# Help with an application of Leibnitz's Rule

I'm trying hard to understand and solve the following:

$$f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dy}\int_{-\sqrt{y}}^{\sqrt{y}}{f_X(x)}dx=?$$ The background information is that $$f_X(x)$$ is the pdf of random variable $$X$$ which follows the standard normal distribution. $$Y$$ is defined as $$Y=X^2$$. It is noted that $$x$$ is therefore $$-\sqrt{y}$$ and $$\sqrt{y}$$. The problem says "hint: use Leibnitz's rule." The main problem I'm having is that in this question, the format of the question is unlike the format in any examples I've seen. Leibnitz's Theorem in our textbook, as well as in every example I could find in other sources, is demonstrated to solve problems of the form $$\frac{d}{d\theta}\int_{a(\theta)}^{b(\theta)}{f(x, \theta)}dx$$ EDIT: By different formats, I meant that the examples have integrands that are multivariate and the problem I was given has a univariate integrand. My confusion came from trying to determine how to apply a theorem explained to me in terms of multiple variables to a univariate problem without accidentally invalidating the result. The use of $$\theta$$ or $$y$$ was not a source of confusion.

In such a form, I could write the canonical formula and just "plug and chug," but since the problem I'm trying to work out is univariate, I'm not sure how to apply the theorem. It seems like everything can one way or another be changed into terms of $$x$$ (which may be the point and it's sort of a trick question?). This is compounded by the fact that I'm new to statistical theory, so I'm both trying to grapple with a poor understanding of transformations plus trying to interpret questions like this in non-standard formats.

I'm not asking for the answer to the problem. I'm asking how to interpret such a problem, how to approach it, and why the approach works. My inclination would be to say that it equals zero, but I have no confidence in that answer since I'm not even sure that I know the meaning of the question. Even if it is zero, I'd like to see someone else's train of thought in finding that.

• This one is easy to do with a direct application of the definition of derivative. Alternatively, write the integral as the sum of integrals over $[0,\sqrt{y}]$ and $[-\sqrt{y},0]$ and apply the sum rule. To see that the derivative will not be zero, draw a graph of the function $F.$ This is no "trick question:" it's a basic calculation we all encounter from time to time and need to understand due to the two-to-one mapping involved.
– whuber
Oct 13, 2020 at 22:14
• I don't see how just knowing the definition of the derivative can solve this if I don't evaluate the integrand first, but when I try to integrate over $[0, \sqrt{y}]$ or $[-\sqrt{y}, 0]$, I just hit the erf (or something that looks like it) with no clean solution. And I'm still not sure why the professor said "Leibnitz's rule" is a hint.
– AJV
Oct 13, 2020 at 23:06
• Use the Fundamental Theorem of Calculus.
– whuber
Oct 14, 2020 at 13:25

Hints:

As you know and state clearly, Leibniz's rule says something like

If $$F(\theta) = \displaystyle\int_{a(\theta)}^{b(\theta)} f(x; \theta) \,\mathrm dx$$ where $$a(\theta), b(\theta)$$, and $$f(x; \theta)$$ are differentiable functions of $$\theta$$, then \begin{align}\frac{\mathrm dF(\theta)}{\mathrm d\theta} &= \frac{\mathrm d}{\mathrm d\theta}\int_{a(\theta)}^{b(\theta)} f(x; \theta) \,\mathrm dx\\ &= \int_{a(\theta)}^{b(\theta)} \frac{\partial f(x; \theta)}{\partial\theta} \,\mathrm dx + f(b(\theta); \theta)\frac{\mathrm db(\theta)}{\mathrm d\theta} - f(a(\theta); \theta)\frac{\mathrm da(\theta)}{\mathrm d\theta}\tag{1}\end{align}

but some weirdos like myself go so far as to replace $$\theta$$ by $$y$$ everywhere in $$(1)$$ and claim that it is also true that

$$\frac{\mathrm d}{\mathrm dy}\int_{a(y)}^{b(y)} f(x; y) \,\mathrm dx = \int_{a(y)}^{b(y)} \frac{\partial f(x; y)}{\partial y} \,\mathrm dx + f(b(y); y)\frac{\mathrm db(y)}{\mathrm dy} - f(a(y); y)\frac{\mathrm da(y)}{\mathrm dy}\tag{2}.$$ So, if I choose $$a(y)=-\sqrt{y}, b(y) = +\sqrt{y}$$, and define $$f(x;y) = f_X(x)$$ (that is, being a constant function of $$y$$ and thus having a partial derivative with respect to $$y$$ of $$0$$), then $$(2)$$ would simplify to $$\frac{\mathrm d}{\mathrm dy}\int_{-\sqrt{y}}^{+\sqrt{y}} f_X(x) \,\mathrm dx = \int_{a(y)}^{b(y)} 0 \,\mathrm dx + f_X\left(+\sqrt{y}\right)\frac{\mathrm d\sqrt{y}}{\mathrm dy} + f_X\left(-\sqrt{y}\right)\frac{\mathrm d\sqrt{y}}{\mathrm dy}$$ which could be worked out by plugging and chugging via replacing $$f_X$$ with the standard normal density and figuring out the derivative of $$\sqrt{y}$$ etc., but since you apparently don't agree that $$(2)$$ follows from $$(1)$$, I can't help you.

The easier way to do this problem (avoiding all the fuss about Leibniz's rule) is to do what your professor suggests and find $$F_Y(y)$$ explicitly for $$y \geq 0$$ in terms of the standard Gaussian CDF function $$\Phi(\cdot)$$: $$F_Y(y) = P\left\{X^2 \leq y\right\} = P\left\{-\sqrt{y} \leq X \leq \sqrt{y}\right\} = \Phi\left(\sqrt{y}\right) - \Phi\left(-\sqrt{y}\right)$$ and find the derivative of that with respect to $$y$$, recalling the chain rule for differentiation from the first calculus course, and remembering that the derivative of $$\Phi(x)$$ is $$\phi(x)$$, the standard Gaussian density function.

• I had already written the problem as $(2)$ and worked out the last two terms, but could not get that the partial derivative should be 0. I should have emphasized that my confusion with the "format" was that the examples I was taught with are all multivariate and that problem I was given is univariate. "...define $f(x;y)=f_X(x)$ (that is, being a constant function of y and thus having a partial derivative with respect to y of 0)" is helpful, but why is the partial derivative with respect to $y$ of $f_X(x)$ zero IF $Y$ itself is a function of $X$? Thank you for the help.
– AJV
Oct 14, 2020 at 12:50
• $Y$ may be a function of $X$ but $y$ is not a function of $x$. $X$ and $Y$ are random variables and $Y=X^2$ in this case, but $x$ and $y$ are arbitrary real numbers (real variables) and it is not the case that $y=x^2$ always. Oct 14, 2020 at 13:02