Variance of independent, identical distributions with random sample size I found this equation in wikipedia (https://en.wikipedia.org/wiki/Variance#Sum_of_correlated_variables), and I was wondering if someone had a proof for it? I have checked the textbook in the references but cannot find this formulation
$$
\large Var(\Sigma \mathcal{X}) = \mathbb{E}(\mathcal{N}) Var(\mathcal{X}) + Var(\mathcal{N}) \mathbb{E}^2(\mathcal{X})
$$

As the wikipedia page you cite in the OP is pointing out, this follows from Eve's law.
Eve's law states:
$$
Var(Y) = \mathbb{E}_X(Var_Y(Y|X)) + Var_X(\mathbb{E}_Y(Y|X)),
$$
where $X$ and $Y$ are arbitrary random variables. Now, in the formula, you are interested in, we can set $Y=\sum X$ and $X=N$. Then:
$$
\begin{align}
Var(\sum X) &= \mathbb{E}_N(Var_{\sum X}(\sum X|N)) + Var_N(\mathbb{E}_{\sum X}(\sum X|N))\\
            &= \mathbb{E}_N(N\,Var(X)) + Var_N(\sum \mathbb{E} (X))\\
            &= \mathbb{E}(N) Var(X) + Var_N(N\mathbb{E}(X))\\
            &= \mathbb{E}(N) Var(X) + Var(N) (\mathbb{E}(X))^2.
\end{align}
$$
Let's go through this step by step: The first equation is just Eve's law. The second equation uses for the first term the fact that, for independent random variables $X_i$, the variance of the sum is the sum of the variances. Furthermore, the $X_i$ are iid, i.e. in particular of the same variance, so the sum of variances just becomes $N\,Var(X)$. The second term in the second equation uses the fact that expectation is always linear, i.e., in particular, it commutes with sums. The third equation uses for the first term again the fact that the expectation operator is linear and furthermore, that $Var(X)$ is a constant w.r.t. $N$. For the second term I use that the expectations of the $X_i$ are all the same, because, again, they are iid. The final equation only uses that $\mathbb{E}(X)$ is independent of $N$, i.e. a constant, and can thus be extracted from the $Var$-operator by attaching the power of two to it.
And this is the equation you were looking for.
