I was trying to prove that the gaussian distribution is "symmetric",
which means that given a standard gaussian variable $N$,
$P(N\in R)=P(N\in-R)$ for all $R\subset\mathbb R$,
where $-R=\{-x:x\in R\}$.
To this end, my idea was to proceed as follows:
$$P(N\in-R)=\int_{-R}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx,$$
then use the change of variable $y=-x$,
which yields
$$P(N\in-R)=\int_{R}\frac{e^{-y^2/2}}{\sqrt{2\pi}}-dy=-P(N\in R),$$
but this is a complete nonsense.
Maybe I'm doing something really dumb here but I can't seem to figure out where I'm going wrong.
Best Answer
When you are considering $-R$, the integration is in the reverse order, so you get an additional sign change.
For example, if $R = [a, b]$, then $P(N\in R) =\int_a^b\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx$ so $P(N\in-R) =\int_{-b}^{-a}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx =\int_{b}^{a}\frac{e^{-x^2/2}}{\sqrt{2\pi}}d(-x) =-\int_{b}^{a}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx =\int_{a}^{b}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx $.
I do find the notation and terminology somewhat unusual.