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Discussion Starter · #1 ·
I'm writing about income distribution, and every time the wealth of a group doubles, the size of the group decreases with two-thirds.
So when there are 24,000 people that earn 400,000$....
24,000*1/3=8,000
400,000*2=800,000
There are 8,000 people that earn 800,000$.

I want to know the formula, so I can plot it in a graph and calculate stuff like "how many people earned more than 1,000,000' and 'how much does the richest guy earn'. Could you help me?

Thanks for the help! :D

I've posted this in the 'advice center' and asked the INTJs too. Sorry for the spam, but I really need an answer and I wonder where the mathematicians are.
 

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Okay let's see, you got two variables exponentially related... Yeah I think I can do this. Income as a function of people:

I = I_0*2^(ln(P_0/P)/ln(3))

Where I_0 and P_0 are an arbitrary point.

And people as a function of income:

P = P_0*1/3^(ln(I/I_0)/ln(2))

Using the wonderful proof that log base a of b equals ln(b)/ln(a).

Sorry, Ln, natural log. Though you can use log() too, it actually doesnt matter in this case :)

Edit 2: P is any given population, it's your dependent variable in the first equation (where as P_0 is just a constant you compare to to say "oh, this population is a third as large, so income must be twice as high").
 

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Discussion Starter · #3 ·
Okay let's see, you got two variables exponentially related... Yeah I think I can do this. Income as a function of people:

I = I_0*2^(ln(P_0/P)/ln(3))

Where I_0 and P_0 are an arbitrary point.
Thank you! :D

But... I'm trying to plot it using fooplot (Google it, I can't post links..), and...
400000*2^(In(24000/P)/In(3))
In=Inverse? How can I do that in for example fooplot..? And what is P?

Sorry, I suck at math :(
 

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Discussion Starter · #4 ·
Okay let's see, you got two variables exponentially related... Yeah I think I can do this. Income as a function of people:

I = I_0*2^(ln(P_0/P)/ln(3))

Where I_0 and P_0 are an arbitrary point.

And people as a function of income:

P = P_0*1/3^(ln(I/I_0)/ln(2))

Using the wonderful proof that log base a of b equals ln(b)/ln(a).

Sorry, Ln, natural log. Though you can use log() too, it actually doesnt matter in this case :)

Edit 2: P is any given population, it's your dependent variable in the first equation (where as P_0 is just a constant you compare to to say "oh, this population is a third as large, so income must be twice as high").
Thanks again! :D And I feel very stupid now, because I still don't properly understand it. Not your fault, I suck at math.

I want to plot it in 'Fooplot'. I have to enter y(x)=..........
What do I write at the ........?
Could you give me an example? I have no idea what I should enter as a 'constant to compare to'.
 

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Sure! I opened it and played with it, so let's say you know that 20000 people make $40000 - that is your arbitrary point, meaning P_0 = 20000 and I_0 = $40000. If you're solving for income I (y), then P is your dependent variable (x), so you'd put in "40000*2^(ln(20000/x)/ln(3))".

Edit: Hope this helps! Lap top is about to die. I can help with other math like summing up people who make above so and so later if you want.

Edit2: I'll do (or at least attempt) the calculations tomorrow, it will involve calculus, specifically integration.

Edit3: Finding the richest is easy, you just plug in 1 for x. :)
 

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Discussion Starter · #6 ·
Sure! I opened it and played with it, so let's say you know that 20000 people make $40000 - that is your arbitrary point, meaning P_0 = 20000 and I_0 = $40000. If you're solving for income I (y), then P is your dependent variable (x), so you'd put in "40000*2^(ln(20000/x)/ln(3))".
Great! I nearly have to go now so I won't reply for a few hours, but thank you really much!

I've just plotted it and I do get a beautiful graph. I've plotted X from 0 to 50,000 and Y from 0 to 1,000,000. Is that the part of the graph I should look at or am I doing it wrong? :p
And 8000 people earn 800,000$, so 1/3 of 8000 (2667 or something) people earn 1,600,000$. If the formula was simpler I'd do it myself, but could you tell me how many people earn more than 1,000,000$? And could you post how you calculated that? Thank you very much! :D I'm forever grateful.

Edit: And I'd like to know what the highest possible income is, so when population= 1 :) If you could do that... :D
 

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Okay, so you want to find the number of people N between incomes a and b, you would integrate the second function between those points, like this:

P = P_0*1/3^(ln(I/I_0)/ln(2))

= P_0*1/3^(ln(I)/ln(2))/(1/3^(ln(I_0)/ln(2)))

= P_0*I^(ln(1/3)/ln(2))/(1/3^(ln(I_0)/ln(2)))

==> Integrate

N = P_0/((ln(1/3)/ln(2)+1)*1/3^(ln(I_0)/ln(2)))*(b^(ln(1/3)/ln(2)+1)-a^(ln(1/3)/ln(2)+1))

So you can just plug this into a calculator for any range (a to b) you're interested in, just put in the values and hit enter. You can actually copy/paste this in google search and replace the variables and it will calculate it.

Edit: This formula assumes a continuous distribution between a and b i.e. there is a group of people making every dollar between $400,000 and $500,000. This would be a massive overestimate, so I would use denominations of thousands instead, e.g. put in 400 instead of 400000. In any case this is more of an estimate than an exact value.
 

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Edit: Turn this into a useful post.

When I plug in 1 for x, I get 400000*2^(ln(24000)/ln(3)) = $232,094,896.

I have a feeling this rule breaks down the closer you get to either extreme.
 

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@PWarren:

Can you state what values you got for X when S=1 using your equation?
 

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Discussion Starter · #10 ·
Edit: Turn this into a useful post.

When I plug in 1 for x, I get 400000*2^(ln(24000)/ln(3)) = $232,094,896.

I have a feeling this rule breaks down the closer you get to either extreme.
Nope, it doesn't :)

This rule applies to the Roman elite, and we do indeed have examples of Roman senators having to count their possessions in hundreds of millions. In fact, Emperor Augustus possessed more than HS 1000 million...
 

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@PWarren:

Since we got the same answer despite using different methods, the figure should be correct. Figuring out the equations was fun.
 
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