Generating two correlated random numbers: Why does volatility be 1 by using Cholesky decomposition? I am trying to use Cholesky decomposition to generate two correlated random numbers by simulating two uncorrelation distributions.
The covariance matrix should be $$ C=
    \left[\begin{matrix}
    \sigma_1^2 & \rho\cdot\sigma_1\cdot\sigma_2 \\
    \rho\cdot\sigma_1\cdot\sigma_2 & \sigma_2^2 \\
    \end{matrix}\right]
$$
where $\rho$ is the correlation between the two correlated distribution.
Let $$ C= LL^T $$, solving the equation, I got:
$$ L=
    \left[\begin{matrix}
    \sigma_1 & 0 \\
    \sigma_2\cdot\rho & \sigma_2\cdot\sqrt{1-\rho^2} \\
    \end{matrix}\right]
$$
But in practice, we use $$ L=
    \left[\begin{matrix}
    1 & 0 \\
    \rho & \sqrt{1-\rho^2} \\
    \end{matrix}\right]
$$ to generate the two correlated distribution, and it makes sense.
Why this happen? In this case, do we just assume $\sigma_1$ and $\sigma_2$ to be 1? but in fact, they are not equal to 1 in my application.
 A: You may wish to consider this approach, if I understand your question, to help work out issues.
Define the following random variables
$$
x_{1} = \eta_{1} \\
x_{2} = \alpha x_{1} + \eta_{2}  \\
$$
with normally distributed uncorrelated noise terms
$$
\eta_{1} \sim \mathscr{N}\left(0,\sigma_{1}\right) \\
\eta_{2} \sim \mathscr{N}\left(0,\sigma_{2}\right) \\
$$
and $\alpha$ a constant scaling factor which makes $x_{1}$ and $x_{2}$ linearly correlated.  Note that $\alpha$ is not the same as the correlation coefficient, $\rho$.
Compute the entries of the covariance matrix:
$$
E\left[x_{1}x_{1}\right]=E \left[\eta_{1}^{2} \right]=\sigma_{1}^{2}
$$
$$
E\left[x_{1}x_{2}\right]=E \left[\eta_{1}\left(\alpha \eta_{1} + \eta_{2}\right)\right]
$$
$$
E\left[x_{1}x_{2}\right]=E \left[\alpha \eta_{1}^{2} + \eta_{1}\eta_{2}\right]=\alpha \sigma_{1}^{2}
$$
$$
E\left[x_{2}x_{2}\right]=E \left[\left(\alpha^{2} \eta_{1}^{2} + 2\alpha\eta_{1}\eta_{2}+\eta_{2}^{2}\right)\right]=\alpha^{2}\sigma_{1}^{2}+\sigma_{2}^{2}
$$
Now the correlation coefficient
$$
\rho = 
\frac{E\left[x_{1}x_{2}\right]}{\sqrt{E\left[x_{1}x_{1}\right]E\left[x_{2}x_{2}\right]}}=\frac{\alpha \sigma_{1}^{2}}{\sqrt{\sigma_{1}^{2}\left(\alpha^{2}\sigma_{1}^{2}+\sigma_{2}^{2}\right)}}=\frac{\alpha \sigma_{1}}{\sqrt{\left(\alpha^{2}\sigma_{1}^{2}+\sigma_{2}^{2}\right)}}
$$
The covariance matrix is then
$$
\Sigma=
\begin{bmatrix}
E\left[x_{1}x_{1}\right] & E\left[x_{1}x_{2}\right] \\
E\left[x_{1}x_{2}\right] & E\left[x_{2}x_{2}\right] \\
\end{bmatrix}
$$
Making the substitution for $E\left[x_{1}x_{2}\right]$ one obtains:
$$
\Sigma=
\begin{bmatrix}
E\left[x_{1}x_{1}\right] & \rho \sqrt{E\left[x_{1}x_{1}\right]E\left[x_{2}x_{2}\right]} \\
\rho \sqrt{E\left[x_{1}x_{1}\right]E\left[x_{2}x_{2}\right]} & E\left[x_{2}x_{2}\right] \\
\end{bmatrix}
$$
With the provided decomposition for $L$:
$$
L=
\begin{bmatrix}
\sqrt{E\left[x_{1}x_{1}\right]} & 0 \\
\rho \sqrt{E\left[x_{2}x_{2}\right]} & \sqrt{E\left[x_{2}x_{2}\right]}\sqrt{1-\rho^{2}} \\
\end{bmatrix}
$$
Comparing to the original model for the random variables, we have:
$$
\begin{bmatrix}
x_{1} \\
x_{2} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\ 
\alpha & 1 \\
\end{bmatrix}
\begin{bmatrix}
\eta_{1} \\
\eta_{2} \\
\end{bmatrix}
$$
Using $
E\left[x_{1}x_{2}\right]=E \left[\alpha \eta_{1}^{2} + \eta_{1}\eta_{2}\right]=\alpha \sigma_{1}^{2}
$ and related from above, we have that
$$
\alpha E\left[x_{1}x_{1}\right] = \rho \sqrt{E\left[x_{1}x_{1}\right]E\left[x_{2}x_{2}\right]} 
$$
which leads to:
$$
\alpha  = \rho \sqrt{\frac{E\left[x_{2}x_{2}\right]}{E\left[x_{1}x_{1}\right]}}
$$
Rewrite the model in terms of $\eta$ now with unit variances, $\hat{\eta}$, scaled by standard deviations $\sigma_{1}$ and $\sigma_{2}$ via the following matrix expression:
$$
\begin{bmatrix}
x_{1} \\
x_{2} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\ 
\alpha & 1 \\
\end{bmatrix}
\begin{bmatrix}
\sigma_{1} & 0 \\
0 & \sigma_{2} \\
\end{bmatrix}
\begin{bmatrix}
\hat{\eta_{1}} \\
\hat{\eta_{2}} \\
\end{bmatrix}
$$
using the results from the earlier derivations:
$$
\begin{bmatrix}
x_{1} \\
x_{2} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\ 
\rho \sqrt{\frac{E\left[x_{2}x_{2}\right]}{E\left[x_{1}x_{1}\right]}} & 1 \\
\end{bmatrix}
\begin{bmatrix}
\sqrt{E\left[x_{1}x_{1}\right]} & 0 \\
0 & \sqrt{E\left[x_{2}x_{2}\right]-\rho^{2}E\left[x_{2}x_{2}\right]} \\
\end{bmatrix}
\begin{bmatrix}
\hat{\eta_{1}} \\
\hat{\eta_{2}} \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
x_{1} \\
x_{2} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\ 
\rho \sqrt{\frac{E\left[x_{2}x_{2}\right]}{E\left[x_{1}x_{1}\right]}} & 1 \\
\end{bmatrix}
\begin{bmatrix}
\sqrt{E\left[x_{1}x_{1}\right]} & 0 \\
0 & \sqrt{E\left[x_{2}x_{2}\right]}\sqrt{1-\rho^{2}} \\
\end{bmatrix}
\begin{bmatrix}
\hat{\eta_{1}} \\
\hat{\eta_{2}} \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
x_{1} \\
x_{2} \\
\end{bmatrix}
=
\begin{bmatrix}
\sqrt{E\left[x_{1}x_{1}\right]} & 0 \\ 
\rho \sqrt{E\left[x_{2}x_{2}\right]} & \sqrt{E\left[x_{2}x_{2}\right]}\sqrt{1-\rho^{2}} \\
\end{bmatrix}
\begin{bmatrix}
\hat{\eta_{1}} \\
\hat{\eta_{2}} \\
\end{bmatrix}
$$
which is
$$
\begin{bmatrix}
x_{1} \\
x_{2} \\
\end{bmatrix}
=
L
\begin{bmatrix}
\hat{\eta_{1}} \\
\hat{\eta_{2}} \\
\end{bmatrix}
$$
Please note: $\sqrt{E\left[x_{2}x_{2}\right]}\ne \sigma_{2}$.  In other words, the variance of $\eta_{2}$ (the seoond noise random variable) is not the same as the $\left(2,2\right)$ entry of the covariance matrix $\Sigma$.
I hope this helps.
