let $Y_ k = Z_ k + \mu _ k$; that is, you add a mean term to each variable $Z_ k$. The covariance structure of the $Y_ k’s$ remains the same. Unconditional simulation is done by simply repeatedly generating k $N(0,1)$ random numbers, stacking them, and performing the transformation

\[  \bW \longmapsto \bZ = \bL \bW \longmapsto \bY = \bZ + \bmu  \]