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notmeatall · Apr 29, 2016, 07:11 AM

I'm confused about if it is possible to determine the sign of the following derivative:

$\frac{\partial y\alpha k}{\partial y}=\alpha k+\frac{\partial\alpha k}{\partial y}y$

where $\frac{\partial\alpha k}{\partial y}<0$

Is it possible to say that the derivative is positive for small ${\partial y}$ ?

Half my brain think the following must be true for positivity: $\frac{\partial\alpha x}{\partial y}\frac{y}{\alpha x}<-1 $
that is, the percentage decline in ${\partial\alpha x}$ should exceed the increase in $\frac{\partial y}{\partial y}$

Thanks for any input in advance :)

ebaines · May 5, 2016, 05:18 AM

Sorry, but I think you've made a number of errors in your formulas, so it's impossible to follow along. For example, from the first equation it appears that there is a variable called "ak," and you are taking it's partial derivative with respect to y. Which implies that the ak is a function of more than one variable - perhaps y and x? But in that first equation you also refer to the partial derivative of "yak" with respect to y - are you referring to a variable y times a variable ak? Then later on you refer to the partial derivative of ax with respect to y - so are there three variable here: x, y, and ak? Finally at the end you refer to the partial derivative of y with respect to y, which doesn't make much sense. Please clarify.

notmeatall · May 5, 2016, 05:31 AM

Originally Posted by ebaines

Sorry, but I think you've made a umber of errors in your formulas, so it's impossible to follow along. For example, from the first equation it appears that there is a variable called "ak," and you are taking it's partial derivative with respect to y. Which implies that the ak is a function of more than one variable - perhaps y and x? But in that first equation you also refer to the partial derivative of "yak" with respect to y - are you referring to a variable y times a variable ak? Then later on you refer to the partial derivative of ax with respect to y - so are there three variable here: x, y, and ak? Finally at the end you refer to the partial derivative of y with respect to y, which doesn't make much sense. Please clarify.

Thank you for answering, I will try to make it more clear what I am asking.

$\frac{\partial[y*a(y)*k(y)]}{\partial y}$
That is, a(y) and k(y) are functions of y.
Is it possible to unambiguously determine the sign of the derivate wrt. y of the product: y*a*k?

What I know is that we have:
$\frac{\partial\alpha(y)*k(y)}{\partial y}<0 $

$ \frac{\partial k(y)}{\partial y}<0$

$ \frac{\partial\alpha(y)}{\partial y}<0 $

I think the answer is no, but under what circumstances is it positive?
We must have that the percentage decrease in ak is smaller than unity for the increase in y to dominate right?

ebaines · May 5, 2016, 05:37 AM

OK, this helps. But one more question - in this equation:

$\frac{\partial\alpha(y)*k(y)}{\partial y}<0 $

do you mean

$\frac{\partial [\alpha(y)*k(y)]}{\partial y}<0 $

or do you mean

$\frac{\partial\alpha(y)}{\partial y} * k(y)<0 $

notmeatall · May 5, 2016, 05:45 AM

I mean this:

$\frac{\partial [\alpha(y)*k(y)]}{\partial y}<0 $

ebaines · May 5, 2016, 06:08 AM

No, the sign of $\frac {\partial [y * a(y) * k(y)]}{ \partial y}$ can not be determined. Here's an existence proof:

Suppose a(y) = 10 - y and k(y) = 10 -y^2. Both of these have derivatives <0 at both y = 0.1 and y = 1. And $\frac { \partial [a(y) * k(y)]}{\partial y}$ is less than 0 at both those points as well. But the value of

$\frac {\partial [y* a(y) * k(y)]}{\partial y} = 100 - 20 y -300 y^2 +4 y^3$

is negative for y=1 and positive for y = 0.1.

notmeatall · May 5, 2016, 06:28 AM

Thank you, that was really helpful.
But I really want to say as much about the sign as possible :)
Is the following correct?

assuming y close to unity.

$\frac{\partial[y*\alpha(y)*k(y)]}{\partial y}>0 $
if
$\frac{\partial[\alpha(y)*k(y)]}{\partial y}\frac{y}{\alpha k}>-1 $
that is, the elasticity is not too large to a change in y. ak changes less than y.

The direct increase in y then dominates the indirect effect from a*k right?