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I was reading this: Characteristic function of Normal random variable squared

And there is a step, where I do not manage to understand the equality marked as "$\stackrel{???}{=}$"

We have $$ F_X\left(\sqrt{y}\right)= \int_{-\infty}^\sqrt{y} \frac{1}{\sqrt{2\pi}} \operatorname{e}^{-\frac{t^2}{2}} \operatorname{d}t$$ and the pdf is $$f_Y(y) = \frac{\operatorname{d}F_Y(y}{\operatorname{d} y}=2\frac{\operatorname{d}F_X(\sqrt{y})}{\operatorname{d} y} \stackrel{???}{=} 2 \frac{1}{\sqrt{2}\sqrt{\pi}} e^{-\frac{y}{2}} \left( \frac{1}{2} y^{-\frac{1}{2}} \right) = \frac{1}{2^{\frac{1}{2}} \Gamma(\frac{1}{2})}y^{\frac{1}{2}-1}e^{-\frac{y}{2}} \Bbb{I}_{[0,+\infty)}(y) $$

Thank you for your insight!

tortue
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rannoudanames
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    I would suggest that you take a look at the Leibniz integral rule https://en.m.wikipedia.org/wiki/Leibniz_integral_rule and the fundamental theorem of calculus https://en.m.wikipedia.org/wiki/Fundamental_theorem_of_calculus – user585522 Oct 03 '18 at 15:27
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    $$\frac{d}{dy}F_X(\sqrt y)=f_X(\sqrt y)\frac{d}{dy}(\sqrt y)$$ , where $f_X$ is the pdf of $X$. – StubbornAtom Oct 03 '18 at 18:50

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First, let's differentiate $F_X$ with respect to its argument $\sqrt{y}$, giving $\frac{1}{\sqrt{2\pi}}\exp -\frac{(\sqrt{y})^2}{2}=\frac{1}{\sqrt{2\pi}}\exp -\frac{y}{2}$. Now we use the chain rule to get the derivative with respect to $y$ instead, by multiplying by $\frac{d}{dy}\sqrt{y}=\frac{1}{2}y^{-1/2}$.

J.G.
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