Derivative of integral of function with respect to the function

[DCG]

Hi all,

I have the following quantity:

f = \int_{\mu}^{\infty} (1-F(x))a(x)dx

I want to claim that by increasing the following quantity:

g = \int_{\mu}^{\infty} a(x)f(x)dx

then f can only increase. Can I differentiate f with respect
to g? Is the following correct?

\frac{\partial (\int_{\mu}^{\infty} (1-F(x))a(x)dx)}{\partial (\int_{\mu}^{\infty} a(x)f(x)dx)}

= \frac{\frac{\partial (\int_{\mu}^{\infty} (1-F(x))a(x)dx)}{\partial x}}{\frac{\partial (\int_{\mu}^{\infty} a(x)f(x)dx)}{\partial x}}

= \frac{(1-F(x))a(x)}{f(x)a(x)}

= \frac{1-F(x)}{f(x)}

I already know that a(x) > 0 for x>\mu and that \frac{1-F(x)}{f(x)} is positive. Therefore f increases when g is increased. Does this resolve the problem?

Thank you for taking the time to read and answer!

 

[HallsofIvy]

No, because you cannot differentiate f and g with respect to x- they are not functions of x.

 

[DCG]

Oh I see. Any pointers on how I could approach this though? Is there a name for this kind of problem? Anything I could read that could help me?

Thanks a lot!

 

[Mute]

It's not clear what you're trying to do. You have an expression for f that doesn't depend on g at all. Why are you talking about increasing g? Is F(x) given, or is the problem to determine F(x) in terms of g?

 

[haruspex]

Hi all,

I have the following quantity:

f = \int_{\mu}^{\infty} (1-F(x))a(x)dx

I want to claim that by increasing the following quantity:

g = \int_{\mu}^{\infty} a(x)f(x)dx

In the second line, you have f as a function of one variable, x. But the first line doesn't make it a function of x. Rather, it appears to be a function of μ, and/or a functional of F(), a().

 

[DCG]

I am sorry, my choice of variable names is terrible. Let:

q = \int_{\mu}^{\infty} (1-F(x))a(x)dx

In my first post I named it f and it has nothing to do with f(x).

g = \int_{\mu}^{\infty} a(x)f(x)dx

Here F(x),f(x) are fixed. They are the cdf and pdf of a regular distribution, \mu is also fixed. It is the mean of that distribution. The only thing not fixed here is a(x). This function is positive for x>\mu and is also monotone non-decreasing. My problem is the following:

Suppose I have some specific a(x) and I want to replace it by another function (a_2) with the same properties. Let:

q_2 = \int_{\mu}^{\infty} (1-F(x))a_2(x)dx

and

g_2 = \int_{\mu}^{\infty} a_2(x)f(x)dx

I want to prove that if g_2 \geq g then q_2 \geq q. Thanks your time.

 

[DCG]

So I read a bit about functional derivatives. It seems like q=q(a) is a functional. It maps functions to numbers. So I can try and find its derivative with respect to a. According to what I read:

\int \frac{\partial q(a)}{\partial a(x)}h(x)dx = {lim}_{\epsilon \rightarrow 0} \frac{q(a + \epsilon h) - q(a)}{\epsilon}

= {lim}_{\epsilon \rightarrow 0} \frac{q(a) + \epsilon q(h) - q(a)}{\epsilon} = q(h) = \int_{\mu}^{\infty}(1-F(x))h(x)

Now if I assume that h(x) is defined only on [\mu ,\infty ] I have that:

\int_{\mu}^{\infty} \frac{\partial q(a)}{\partial a(x)}h(x)dx = \int_{\mu}^{\infty}(1-F(x))h(x)

which means that (I am not sure about this):

\frac{\partial q(a)}{\partial a(x)} = (1-F(x))

In the same way it turns out that \frac{\partial g(a)}{\partial a(x)} = f(x)

Using the chain rule I get:

\frac{\partial q(a)}{\partial g(a)} = \frac{\frac{\partial q(a)}{\partial a(x)}}{\frac{\partial g(a)}{\partial a(x)}} = \frac{1-F(x)}{f(x)} \geq 0

Does this even resemble anything valid?

 

[Mute]

That's more or less right, but you might run into problems trying to do it that way when your functionals aren't linear in your function a(x). When doing variational derivatives it's helpful to use the identity

$$\frac{\delta a(x)}{\delta a(y)} = \delta(x-y),$$
where $\delta(x-y)$ is the Dirac delta function (I will assume you know what this is, but ask if you don't). If we take the variational derivative of, say, g = g[a], we get

$$\frac{\delta g[a]}{\delta a(y)} = \int_{\mu}^\infty dx~f(x) \frac{\delta a(x)}{\delta a(y)} = \int_{\mu}^\infty dx~f(x) \delta(x-y) = f(y)\Theta(y-\mu),$$
where $\Theta(x)$ is the Heaviside step function, which appears because the delta function doesn't contribute anything to the integral unless y is in the integration range.

I'll let you try the variational derivative of q[a] for yourself.

What the derivative tells you is that if you have some fixed $a(x)$ and you perturb it to $a(x) + \epsilon \delta a(x),$ where $\epsilon$ is some small number and $\delta a(x)$ is O(1), then

$$g[a + \epsilon \delta a] \approx g[a] + \epsilon \int_{\mu}^\infty dy f(y) \delta a(y);$$
(well, it's exactly equal to that in this case because g was linearly dependent on a, but this is basically the functional equivalent of the tangent approximation to a function).

That said, I'll have to think for a second if the chain rule would work like that for variational derivatives.

 

[chiro]

Hey DCG and welcome to the forums.

You have a PDF so it means that the CDF is monotonic increasing. This implies 1 - F(x) is monotonic decreasing.

If g2 > g then this means integral of (a2(x) - a(x))f(x)dx over the appropriate interval is > 0. If q2 > q then this means integral (a2(x) - a(x))(1 - F(x))dx over the appropriate interval is also > 0.

But (a2(x) - a(x))(1 - F(x)) = (a2(x) - a(x)) - F(x)(a_2(x) - a(x)) which implies integral (a_2(x) - a(x))dx > F(x)(a_2(x) - a(x))dx over the right region (which makes sense).

Now we know (a_2(x) - a(x))dx > 0 and we know that F(x) <= 1 for any x by the properties of the CDF so the above inequality holds.

 

[Mute]

To push the functional way of doing it a little bit further (although DCG might be happy with chiro's solution, which is less technically demanding than what I'm about to do), I think the way you would need to think about things is like this: suppose g is just a variable, not a functional. Then we could interpret a(x) as a function of both x and g, which is determined by solving the integral equation

$$g = \int_{\mu}^\infty dx~f(x)a(x|g).$$

Differentiating implicitly with respect to g gives

$$1 = \int_{\mu}^\infty dx~f(x)\frac{\partial a(x|g)}{\partial g}.$$
Notice that if we set g = 1 in the first equation we would have a similar equation to this second one, suggesting that we can state

$$a(x|1) = \frac{\partial a(x|g)}{\partial g},$$
where a(x|1) is the solution to the original integral equation with g = 1.

Now, we can also consider q to be a function of g. (q is no longer considered to be a functional of a because a is determined by solving the integral equation). So,

$$q(g) = \int_{\mu}^\infty dx~(1-F(x))a(x|g),$$
and
$$\frac{\partial q(g)}{\partial g} = \int_{\mu}^\infty dx~(1-F(x))a(x|1).$$

One could then presumably solve the integral equation for a(x|1), and could then insert the solution into the equation for q'(g). Of course, the problem is that I don't think you get a unique solution. Suppose I set $a(x|1) = h(x)/f(x)$, assuming f(x) is never zero on $(\mu,\infty)$. Then all I need to do is pick any function h(x) that's normalizable on $(\mu,\infty)$.

 

[DCG]

Hey DCG and welcome to the forums.

You have a PDF so it means that the CDF is monotonic increasing. This implies 1 - F(x) is monotonic decreasing.

If g2 > g then this means integral of (a2(x) - a(x))f(x)dx over the appropriate interval is > 0. If q2 > q then this means integral (a2(x) - a(x))(1 - F(x))dx over the appropriate interval is also > 0.

But (a2(x) - a(x))(1 - F(x)) = (a2(x) - a(x)) - F(x)(a_2(x) - a(x)) which implies integral (a_2(x) - a(x))dx > F(x)(a_2(x) - a(x))dx over the right region (which makes sense).

Now we know (a_2(x) - a(x))dx > 0 and we know that F(x) <= 1 for any x by the properties of the CDF so the above inequality holds.

Thank you for your answer but I do not understand what happened to f(x) in the integral of
(a2(x)-a1(x))f(x)dx >0
this does not imply that integral(a2(x)-a1(x))dx > 0
since f(x) could allocate more probability mass to intervals where a2(x)>a1(x) and less to opposite ones.

 

[DCG]

To push the functional way of doing it a little bit further (although DCG might be happy with chiro's solution, which is less technically demanding than what I'm about to do), I think the way you would need to think about things is like this: suppose g is just a variable, not a functional. Then we could interpret a(x) as a function of both x and g, which is determined by solving the integral equation

$$g = \int_{\mu}^\infty dx~f(x)a(x|g).$$

Differentiating implicitly with respect to g gives

$$1 = \int_{\mu}^\infty dx~f(x)\frac{\partial a(x|g)}{\partial g}.$$
Notice that if we set g = 1 in the first equation we would have a similar equation to this second one, suggesting that we can state

$$a(x|1) = \frac{\partial a(x|g)}{\partial g},$$
where a(x|1) is the solution to the original integral equation with g = 1.

Now, we can also consider q to be a function of g. (q is no longer considered to be a functional of a because a is determined by solving the integral equation). So,

$$q(g) = \int_{\mu}^\infty dx~(1-F(x))a(x|g),$$
and
$$\frac{\partial q(g)}{\partial g} = \int_{\mu}^\infty dx~(1-F(x))a(x|1).$$

One could then presumably solve the integral equation for a(x|1), and could then insert the solution into the equation for q'(g). Of course, the problem is that I don't think you get a unique solution. Suppose I set $a(x|1) = h(x)/f(x)$, assuming f(x) is never zero on $(\mu,\infty)$. Then all I need to do is pick any function h(x) that's normalizable on $(\mu,\infty)$.

I like this approach. Since the only thing that I want to prove is that if g increases then
q can only increase or stay the same, it suffices to prove that:

\frac{\partial q(g)}{\partial g} \geq 0

Thank you all for the answers.This was really helpful. I think that I am in the right direction now, so I will try to work it out.

 

[chiro]

Thank you for your answer but I do not understand what happened to f(x) in the integral of
(a2(x)-a1(x))f(x)dx >0
this does not imply that integral(a2(x)-a1(x))dx > 0
since f(x) could allocate more probability mass to intervals where a2(x)>a1(x) and less to opposite ones.

What does g2 - g >= 0 imply in terms of the integral?