Simulate two centered normal random variables with given variances and given covariance How can I, by the central limit theorem, simulate two random variables $Z_{1}$ and $Z_{2}$ such that $$Z_{1}\sim N(0,\sigma^{2})\qquad Z_{2}\sim N\left(0,\dfrac{(\sigma^{2})^{3}}{3}\right)\qquad\mathrm{cov}(Z_{1},Z_{2})=\dfrac{1}{2}(\sigma^{2})^{2}$$ with $\sigma^{2}=4$?
I take two random variable $U$ and $V$ with distribution $N(0,1)$ and generated the random variable $Z_{1}=\sigma U$ and $Z_{2}=\dfrac{1}{2}\sigma^{3}\left(U+\dfrac{1}{\sqrt{3}}V\right)$, but I'm not sure this really works. Thanks, any hint is really appreciated!
 A: For ordinary simulation, a box-muller/polar marsaglia/ziggurat algorithm with cholesky decomposition allows you to simulate a bivariate normal very efficiently. However as you said you want to apply the central limit theorem, which seems that you want to simulate a lot of random variables and show the convergence.
Assume you already have a random number generator which generate $\text{Uniform}(0,1)$. From the given condition, we know that 
$$Corr[Z_1, Z_2] = \frac {1} {2} (\sigma^2)^2 \sqrt{\frac {3} {\sigma^2(\sigma^2)^3 } }
= \frac {\sqrt{3}} {2}$$
So we may try to simulate some easy bivariate distribution satisfying with this correlation requirement, and having mean zero. Let $X_1, X_2$ be a pair of discrete random variables, with
$$ \Pr\{X_i = 1\} = \Pr\{X_i = -1\} = \frac {1} {2}, i = 1, 2$$
Let $\displaystyle \Pr\{X_1 = 1, X_2 = 1\} = p < \frac {1} {2}$. Then
$$ \Pr\{X_1 = 1, X_2 = -1\} = \Pr\{X_1 = -1, X_2 = 1\} = \frac {1} {2} - p$$
$$ \Pr\{X_1 = -1, X_2 = -1\} = 1 - p - (1 - 2p) = p$$
and thus
$$ Var[X_i] = E[X_i^2] = 1, Corr[X_1, X_2] = Cov[X_1, X_2] = E[X_1X_2] = 2p - (1 - 2p) = 4p - 1$$
So by setting $\displaystyle 4p - 1 = \frac {\sqrt{3}} {2} \Rightarrow p = \frac {\sqrt{3} + 2} {8} $, we can satisfy the correlation requirement. Now we have the algorithm.
The rest of the algorithm is left to you.
Edit: Ok it seems that I need to post the full algorithm out:


*

*Generate $U \sim \text{Uniform}(0,1)$

*Set 
$$ (X_{1,i}, X_{2,i}) = \begin{cases} 
(1, 1) & \text{if} & 0 < U < p \\
(-1, -1) & \text{if} & p < U < 2p \\
(1, -1) & \text{if} & \displaystyle 2p < U < \frac {1} {2} + p \\
(-1, 1) & \text{if} & \displaystyle \frac {1} {2} + p < U < 1 \\
\end{cases} $$
where $\displaystyle  p = \frac {\sqrt{3} + 2} {8} $.

*Repeat step 1 and 2 for $i = 1, 2, \ldots, n$ times and obtain the following pair
$$ \frac {1} {\sqrt{n}} \sum_{i=1}^n X_{1,i}, \frac {1} {\sqrt{n}} \sum_{i=1}^n X_{2,i} $$
By multivariate central limit theorem, this vector converges to the bivariate normal distribution
$$ \mathcal{N}\left(\begin{bmatrix} 0 \\ 0\end{bmatrix},
\begin{bmatrix} 1 & \frac {\sqrt{3}} {2} \\ \frac {\sqrt{3}} {2} & 1 \end{bmatrix}\right)$$

*Finally, we scale this vector by setting 
$$Z_1 = \sigma\frac {1} {\sqrt{n}} \sum_{i=1}^n X_{1,i}, 
Z_2 =  \frac {\sigma} {\sqrt{3}} \frac {1} {\sqrt{n}} \sum_{i=1}^n X_{2,i}$$
and this is one way to use simulation to illustrate the CLT.

