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Given a linear function such as

$y = 1.62*x - 0.49$

Scenario 1

If $x = .5$ then $y = .32$

If we then increase $x$ by $10$% ($x=.55$), then $y=.401$, which results in $y$ increasing by $\approx .25$%

Scenario 2

If $x = .6$ then $y = .482$

If we then increase $x$ by $10$% ($x=.66$), then $y=.579$, which results in $y$ increasing by $\approx .20$%

Question

I am looking to explain how an improvement in x results in an improvement in y. Clearly, the relationship is not linear. I do not remember enough calc to model this relationship, if that is possible. I am looking for a mathematical way to say "as you improve x, you will see an improvement in y; however, if x is already high then an improvement in y is less impactful. Therefore, increasing x has the affect of improving y, but the return on investment is greater if x is initially very low." or "if x is bad and you improve it you can expect this return in y, but is diminishes as the starting x is higher"

I will also allow that my original premise is faulty. Thanks!

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    @GerryMyerson thanks for the point on percentage increase in y. The percentage increase in$y$and$x$goes down as initial$x$grows.2012-06-22

2 Answers 2

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Let $f(x) = mx + b$ be any linear function. Suppose we start from two points $x_1$, and $x_2$, with images $f(x_1) = mx_1 + b$ and $f(x_2) = mx_2 + b$. Then we consider a proportional increment $\Delta$ in each case. That is, we transform $x_1 \mapsto (1 + \Delta)x_1$ and $x_2 \mapsto (1 + \Delta)x_2$.

The images of these values will be: $f\big((1 + \Delta)x_1\big) = m(1 + \Delta)x_1 + b$ and $f\big((1 + \Delta)x_2\big) = m(1 + \Delta)x_2+b$. If we want to find the proportional increase in the images, we have:

$\displaystyle \frac{f\big((1 + \Delta)x_1\big)}{f(x_1)}-1 = \frac{m(1 + \Delta)x_1 + b}{mx_1 + b}-1$ and $\displaystyle \frac{f\big((1 + \Delta)x_1\big)}{f(x_1)} -1= \frac{m(1 + \Delta)x_2 + b}{mx_2 + b}-1$

Now, these numbers don't have to be equal at all, even though $\Delta$, $m$ and $b$ are the same in both equations.

The proportional increase in $y$, for a proportional increase in $x$ of $\Delta$ is given by:

$\displaystyle P_\Delta(x)= \frac{f\big((1 + \Delta)x\big)}{f(x)} -1= \frac{m(1 + \Delta)x + b}{mx + b}-1 = \frac{m x\Delta}{mx + b}= \frac{1.62x\Delta}{1.62x-0.49}$

What you have observed is that $P_\Delta(x)$ is decreasing somewhere. We can verify this by taking the derivative and finding where it's negative, and also note that it will tend to $0$ with bigger and bigger $x$, by taking the limit as $x \to \infty$.

If we want to know when $P_\Delta(x) \geq \Delta$, i.e., the increase in $y$ is bigger or equal than the increase in $x$:

$\displaystyle P_\Delta(x) = \frac{m x\Delta}{mx + b} \geq \Delta \Longrightarrow \frac{mx\Delta - (mx+b)\Delta}{mx+b} = -\frac{b\Delta}{mx+b} \geq 0$

This happens when:

  1. $b\Delta \leq 0$ (i.e. $b$ and $\Delta$ have different signs) and $mx+b \gt 0 \Longleftrightarrow x \gt -\frac{b}{m}$; and when
  2. $b \Delta \geq 0$ (i.e. $b$ and $\Delta$ have the same sign) and $mx+b \lt 0 \Longleftrightarrow x \lt -\frac{b}{m}$.
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    @WebDesignHero See my edit.2012-06-22
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Sounds like the function you are looking for is (x/y)(dy/dx), or equivalently d(log y)/d(log x). It is called the elasticity of the function.

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    Can you please expand on this and how I can explain the impact in English?2012-06-22