If X is standard normally distributed, what does -X follow?

Question

What is the most intuitive proof of the fact that 'if X follows standard normal distribution, then -X also follows standard normal distribution'? Can one use characteristic function to establish this fact? That is, $$\phi_X(t) = \phi_{-X}(t) = e^{-0.5*t^2} $$ Does this simple proof make sense?

Answer 1

Recall $a\mathcal{N}(\mu,\sigma^2) +b = \mathcal{N}(a\mu + b,a^2\sigma^2)$.

Here, $\mu = 0$, $\sigma^2=1$, $a=-1$ and $b=0$, so you have $-\mathcal{N}(0,1) = \mathcal{N}(0,1)$ as desired.

Answer 2

You are right that $\phi_{-X}(t)=\phi_X(-t)=\phi_X^\ast(t)$ can be used to prove this result, or indeed the more general result $X\sim N(\mu,\,\sigma^2)\implies -X\sim N(-\mu,\,\sigma^2)$ (since then $\phi_X(t)=\exp(i\mu t-\tfrac12\sigma^2t^2)$).