I'm reading a paper about kernel estimation and struggling with this part, which constructs the pdf of logZ^2 (Z has Standard normal distribution)
I don't understand how they compute the function like this. Can someone provide the detailed computation from step 1 to step 2?
Many thanks in advance.
Let $f_Z$ denote the density of $\log Z_i^2$. We obtain, $\newcommand{\brp}[1]{{\left(#1\right)}}$ $\newcommand{\brs}[1]{{\left[#1\right]}}$ $\newcommand{\dx}{\mathrm{dx}}$ $\newcommand{\deriv} [2] {{\frac{\mathrm{d}#1}{\mathrm{d}#2} }}$ \begin{align*} f_Z(x) &= \deriv{}{x}\brp{\frac{2}{\sqrt{2\pi}}\int_{-\infty}^{e^{x/2}} e^{-s^2/2}ds-1} \\&= \frac{2}{\sqrt{2\pi}}\brs{\brp{\deriv{}{x}e^{x/2}}e^{-e^x/2}} \\&= \frac{2}{\sqrt{2\pi}}\brs{\frac{1}{2}e^{x/2}e^{-e^x/2}} \\&= \frac{1}{\sqrt{2\pi}}e^{x/2}e^{-e^x/2} \end{align*}
Sorry but reading it with my cell phone I cannot see your picture. That is one of the reason why it is not a good idea to substitute formulas with pictures.
Anyway, just using the Fundamental Transformation theorem, Here in the paragraph "Functions of Random Variables, you will easy find that
$$y=\log z^2$$
$$z^2=e^y$$
$$|z|=e^{y/2}$$
and
$$\Bigg|\frac{d}{dy}z\Bigg|=\frac{1}{2}e^{y/2}$$
thus
$$f_Y(y)=\frac{2}{\sqrt{2\pi}}e^{-e^{y}/2}\frac{1}{2}e^{y/2}=\frac{1}{\sqrt{2\pi}}e^{-e^{y}/2}e^{y/2}$$
At present I cannot check if my result matches with yours but I think it does.