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080-appendix.Rmd
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080-appendix.Rmd
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# Appendix {.unnumbered #appendix}
\noindent Consider a bivariate example where we have a correlated $(Y_1, Y_2)^\top$ with $Y_1\sim N(\mu_1, \sigma_1^2)$ (normal) and $Y_2\sim LN(\mu_2, \sigma_2^2)$ (lognormal). First, let us note that when we get such a vector from the `bigsimr` package then in fact it can be represented as
\begin{equation}
\label{kram1}
(Y_1, Y_2)^\top \stackrel{d}{=} \left(X_1, e^{X_2}\right)^\top,
\end{equation}
where $(X_1, X_2)^\top \sim N_2(\boldsymbol \mu, \boldsymbol \Sigma)$, which is bivariate normal with mean vector $\boldsymbol \mu = (\mu_1, \mu_2)^\top$ and variance-covariance matrix
\begin{equation}
\label{kram2}
\boldsymbol \Sigma =
\left[
\begin{array}{cc}
\sigma_1^2 & \rho \sigma_1\sigma_2\\
\rho \sigma_1\sigma_2 & \sigma_2^2
\end{array}
\right].
\end{equation}
To see this, consider the three steps of the NORTA construction:
\begin{enumerate}
\item Generate $(Z_1, Z_2)^\top \sim N_2(\boldsymbol 0, \boldsymbol R)$, where
\begin{equation}
\label{kram3}
\boldsymbol R =
\left[
\begin{array}{cc}
1 & \rho \\
\rho & 1
\end{array}
\right].
\end{equation}
\item Transform $(Z_1, Z_2)^\top$ to $(U_1, U_2)^\top$ viz. $U_i =\Phi(Z_i)$, $i=1,2$, where $\Phi(\cdot)$ is the standard normal CDF.
\item Return $(Y_1, Y_2)^\top$, where $Y_i=F_i^{-1}(U_i)$, $i=1,2$, and $F_i(\cdot)$ is the CDF of $Y_i$ and $F_i^{-1}(\cdot)$ is its inverse (the quantile function).
\end{enumerate}
In this example, $F_1$ is the CDF of $N(\mu_1, \sigma_1^2)$ while $F_2$ is the CDF of $LN(\mu_2, \sigma_2^2)$, so that the two required quantile functions are
\begin{equation}
\label{kram4}
F_1^{-1}(u) = \mu_1+\sigma_1 \Phi^{-1}(u)\,\,\, \mbox{and} \,\,\, F_2^{-1}(u) = e^{\mu_2+\sigma_2 \Phi^{-1}(u)},
\end{equation}
as can be seen by standard calculation. Thus, when we apply Step 3 of the NORTA algorithm, we get
\begin{equation}
\label{kram5}
F_1^{-1}(U_1) = \mu_1+\sigma_1 \Phi^{-1}(\Phi(Z_1)) = \mu_1+\sigma_1 Z_1 \,\,\, \mbox{and} \,\,\, F_2^{-1}(U_2) = e^{\mu_2+\sigma_2 \Phi^{-1}(\Phi(Z_2))} = e^{\mu_2+\sigma_2 Z_2},
\end{equation}
where $(X_1, X_2)^\top = (\mu_1+\sigma_1 Z_1, \mu_2+\sigma_2 Z_2)^\top \sim N_2(\boldsymbol \mu, \boldsymbol \Sigma)$. Consequently, the vector $(Y_1, Y_2)^\top$ obtained in Step 3 of the algorithm has the structure provided by (\ref{kram1}).
\vspace{0.1in}
\noindent Next, we provide the exact covariance structure of the random vector $(Y_1, Y_2)^\top$ given by (\ref{kram1}), and relate the correlation of $Y_1$ and $Y_2$ to $\rho$, which is the correlation of the normal variables $Z_1$, $Z_2$ (and also $X_1$ and $X_2$). A straightforward albeit somewhat tedious algebra produces the following result.
\begin{lemma}
Let ${\bf Y} = (Y_1, Y_2)^\top$ admit the stochastic representation (\ref{kram1}), where ${\bf X} = (X_1, X_2)^\top \sim N_2(\boldsymbol \mu, \boldsymbol \Sigma)$ with $\boldsymbol \mu$ and $\boldsymbol \Sigma$ as above. Then the mean vector and the variance-covariance matrix of ${\bf Y}$ are given by
\begin{equation}
\label{kram6}
\boldsymbol \mu_{{\bf Y}} = \left(\mu_1, e^{\mu_2+\sigma_2^2/2}\right)^\top
\end{equation}
and
\begin{equation}
\label{kram7}
\boldsymbol \Sigma =
\left[
\begin{array}{cc}
\sigma_1^2 & \rho \sigma_1\sigma_2 e^{\mu_2+\sigma_2^2/2} \\
\rho \sigma_1\sigma_2 e^{\mu_2+\sigma_2^2/2} & \left[ e^{\mu_2+\sigma_2^2/2}\right]^2 \left[e^{\sigma_2^2} -1 \right],
\end{array}
\right].
\end{equation}
respectively.
\end{lemma}
\begin{proof}
The values of the means and the variances are obtained immediately from normal marginal distribution of $Y_1$ and lognormal marginal distribution of $Y_2$. It remains to establish the covariance of $Y_1$ and $Y_2$,
\begin{equation}
\label{covy1y2}
\mbox{Cov}(Y_1, Y_2) = \mathbb E(Y_1 Y_2) - \mathbb E(Y_1)\mathbb E(Y_2),
\end{equation}
and in particular the first expectation on the right-hand-side above. By using the tower property of expectations, the latter expectation can be expressed as
\begin{equation}
\label{kre1}
\mathbb E(Y_1 Y_2) = \mathbb E\left(X_1 e^{X_2}\right) = \mathbb E \left\{ \mathbb E \left(X_1 e^{X_2} | X_1 \right) \right\} = \mathbb E \left\{ X_1 \mathbb E \left(e^{X_2} | X_1 \right) \right\}.
\end{equation}
Further, by using the fact that the conditional distribution of $X_2$ given $X_1=x_1$ is normal with mean $\mathbb E(X_2|X_1=x_1) = \mu_2+\rho\sigma_2(x_1-\mu_1)/\sigma_1$ and variance $\sigma_2^2(1-\rho^2)$, one can relate the inner expectation on the far right in (\ref{kre1}) to the moment generating function $M(t)$ of this conditional distribution evaluated at $t=1$, leading to
\begin{equation}
\label{kre2}
E \left(e^{X_2} | X_1 \right) = e^{ \mu_2+\rho\sigma_2(x_1-\mu_1)/\sigma_1 + \sigma_2^2(1-\rho^2)/2},
\end{equation}
so that
\begin{equation}
\label{kre3}
\mathbb E(Y_1 Y_2) = e^{ \mu_2+ \frac{1}{2} \sigma_2^2(1-\rho^2)} \mathbb E \left\{ X_1 e^{ \rho\frac{\sigma_2}{\sigma_1} (X_1-\mu_1)} \right\}.
\end{equation}
Since $X_1$ is normal with mean $\mu_1$ and variance $\sigma_1^2$, the expectation in (\ref{kre3}) becomes
\begin{equation}
\label{kre4}
\mathbb E \left\{ X_1 e^{ \rho\frac{\sigma_2}{\sigma_1} (X_1-\mu_1)} \right\} = \int_{-\infty}^\infty x_1 e^{ \rho\frac{\sigma_2}{\sigma_1} (x_1-\mu_1)} \frac{1}{\sqrt{2\pi}\sigma_1} e^{-\frac{1}{2\sigma_1^2}(x_1-\mu_1)^2}.
\end{equation}
Upon substituting $u=x_1-\mu_1$, followed by some algebra, we arrive at
\begin{equation}
\label{kre5}
\mathbb E \left\{ X_1 e^{ \rho\frac{\sigma_2}{\sigma_1} (X_1-\mu_1)} \right\} = e^{\frac{1}{2}\sigma_2^2\rho^2} \int_{-\infty}^\infty (u+\mu_1) \frac{1}{\sqrt{2\pi}\sigma_1} e^{-\frac{1}{2\sigma_1^2}(u-\rho\sigma_1\sigma_2)^2}.
\end{equation}
Since the integral in (\ref{kre5}) is the expectation $\mathbb E(X+\mu_1)$, where $X$ is normal with mean $\rho\sigma_1\sigma_2$ and variance $\sigma_1^2$, we conclude that
\begin{equation}
\label{kre6}
\mathbb E \left\{ X_1 e^{ \rho\frac{\sigma_2}{\sigma_1} (X_1-\mu_1)} \right\} = e^{\frac{1}{2}\sigma_2^2\rho^2} (\rho\sigma_1\sigma_2 +\mu_1),
\end{equation}
which, in view of (\ref{kre3}), leads to
\begin{equation}
\label{kre7}
\mathbb E (Y_1Y_2) = e^{\mu_2 + \frac{1}{2}\sigma_2^2} (\rho\sigma_1\sigma_2 +\mu_1).
\end{equation}
Finally, (\ref{covy1y2}), along with the expressions for the means of $Y_1$ and $Y_2$, produce the covariance of $Y_1$ and $Y_2$:
\begin{equation}
\label{kre8}
\mbox{Cov}(Y_1, Y_2) = e^{\mu_2 + \frac{1}{2}\sigma_2^2} (\rho\sigma_1\sigma_2 +\mu_1) - \mu_1 e^{\mu_2 + \frac{1}{2}\sigma_2^2} = \rho\sigma_1\sigma_2 e^{\mu_2 + \frac{1}{2}\sigma_2^2}.
\end{equation}
This concludes the proof
\end{proof}
Using the above result, we can directly relate the correlation coefficient of $Y_1$ and $Y_2$ with that of $X_1$ and $X_2$,
\begin{equation}
\label{kram8}
\rho_{Y_1, Y_2} = \rho \frac{\sigma_2}{\sqrt{e^{\sigma_2^2} -1}},
\end{equation}
where $\rho$ is the correlation of $X_1$ and $X_2$. The above result is useful when studying the possible range of correlation of $Y_1$ and $Y_2$ in the above example. It can be shown that the factor on the far right in (\ref{kram8}) is a monotonically decreasing function of $\sigma_2$ on $(0,\infty)$, with the limits of 1 and 0 at zero and infinity, respectively. Thus, in principle, the range of correlation of $Y_1$ and $Y_2$ is the same as that of $X_1$ and $X_2$, as the factor on the far right in (\ref{kram8}) can be made arbitrarily close to 1. However, by changing this factor we may affect the marginal distributions of $Y_1$ and $Y_2$. It can be shown that if the marginal distributions of $Y_1$ and $Y_2$ are fixed, then the relation (\ref{kram8}) becomes
\begin{equation}
\label{kram9}
\rho_{Y_1, Y_2} = \rho \sqrt{\frac{\log(1+c_2^2)}{c_2^2}},
\end{equation}
where $c_2=\sigma_{Y_2}/\mu_{Y_2}$ is the *coefficient of variation* (CV) of the variable $Y_2$. Thus, the range of possible correlations in this model is not affected by the distribution of $Y_1$, and is determined by the CV of $Y_2$ as follows:
\begin{equation}
\label{kram10}
- \sqrt{\frac{\log(1+c_2^2)}{c_2^2}} \leq \rho_{Y_1, Y_2} \leq \sqrt{\frac{\log(1+c_2^2)}{c_2^2}}.
\end{equation}
Plugging in the estimates of these quantities from Section \@ref(package), we see that
\begin{equation}
\label{kram11}
\frac{\hat{\sigma}_2}{\sqrt{e^{\hat{\sigma}_2^2} -1}} = 0.881.
\end{equation}
Thus, the possible range of correlation becomes $(-0.881, 0.881)$. This approximately agrees with the MC results provided.