How to maximize when two variables depend on each others

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In summary, the conversation discusses how to maximize a function of the form F(y, x1, x2,..., xn) with y=f(x1). The use of total derivative, gradient, and Hessian in this context is questioned. It is suggested to use the chain rule and knowledge of \frac{\partial F}{\partial y} when calculating the gradients and hessians. An example is given where F(y, x, z) is a utility function and y=f(x) is an endogenous probability function. The conversation concludes with a clarification on the notation used and a question on how to maximize a function with a hypothetical formula.
  • #1
Kolmin
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Hi all.

I was thinking on how we can maximize a function with the following form:

F(y, x1, x2,..., xn) with y=f(x1)

I know that I should use the total derivative to find out the effective rate of y, but I don't see how gradient and Hessian should be organized in this context.

For example, if we have F(y, x1, x2) with y=f(x1), how should we set gradient and Hessian?
Should the gradient be [itex]\nabla[/itex]F= (F1, F2)?
Or [itex]\nabla[/itex]F= (Fy, F1, F2)
What about the Hessian?

Sorry for the question (maybe not exactly challenging...:redface:), but this problem is not explicitly mentioned in any book.

Thanks. :smile:
 
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  • #2
Just to put me on the right path, is the question completely meaningless or there is some sort of amazing mistake in it that I cannot see that makes the answer trivial?
 
  • #3
F is really only a function of x1,...,xn so you would only calculate derivatives with respect to these variables

To calculate the gradients and hessians you may have to use the chain rule and knowledge of [itex] \frac{\partial F}{\partial y}[/itex] depending on the problem.

It would probably be clearer if you tried to choose specific functions F and f which you think are confusing and we can work through it
 
  • #4
Office_Shredder said:
F is really only a function of x1,...,xn so you would only calculate derivatives with respect to these variables

To calculate the gradients and hessians you may have to use the chain rule and knowledge of [itex] \frac{\partial F}{\partial y}[/itex] depending on the problem.

It would probably be clearer if you tried to choose specific functions F and f which you think are confusing and we can work through it

First of all, thanks. :smile:

Let's put that F(y,x,z) is a Utility function and y=f(x) is an endogenous probability functions that depends on whatever you want (the important thing is that it is endogenous).
The decision maker has to choose the variables in order to maximize is problem (without constraints).
This is not the typical Von Neumann-Morgenstern Utility function, so how do we maximize it?

Do we set gradient [itex]\nabla[/itex]F= ( [itex]\frac{\partial F}{\partial y}[/itex], [itex]\frac{\partial F}{\partial x}[/itex], [itex]\frac{\partial F}{\partial z}[/itex] ) or do we set it [itex]\nabla[/itex]F= ([itex]\frac{\partial F}{\partial x}[/itex], [itex]\frac{\partial F}{\partial z}[/itex] ), which honestly doesn't make much sense to me?

Hessian is the next step, so gradient is the main problem I guess.
 
  • #5
If we re-write this as F(f(x),x,z) then it's clear that F is a function of two variables. To calculate the gradient we need to know what [itex] \frac{\partial F}{\partial x}[/itex] is (hopefully it's clear what dF/dz is). Unfortunately the notation is a bit unclear so I am going to re-define things. We will assume z is a constant and what we want to calculate is
[tex] \frac{dF(y(x),x,z)}{dx} [/tex]
where this is the derivative of F as a function of x (which is what you normally think of as the partial derivative. The reason for this is that I will use [itex] \frac{\partial F(y,x,z)}{\partial x} [/itex] to denote the partial derivative of F with respect to the second variable, treating y as a constant value independent of x. Then the chain rule for multivariable calculus says that
[tex] \frac{dF}{dx} = \frac{\partial F}{\partial y} \frac{ dy}{dx} + \frac{\partial F}{\partial x}[/tex]

So if you just have a function F(y,x,z) whose partial derivatives (as a function of 3 variables) are known, you can calculate the derivative of F with respect to x even when y is a function of x.
 
  • #6
Great!
Thanks a lot and sorry for the (now I see why...) trivial question. :smile:
 
  • #7
Another trivial question on this topic after that I started to play around a bit with it (assuming that hopefully I got the point after the last answer…).

Following the notation introduced by the fourth post I guess the gradient should look like

[itex]\nabla[/itex]F = ( [itex]\frac{dF}{dx}[/itex],[itex]\frac{\partial F}{\partial z}[/itex])

right? Then set it equal zero and it should work. But still I find some problems in my understanding of the situation.

What I realize lately is that the way in which I previously stated the question is really bad and doesn't emphasize one thing that is probably the most important to me. The fact is that in my problem not only s influences p, but at the same time s stands in itself in the hypothetical formula (with which I mean that you cannot simply decompose y(x) and that's it). Hopefully an example will make my terms of my problem more clear.

[itex]F(y(x),x,z)=(1-y)(x+z)+y(x-z)[/itex]

[This is what I mean with my peculiar - and probably wrong - jargon "not only s influences p, but at the same time s stands in itself"]

How do we maximize something like this?
Still with the gradient I wrote few lines above?

And then, if the gradient is that one, how the heck does the Hessian look like?

Many thanks to the one who will help this soul in the reign of shadows. :smile:
 

Related to How to maximize when two variables depend on each others

1. How do I find the maximum value when two variables are dependent on each other?

The best way to find the maximum value is by using a graphing calculator or by creating a graph of the two variables and visually identifying the point where the graph reaches its peak. You can also use calculus techniques, such as taking the derivative and setting it equal to zero, to find the maximum value analytically.

2. Can I use linear regression to maximize two dependent variables?

No, linear regression is only useful for finding the relationship between two variables and predicting future values. It cannot be used to find the maximum value when two variables are dependent on each other.

3. How can I determine the relationship between two dependent variables?

You can determine the relationship between two dependent variables by creating a scatter plot of the data and observing the pattern. If the points form a straight line, the variables have a linear relationship. If the points form a curve, the variables have a non-linear relationship.

4. Is there a specific formula for maximizing two dependent variables?

There is no specific formula for maximizing two dependent variables. It depends on the specific relationship between the variables and may require different techniques, such as graphing or calculus, to find the maximum value.

5. Can I use the same method to maximize two dependent variables in different scenarios?

Not necessarily. The method for maximizing two dependent variables may vary depending on the specific relationship between the variables and the nature of the data. It is important to assess each scenario individually and determine the most appropriate method for finding the maximum value.

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