TITLE: Scaling the normal distribution?
QUESTION [27 upvotes]: I might just be slow (or too drunk), but I'm seeing a conflict in the equations for adding  two normals and scaling a normal. According to page 2 of this, if $X_1 \sim N(\mu_1,\sigma_1^2)$ and $X_2 \sim N(\mu_2,\sigma_2^2)$, then $X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$, and for some $c \in \mathbb{R}$, $cX_1 = N(c\mu_1, c^2\sigma_1^2)$.
Then for $X \sim N(\mu,\sigma^2)$, we have $X + X = N(\mu + \mu,\sigma^2 + \sigma^2) = N(2\mu,2\sigma^2)$, but also $X + X = 2X = N(2\mu,2^2\sigma^2) = N(2\mu,4\sigma^2)$ ? Ie, the variances disagree.
edit: Oh, am I mistaken in saying that $2X = X + X$? Is the former "rolling" $X$ just once and doubling it while the latter "rolls" twice and adds them?

REPLY [2 votes]: Suppose $X_1, X_2,..., X_k$ are any k random variables (not necessarily independent).
Then variance of $X_1+X_2+...+X_n$ is calculated by:
$Var(X_1+X_2+...+X_n)=\sum_{i=1}^kVar(X_i)+2*\sum_{1\le i\le k}Cov(X_i,X_j)$
When $X_1, X_2,...X_k$ are independent, then $Cov(X_j, X_j)=0$
$\therefore$ $Var(X_1+X_2+...+X_n)=\sum_{i=1}^kVar(X_i)$
When you have $X, X,...,X$, i.e $kX$ variables, $X$ and $X$ are no more independent then $Cov(X_i,X_j)$ will not be zero.
For example, you want to calculate $Var(X+X+X)$ it is not $3*Var(X)$ it will be $9*Var(X)$
Since $Var(X+X+X)=\sum_{i=1}^3Var(X)+2*\sum_{1 \le i \le j} Cov(X,X) =3*Var(X)+6*Cov(X,X)$
and $Cov(X,X)=Var(X,X)$
Therefore, $Var(X+X+X)=9*Var(X)$
The key is "When you consider 2X=X+X, you are not dealing with two independent variables" as David K points out.