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I found a couple of similar question, but I am struggling applying their logic to my example.

Derivative of double integral with respect to upper limits

Differentiation under the double integral sign

Derivative of double integral with respect to upper limits

Derivative of double integral using Leibniz integral rule

I have the following differentiation of the double integral:

$$\frac{d}{dc}\int_{z^{-1}(c)}^{1}\int_{z(x)}^{z^{-1}(c)} (v-y)f(x)f(y)dydx$$

where $f(x)$ and $f(y)$ and probability density functions.

Is it possible to apply Leibniz rule right away here to somehow simplify it? Substituting the internal integral for anti-derivatives as in the examples does not seem possible because that's a product of functions.

The best I can do is to "open up" the internal integral by integrating by parts and then apply the rule:

$$\int_{z(x)}^{z^{-1}(c)} (v-y)f(y)dy=(v-z^{-1}(c))F(z^{-1}(c))-(v-z(x))F(z(x))+\int_{z(x)}^{z^{-1}(c)}F(y)dy$$

Then my expression becomes something like:

$$\frac{d}{dc}\int_{z^{-1}(c)}^{1}(v-z^{-1}(c))F(z^{-1}(c))f(x)dx-\frac{d}{dc}\int_{z^{-1}(c)}^{1}(v-z(x))F(z(x))f(x)dx+$$ $$\frac{d}{dc}\int_{z^{-1}(c)}^{1}\int_{z(x)}^{z^{-1}(c)}F(y)f(x)dydx$$

where the first term seems to be feasible to take, and even the second one, but the third one - back to square one. Is there any way around it?

1 Answers 1

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Let's look at the Leibnitz rule: If $g(r, t)$ and $g_r(r,t)$ are continuous, and $a(r), b(r)$ have continuous derivatives, then $$\frac{d}{dr}\left(\int_{a(r)}^{b(r)} g(r, t)\,dt\right) = g(r, b(r))\frac{db}{dr} - g(r,a(r))\frac{da}{dr} + \int_{a(r)}^{b(r)}\frac{\partial}{\partial r}g(r, t)\,dt$$

For your problem, $r = c, t= x, b = 1, a(c) = z^{-1}(c)$ and $$g(c, x) = f(x)\int_{z(x)}^{z^{-1}(c)}(v-y)f(y)\, dy$$

Note that no condition of the Leibnitz rule says anything about products of functions. All that it requires is the $g_c(c,x)$ exist and be continuous. $b$ obviously meets its condition. Assuming that $z^{-1}$ is continuously differentiable and that $f$ and $z$ are at least continuous, that expression will satisfy the conditions. Since $b$ is constant, $\frac{db}{dr}$ term is $0$.

Therefore $$\frac{d}{dc}\int_{z^{-1}(c)}^{1}\int_{z(x)}^{z^{-1}(c)} (v-y)f(x)f(y)dydx\\=-\frac{dz^{-1}}{dc}f(z^{-1}(c))\int_{c}^{z^{-1}(c)}(v-y)f(y)\, dy + (v-z^{-1}(c))f(z^{-1}(c))\frac{dz^{-1}}{dc}\int_{z^{-1}(c)}^{1}f(x)\,dx$$

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    Thank you, I think it makes sense now... So we actually treat the internal integral as a part of the function we are applying Leibnitz rule to, right? Then even the limits of this integral obey the formula, i.e. $z(x)$ becomes $z(z^{-1}(c))=c$. – 2017-02-26
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    However, I am not sure how do you get $(v-z^{-1}(c))f(z^{-1}(c))\frac{dz^{-1}(c)}{dc} \int_{z^{-1}(c)}^{1} f(x)dx$. Is it because $f(x)\frac{d}{dc}\int_{z(x)}^{z^{-1}(c)} (v-y)f(y)dy$, which is equal to $f(x)\left((v-z^{-1}(c))f(z^{-1}(c))\frac{dz^{-1}(c)}{dc} - (v-z(x))f(z(x))\frac{dz(x)}{dc} + \int_{z(x)}^{z^{-1}(c)}\frac{d}{dc}f(y)(v-y)dy \right)$ and since $\frac{dz(x)}{dc}=0$ and $\frac{d}{dc}f(y)(v-y)dy=0$ we are left only with the first term which is constant for the integration by $dx$? – 2017-02-26
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    That is one way of figuring it out. Though what I did was simply to note that the only dependence on $c$ was the upper limit of the integral, then applied the Differentiation theorem ($\frac d{dx}\int_a^x f(t)\, dt = f(x)$, aka 2nd FTC) and the chain-rule. – 2017-02-26
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    OK, thank you, and one last question about $\int_{c}^{z^{-1}(c)}(v-y)f(y)dy$. Since there is no differentiation I guess we need to integrate by parts, but it leaves one integral $\int_{c}^{z^{-1}(c)}F(y)dy$, which is unsolvable. I was wondering is there any other way to solve for this integral? – 2017-02-26
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    Since I don't know what $f$ is, I have no way of knowing how it can be calculated. It is true that for a generic function $f$, integration by parts in either direction does not appear to be useful. For specific functions, though, It could be easy or impossible to calculate, except by numerical methods. – 2017-02-27