# Gaussian Mixture, sampling and interpretation

I came across the following which I have troubles understanding and would appreciate your help.

Suppose we want to fit a Gaussian mixture model (GMM) to a $t$-dstribution with some degrees of freedom. Let the GMM have $K$ components. Using the $EM$-algorithm we found $K$ means $\mu_k$ and variances $\sigma^2_k$ of the Gaussians as well as $K$ mixing coefficients $\pi_k$.

In order to sample from the GMM we use an ancestral sampling scheme with one exception (steps 4 and 5). That is, for one sample, we have to

1. generate $U$ following a standard uniform distribution.
2. compute $k=\sum_{j=1}^K I_{\{g_j>U\}} + 1$, with $g_j=\sum_{i=1}^j\pi_i$, $j=1,...,K$.
3. Given $k$ choose $\mu_k$ and $\sigma^2_k$.
4. Sample $v_t$, $t=1,...,L$, each following to $N(\mu_k, \sigma^2_k)$.
5. Compute $V=\sum_{t=1}^Lv_t$

Note that in step 4 we sample from the same Gaussian $L$ samples which are summed in 5.

Now the claim is that $V/\sqrt{L}$ is GMM-distributed, where the scaling comes from the sum of $L$ Gaussians. In this particular case approximately $t$-distributed. Is this true? I find it odd but cannot grasp why. Thanks for help.

Given that you are taking the sum of $L$ Gaussian $\mathcal{N}(\mu_k,\sigma^2_k)$ variates, the sum $V$ is also Gaussian $\mathcal{N}(L\mu_k,L^2\sigma^2_k)$. Hence your outcome $V/\sqrt{L}$ is still one from a Gaussian mixture $$\sum_{k=1}^K \pi_k \mathcal{N}(\sqrt{L}\mu_k,L\sigma^2_k)$$ I must add that I see little motivation for considering this simulation and average.
Regarding your answer: Isn't the variance of the variance of $V$ equal to $\sum_{k=1}^K \omega_k \left(\sigma_k^2+L(\mu_k-\mu_g)^2\right)$?, with, $\mu_g=\sum_k\omega_k\mu_k$?
Here is how I derived it, with $v_{t,k}$ denoting the random variables form the $K$ components:
$$\sum_k\omega_k E\left(\frac{\sum_t v_{t,k}}{\sqrt{L}}-\sqrt{L}\mu_g\right)^2\\ = \frac{1}{L}\sum_k\omega_kE\left(\sum_t v_{t,k}-L\mu_k+L\mu_k-L\mu_g \right)^2\\ =\frac{1}{L}\sum_k\omega_k\left(E\left(\sum_t (v_{t,k}-\mu_k)\right)^2+2E\left(\sum_tv_{t,k}-L\mu_k\right)\left(L\mu_k-L\mu\right)+L^2\left(\mu_k-\mu_g\right)^2\right)\\ =\frac{1}{L}\sum_k\omega_k\left(L\sigma_k^2+L^2(\mu_k-\mu_g)^2\right)\\ =\sum_k\omega_k\left(\sigma_k^2+L(\mu_k-\mu_g)^2\right)$$