# For $X_i$ iid $Var( \sum_{i=1}^n X_i )= \sum_{i=1}^n Var (X_i)$?

My contention is that it's true. I thought of two ways of proving it, unsure which one is better (and/or correct):

1. Suppose $X_i \sim N(\mu, \sigma^2)$, then it's moment generating function (MGF) is $$\Psi_{X_i}(t) = \exp\{\mu t + \frac12 \sigma^2 t^2\}$$ If we've got the iid property, then MGF of the sum is equal to the product of the MGF's which is: $$\Psi_{\sum_{i=1}^n X_i}(t) = \prod_{i=1}^n \exp\{\mu t + \frac12 \sigma^2 t^2\} = \exp\{n \mu t + n \frac12 \sigma^2 t^2\}$$ Which gives us that $\sum_{i=1}^n X_i \sim N(n \mu, n \sigma^2)$. It should probably hold in general (can't think of an argument, why?)

2. From the definition of variance

\begin{align} Var( \sum_{i=1}^n X_i ) &= \mathbb{E} \Bigg (\sum_{i=1}^n X_i- \mathbb{E} \sum_{i=1}^n X_i \Bigg)^2\\ &= \mathbb{E} \Bigg( (\sum_{i=1}^n X_i)^2- 2 \sum_{i=1}^n X_i \sum_{i=1}^n\mathbb{E} X_i + (\sum_{i=1}^n\mathbb{E} X_i)^2 \Bigg)\\ &=\sum_{i=1}^n\mathbb{E} X_i^2- (\sum_{i=1}^n\mathbb{E} X_i)^2\\ &=n \mathbb{E}X_1^2 - n^2 (\mathbb{E}X_1)^2 \end{align} since $\mathbb{E}[\sum_{i \neq j} X_i X_j ]=0$ and we have the iid property, i.e.: $\mathbb{E}X_1 = \mathbb{E}X_i \space \forall i$.

Whereas \begin{align} \sum_{i=1}^n Var(X_i) &=\sum_{i=1}^n \mathbb{E} \Bigg( X_i - \mathbb{E}X_i \Bigg)^2 \\ &= \sum_{i=1}^n \mathbb{E} \Bigg( X_i^2- 2X_i \mathbb{E}X_i + (\mathbb{E}X_i)^2 \Bigg)\\ &= n \mathbb{E}X_1^2 - n (\mathbb{E}X_1)^2 \end{align} I can't seem to get the $n^2$ here the same way I got (above). This would suggest that my contention is wrong. Is it or did I make a mistake in the calculations?

\begin{align} \operatorname{var}\left( \sum_{i=1}^n X_i \right) &= \mathbb{E} \Bigg (\sum_{i=1}^n X_i- \mathbb{E} \sum_{i=1}^n X_i \Bigg)^2\\ &= \mathbb{E} \Bigg (\sum_{i=1}^n (X_i- \mathbb{E}X_i) \Bigg)^2\\ &= \sum_{i=1}^n \mathbb{E} (X_i- \mathbb EX_i)^2 + \mathbb E\sum_{i=1}^n ~\sum_{j=1, j \neq i}^n (X_i -\mathbb{E} X_i) (X_j -\mathbb{E} X_j)\\ &=\sum_{i=1}^n\operatorname{var}(X_i)+ \sum_{i=1}^n~ \sum_{j=1, j \neq i}^n \mathbb E(X_i -\mathbb{E} X_i)\mathbb E(X_j -\mathbb{E} X_j)\\ &= \sum_{i=1}^n\operatorname{var}(X_i) \end{align}

1. Let us begin with the $n=2$ case to show some explicit calculation. Let me change the notation a bit and consider the iid random variables $X$,$Y$ s.t. $\mathbb E[X]=\mu=\mathbb E[Y]$. Then

$Var(X+Y)=\mathbb E[(X+Y-2\mu)^2]=\mathbb E[(X-\mu)^2]+\mathbb E[(Y-\mu)^2]+\mathbb E[2XY+2{\mu}^2-2\mu(X+Y)]=Var(X)+Var(Y)+2{\mu}^2+2{\mu}^2-2\mu(\mu+\mu)= Var(X)+Var(Y)$,

as $\mathbb E[XY]=\mathbb E[X]\mathbb E[Y]={\mu}^2$ ($X$ and $Y$ are independent). For $n=2$ your statement holds without giving any explicit form for the distribution of $X$.

1. Let us move to the general case ($n>1$). Point 2. in your question is bugged. More precisely, when you compute $Var(\sum_{i=1}^nX_i)$ you arrive at

$\mathbb E[(\sum_{i=1}^nX_i)^2]=\sum_{i=1}^n \mathbb E[ {X}^2_i]$,

which is wrong (check it for $n=2$ to convince yourself). In fact

$\mathbb E[(\sum_{i=1}^nX_i)^2]=\sum_{i=1}^n \mathbb E[ {X}^2_i]+2\sum_{i\neq j}E[X_i]E[X_j]$.

This last equality can be proved by induction and using the fact that $E[X_iX_j]=E[X_i]E[X_j]$ for all $i\neq j$. But then

$2\sum_{i\neq j}E[X_i]E[X_j]=2\sum_{i\neq j}\mu^2=2\mu^2n(n-1)$,

as each of the $n$ indices (let us say $i$) is multiplied with all the remaining $n-1$ indices (i.e. all those indices in $1,\dots,n$ distinct from $i$).

In summary, we have

$\sum_{i=1}^n Var(X_i)=\sum_{i=1}^n\mathbb E[X^2_i]-n\mu^2$ (as you correctly computed)

and

$Var(\sum_{i=1}^nX_i)=$(you did this correctly) $=\mathbb E[(\sum_{i=1}^nX_i)^2]-2n^2\mu^2+n\mu^2=\sum_{i=1}^n \mathbb E[ {X}^2_i]+2\mu^2n(n-1)-2n^2\mu^2+n\mu^2=\sum_{i=1}^n \mathbb E[ {X}^2_i]-n\mu^2,$

as stated.

I hope this can help.