2
$\begingroup$

I have tried the differentiation of an integral below with Mathematica and it gives me the following solution:

$$\frac{d}{dc}\int_{z^{-1}(c)}^{1}z(x)dx = -\frac{c}{z'(z^{-1}(c))}$$

I am not quite sure where it gets it from...I think it can be separated and with differentiation the first part will be zero:

$$\frac{d}{dc}\int_{z^{-1}(c)}^{1} z(x)dx = \frac{d}{dc}\left(\int_{0}^{1} z(x)dx-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx\right)=$$ $$-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx$$

By the formula $\frac{d}{dc}z^{-1}(c) = \frac{1}{z'(z^{-1}(c))}$, so it definitely has to play a role here, but how? and where does $c$ come from? Do we need to somehow use chain rule here?

If the upper limit was just $c$ it would be easy to take it with a Leibniz rule:

$$−\frac{d}{dc}\int_0^cz(x)dx=z(c)$$

However, it seems that when one of the limits is an inverse function, there is something different at work which I have no understanding of and can't find an answer, so any help is appreciated!

UPDATE:

Thanks everyone for different approaches to this problem, I personally prefer the one with the fundamental theorem of calculus, but every approach has enriched my understanding of the problem!

I have to admit that it was just a private case with uniform distribution of a more general problem which Mathematica cannot do. Could anyone please confirm whether the fundamental theorem of calculus may also be applied to generic distribution functions under the integral, i.e. would the following general case be correct (where f and F - are correspondingly generic PDF and CDF)?

$$\frac{d}{dc}\int_{z^{-1}(c)}^{1}f(x)F(z(x))dx = -f(z^{-1}(c))F(z(z^{-1}(c))z'^{-1}(c) = -\frac{f(z^{-1}(c))F(c)}{z'(z^{-1}(c))}$$

3 Answers 3

3

The fundamental theorem of Calculus tells me $$ \frac{\partial}{\partial x} \int_{a(x)}^{b(x)}f(s,x)ds = f(s,b(x))b'(x) - f(s,a(x))a'(x) + \int_{a(x)}^{b(x)}\partial_x f(s,x)ds $$ so we have $$ \frac{\partial}{\partial c} \int_{z^{-1}(c)}^{1}z(x)dx = z(1)\frac{d}{dc}1-z(z^{-1}(c))\frac{d}{dc}z^{-1}(c) + \int_{z^{-1}(c)}^{1}\partial_c z(x)dx $$ so all we have is $$ \frac{\partial}{\partial c} \int_{z^{-1}(c)}^{1}z(x)dx = z(z^{-1}(c))\frac{d}{dc}z^{-1}(c) = c\frac{d}{dc}z^{-1}(c) $$ which gives you the answer you want.

  • 0
    In my experience the limits of integrals were constants and now it makes sense that their derivative would always be 1 and not written... Thanks for the explanation! Could you please also take a look at the equation after the "update"? Did I solve it correctly or something doesn't make sense there? – 2017-02-23
  • 0
    Yes your edit makes sense. Just be careful with the statement the derivative of a constant will always be $1$ - I believe it would be $0$ :). thanks for the tick mark. – 2017-02-23
  • 0
    Yes, sorry I wanted to say variable limits, but not in the functional form; for constants it is indeed zero :) – 2017-02-23
2

Let us assume that the $z$ function is monotonic and $C^1$, and let $w$ be its inverse function.
$$ \int_{w(c)}^{1}z(x)\,dx \stackrel{x\mapsto w(t)}{=} \int_{c}^{z(1)} t\cdot w'(t)\,dt \tag{1}$$ hence by differentiating both sides with respect to $c$ we get: $$ \frac{d}{dc}\int_{w(c)}^{1}z(x)\,dx = -c\cdot w'(c). \tag{2} $$ On the other hand, by differentiating with respect to $u$ both sides of the identity $z(w(u))=u$ we get $w'(u)\,z'(w(u)) = 1$, hence the RHS of $(2)$ can be written as $-\frac{c}{z'(w(c))}$ as wanted.

1

HINT

Let $Z$ be the anti-derivative of $z$. then, $$ \begin{split} \frac{d}{dc} \int_{z^{-1}(c)}^1 z(x)dx &= \frac{d}{dc}\left[ Z(1) - Z(z^{-1}(c))\right] \\ &= 0 - \frac{d}{dc} Z(z^{-1}(c)) \\ &= -z(z^{-1}(c)) \frac{dz^{-1}(c)}{dc} \end{split} $$