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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 $\approx .20$%


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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It's not linear because you are not increasing $x$ by the same amount in both cases: $10\% \text{ of } .6 = .06$ is bigger than $10\% \text{ of } .5 = .05$. If you increased $x$ by the same absolute amount, you'd get a linear relationship. – talmid Jun 22 '12 at 1:17
@talmid, I think if you run the number with same absolute increases in $x$ you still won't get the same percentage increases in $y$. – Gerry Myerson Jun 22 '12 at 1:22
@GerryMyerson No. What I meant to say is that if you start with any $x_0$ and increase a fixed quantity $\Delta$, then independently of what $x_0$ is, the absolute increase in $y$ will be $1.62 \cdot \Delta$. – talmid Jun 22 '12 at 1:48
@talmid, right you are. But OP seems to be more interested in explaining why the percentages work out the way they do, than in what happens if the increase in $x$ is an absolute constant. – Gerry Myerson Jun 22 '12 at 1:58
Correct, I am not expecting the same the % change to be the same or linear. However, perhaps I can mathematical describe the rate of decrease? E.g. a 10% increase decreases by $w$ as initial x increases? At a more high level when you are saying that increasing x increases y and that is a significant increase, the significance of that increase decreases as initial x is larger. This is not a math problem that requires a specific number, I just want to better describe the relationship for the impact of an improvement of x on y. – Eric G Jun 22 '12 at 2:10
up vote 0 down vote accepted

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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Given the 10% increase in $x$, it will approach about a 14% increase $y$ . This equation does seem to capture the idea. Since it is multivariate it would be difficult to come up with any broad statements, but I think this gives me enough to talk about my point by giving a few scenarios e.g., if the score on $x$ is currently a .4 and there is a 20% improvement, we can expect to see an 82% improvement in $y$ whereas even a 10% improvement of $x$ at $x= .7$ provides a 17% improvement in $y$ . I would still wonder if this could be more generalized without specifying delta. – Eric G Jun 22 '12 at 2:42
Would you agree that in simple english it is fair to say that in general an increase in x corresponds to a greater increase in y? I assume for any given delta there is a point when it is no longer "significant" e.g., it takes some amount of work to increase $x$ by delta but if $y$ is already pretty high, better spend that amount of work elsewhere? – Eric G Jun 22 '12 at 2:56
@WebDesignHero Well, $P_\Delta(x)$ certainly depends on both $x$ and $\Delta$. You can keep $\Delta$ fixed, and see what happens, which is the case I considered in the last paragraph; or maybe analyze it as a bivariate function. It depends on what you want to do. Note that, qualitatively your observations do not depend on the value of $\Delta$. I added a simpler expression for $P_\Delta(x)$. – talmid Jun 22 '12 at 2:58
@WebDesignHero We certainly can say that the corresponding increase in $y$ gets smaller when $x$ is big (which is exactly when $y$ is big if $m\gt0$). That is because $\lim_{x \to \infty}P_\Delta(x) = 0$. Now, it isn't so clear that an increase in $x$ always implies a greater increase in $y$. At some points it might be greater, but definitely for big $y$ it will eventually be smaller than the increase in $x$. – talmid Jun 22 '12 at 2:59
I agree thanks. I should also mention that we are dealing with "scores" so $x$ and $y$ are both bound to 0-1.0, any y < 0 is 0 for these purposes outside of the pure math. – Eric G Jun 22 '12 at 3:14

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? – Eric G Jun 22 '12 at 2:43

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