$X$ follows normal distribution $\mathcal{N}(0,1)$

Question

$X$ follows normal distribution $\mathcal{N}(0,1)$. Find distribution of $X^2$. My question is why we are allowed to calculate this as follows:

\begin{align*}P\left(-\sqrt t \le X\le \sqrt t\right) &=F_X(\sqrt{t})-F_X(-\sqrt{t})\\ &=\frac{\partial F_X(\sqrt{t})}{\partial t} - \frac{F_X(-\sqrt{t})}{\partial t}\\ &=f_x(\sqrt{t})(\sqrt{t})^\prime-f_x(\sqrt{t})^\prime(\sqrt{t})\end{align*}

I mean how is it possible that we have derivatives here? Why to apply this strange formula for derivatives at the end (it does not seem to be a normal product rule).

Answer 1

It's not a product rule but a chain rule. You have $(F_X)'(t) = f_x(t)$ so $$(F_X(\sqrt{t}))' = F_X'(\sqrt{t}) (\sqrt{t})' = \frac{f_x(\sqrt{t})}{2\sqrt{t}}. $$

Answer 2

You're asking "why is there a derivative". Well, what if I had some function $F(u)$ and I told you to take the derivative. you might say, $$f(u)\cdot u'$$ right? It's the same concept. As for the approach, it is equivalent to using a "regular" change of variable $$f_Y(y) = \frac{f_X(\sqrt y)}{\left|\frac{dy}{dx}\right|}+\frac{f_X(-\sqrt y)}{\left|\frac{dy}{dx}\right|}= f_X(\sqrt y)\left|\frac{dx}{dy}\right|+f_X(-\sqrt y)\left|\frac{dx}{dy}\right|$$