Subtracting 2 random variables:
Let $X$~$N(0,1)$ and $Y$~$N(0,1)$ be independent random variables. Show that $X-Y$~$N(0,2)$.

Question

Subtracting 2 random variables:
Let $X$~$N(0,1)$ and $Y$~$N(0,1)$ be independent random variables. Show that $X-Y$~$N(0,2)$.

kirbykirby · Answer

I did:
Let $X=X_1$ and 
let $Y=X_2$ 

Then, there is a theorem stating that if $X_i,i=1,2,...n$ are independent $N(\mu_i,\sigma_i ^2)$ r.v.'s  and $a_i,i=1,2,...,n$ are constants, then $\sum_{i=1}^{n}a_i X_i$~$N(\sum_{i=1}^{n}a_i \mu_i\,,\sum_{i=1}^{n}a_i ^2 \sigma_i ^2$).

So,  $X-Y=X_1-X_2$~$N((1)(0)+(-1)(0),(1)^2(1)^2+(-1)^2(1)^2)=N(0,2)$

But I find it odd because I know when we add r.v.'s,(like if we had X+Y.... can we not say X=Y because they have the same distribution, so X+X=2X? 
It's just odd for X-Y because that would yield X-X= 0 ??