Question on sum of normal variable

Question

I have a small doubt.

If X and Y are standard normal variables, is $ Z=(X+Y)/\sqrt { 2 } $ a standard normal variable ?

If I am correct, $X+Y$ follows $N(0, 2)$. So, Z must follow $N(0, 2 / \sqrt { 2 } ) $

and not $N(0, 2)$.

Am I right ?

Answer 1

If $X$ and $Y$ are independent standard normal, then your calculation is correct: the random variable $\frac{X+Y}{\sqrt{2}}$ has variance $1$, and is normal.

Under the assumption of independence, $X+Y$ is indeed normal with variance $2$. Dividing $X+Y$ by $\sqrt{2}$ divides the variance by $(\sqrt{2})^2$, that is, by $2$, giving variance $1$.