Tl;dr: I'm trying to apply the Central Limit Theory (CLT) and Lyapunov CLT to the same data and get different results. Do you know what I am doing wrong?
Lets assume there are 3 samples A, B and C taken out of a (standard) normal distribution, so $\sigma_a = \sigma_b = \sigma_c = 1$. I would like to know the standard deviation (std-dev) of A+B+C. The Central Limit Theory (CLT) tells that the std-dev of $\text{std-dev}(A+B+C) = (\sigma_a + \sigma_b + \sigma_c)*\sqrt{3} = 5.196$, which is indeed the case.
However I could also first calculate the std-dev of A+B and then that of (A+B)+C. This results in $\text{std-dev}((A+B)+C) = ((\sigma_a + \sigma_b)\cdot\sqrt{2} + \sigma_c )\cdot\sqrt{2} = 5.414$. This answer is different from the one before. Note that this is not the standard CLT, but the Lyapunov CLT is used. What I did might not strictly be correct since I am not sure of the Lyapunov conditions are sattisfied in this case (due to a lack of statistics skills). However my professor told me that it is hard to prove these conditions but they are mostly satisfied. Since this is a very basic example I assume that they are therefore statisfied (I know, a bit shabby but I have no other choice). This makes that I am doing something wrong but I don't know what.
Do you know what I am doing wrong or is the cause of this discrepancy that the Lyapunov conditions are not being satisfied?
I found my problem. First of all the std-dev of A+B+C is not $(\sigma_a + \sigma_b + \sigma_c)\cdot \sqrt{3}$. CLT says that $\sigma_t^2 = \sum_n \sigma_n^2$. In my case that results to $\sigma_t = \sqrt(3)\cdot 1$.
In case of the Lyapunov CLT: The combination of A and B has a std-dev of $\sqrt{2}$. so that variance of (A+B)+C is equal to: $\sigma_t^2 = \sqrt{2}^2 + \sqrt{1}^2 = 3$ and thus $\sigma_t = \sqrt{3}$.
What I did wrong was that I interchanged the summing and square root (or something like that). Instead of $\sigma_t = \sqrt{\sum_n \sigma_n^2}$, I calculated $\sigma_t = \sum_n(\sqrt{\sigma_n})$. Anyhow, problem solved!