What is a decrease in denominator called?

[sethmachine]

Hi,

Edit: I think it would actually decrease. I consider 75 / 100 versus 50 / 75. The former is larger than the latter! What kind of "decrease" is this called?

Suppose we have some set of observations whose total count is n, and we need to predict some property for each observation from a set of states Y.

If our initial score is some k / n, where k <= n, what kind of "increase" in score would it be called if say we decide to exclude some subset of observations which amount to p / n of all the observations?

I.e. our score k / n will increase, since the denominator will decrease by some factor which is a percent of it (so n decreases by some percent of it).

Is this called a linear or constant increase?

For example, it is clear that going from n to n^2 is a quadratic increase.

But what is (k / n) to k / (n - (p / n))) called? A linear increase?

 

[GhostWolf]

It's called a bigger number.

 

[Menag]

it's obviously not linear.
If I understand you correctly you want to know how the difference behaves. Therefore we first see that
k/(n - (p/n))) = nk/(n^2 - p)
Taking the difference now yields
nk/(n^2 - p) - n/k = p/(n^2 - np) = 1/(n - p) - 1/n
Hence it's just a 1/x which is a bit moved.

 

[Adiktuz]

It's called a decrease

Also, if it's only the denominator that is decreased, then the resulting value will increase... The EDIT case you presented showed a decrease in value, because you reduced both the numerator and denominator...

so it's actually not k/n to k/(n-x) but rather k/n to (k-x)/(n-x)

 

[sethmachine]

It's called a decrease

Also, if it's only the denominator that is decreased, then the resulting value will increase... The EDIT case you presented showed a decrease in value, because you reduced both the numerator and denominator...

so it's actually not k/n to k/(n-x) but rather k/n to (k-x)/(n-x)

Then it is called a linear decrease, since both the denominator / numerator decrease by some constant value?

 

[Waffle]

definitely not linear.

unless k and n are equal you can have either (k-k)/(n-k) = 0/(n-k) = 0
or (k-n)/(n-n) = (k-n)/0 == +/- infinity.

you can put some numbers in there and see how drastically the ratio between the numbers changes.

 

[Adiktuz]

for it to be linear, there must be a linear relation between the decrease value and the resulting value

Methods to check for linearity

a linear equation has the form

y = mx + b

here y = the result of the fraction, x = the number you subtract from both the numerator and denominator... b = some constant value, which is the value of y when x = 0

so on your equation (k-x)/(n-x)

b = 75/100 = 0.75

now let's plug your 2nd try to find m

50/75 = m(25) + 0.75

m = (0.67-0.75)/25

m = -0.0032

so now our theoretical Linear equation is

y = (-0.0032)x + 0.75

now let's see if it's correct, if it's correct then it is indeed linear

let's consider a reduction of 50, we compare

y = (-0.0032)(50) + 0.75 vs (75-50)/(100-50)

0.59 vs 0.5

not equal but might still be within linear range...

to conclude if it's linear or not, you'd need to check more points then get the r^2 value...

Alternatively, you could do different values and plot them on excel, pick the linear option + show equation... that will show the graph + the equation assuming it is linear + the r^2 value

if the resulting r^2 value is greater than 0.9 (for some it's 0.99), then a linear approximation is fine... so you can say that it is indeed statistically linear

I did the excel test and the result was:

the graph was a curve with increasing slope as the value of X goes up... r^2 for linear was 0.65 so it is definitely non-linear

I tried plotting it as 2nd degree and so on, it shows increasing r^2...

r^2 values (from linear to 6th degree polynomial equation): 0.65,0.897,0.9744,0.9947,0.9991,0.9999

it is acceptable already as a 4th degree polynomial if you ask me, but the "perfect" fit was at 6th degree

 

[Menag]

To be exact. f(x) = m * x + n is an affine transformation.
m * x would be linear. A linear mapping f has the properties:
f(x + y) = f(x) + f(y)
f(ax) = af(x)
or to write it more compact
f(ax+y) = af(x)+f(y)

And i write it again. The function you're asking for is a rational function.

 

[sethmachine]

So what kind of decrease do I call it? Or do I just say the value should be expected to decrease?

 

[Adiktuz]

You can say it's a decrease that follows at least a 4th degree rational/polynomial function