Summation of i.i.d. Normal Random Variables Assuming I have $$X_1,X_2,...,X_{100}\sim N(1,4)$$ and $$Y_1,Y_2,...,Y_{20}\sim N(2,9)$$ where all $X$ are iid, all $Y$ are iid.
Then should $$\text{var}(X_1+X_2+\ldots+X_{100}+Y_1+\ldots + Y_{20}) = 100 \times 4 + 20 \times 9$$ or $$\text{var}(X_1+X_2+\ldots+X_{100}+Y_1+\ldots + Y_{20})=100^2*
 \times 4 + 20^2 \times 9$$ ?
 A: It holds that:
$$
Var \left [ \sum_{i=1}^nX_i \right ] = \sum_{i=1}^n \sum_{j=1}^n Cov \left [ X_i, X_j \right ]
$$
If $X_i$ are independent (identical distribution not needed) (assuming variances exist and are finite) then $Cov[X_i,X_j] = 0$ for all $i \ne j$. Thus the formula above simplifies to:
$$
Var \left [ \sum_{i=1}^nX_i \right ] = \sum_{i=1}^n \sum_{j=1}^n Cov \left [ X_i, X_i \right ] = \sum_{i=1}^n Var \left [ X_i \right ]
$$
Thus, for independent random variables, the variance of the sum is the sum of the variances, i.e. in your case: $4 + 4 + ... + 9 + 9 + ... = 4 * 100 + 9*20$.

Tip:
You are confusing the number of elements in the sum with the weights of the elements. If you have a weighted sum, then the formula for the variance of the sum changes by needing to multiply each individual variance with the squared weight. For example, in 2-variable case (still assuming $X1 \perp \! \! \! \perp X_2$, i.e. independence):
Not weighted sum (your case):
$Var[X_1 + X_2] = Var[X_1] + Var[X_2]$
Weighted sum:
$Var[w_1 X_1 + w_2 X_2] = w_1^2 Var[X_1] + w_2^2 Var[X_2]$
Note how in your example you have no weights (or weights of 1), so no need to square anything.
