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How to make the probability density function of standard normal distribution symmetric?
I have recently been working with the probability density function of the standard normal distribution $\mathcal{N}(0,1)$. The function is asymptotically distributed according to a normal distribution. Thus it's standard to use the characteristic function of a normal distribution to describe this. I have noticed however, that the characteristic function has the property that it is always symmetric about $z = 0$.
\begin{equation*}
\varphi(z) = e^{ -z^2/2}, \text{ }\forall z \in \mathbb{R}.
\end{equation*}
The characteristic function of the standard normal distribution is, $\varphi(z) = e^{ -z^2/2}, \text{ }\forall z \in \mathbb{R}.$
How can one generate a new function that is asymptotically distributed according to a normal distribution, but is symmetric about $z = 0$?
A:
Let $Z\sim\mathcal{N}(0,1)$, and $g(z)=\frac12\mathbf{1}_{ -\pi 0b46394aab
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