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Differential Proportions

In a large population, the proportion with income between x and x+dx is f(x) dx. Express the mean (average) income,

\mu

,as an integral, assuming that any positive income is possible.
Let p=F(x) be the proportion of the population with income less than x, and G(x) be the mean (average) income earned by people with income less than x. Further, let θ(p) be the proportion of the total income which is earned by people with income less than x as a function of the proportion p of the population which has income less than x. Express F(x) and G(x) as integrals and thence derive an expression for θ(p).

I have so far concluded that:

\mu= \displaystyle\int^\infty_0 x f(x)\ dx

p= F(x)= \displaystyle\int^x_0 f(t)\ dt

G(x) = \displaystyle\int^x_0 tf(t)\ dt

However, I am struggling to come up with an expression for θ(p). Any help?

(edited 9 years ago)

Reply 1

Phichi

Original post by newblood

(..)

Before answering any questions, I disagree with

\mu = \displaystyle \int_0^{\infty} f(x) \, dx

Think about what you're calculating with that.

f(x)dx

represents the proportion of the population with an income between

x

and

x + dx

. So what would integrating this from

0 \Rightarrow \infty

, give you?

(edited 9 years ago)

Reply 2

newblood

Original post by Phichi

Before answering any questions, I disagree with

\mu = \displaystyle \int_0^{\infty} f(x) \, dx

Think about what you're calculating with that.

Ah sorry, I seem to have mislaid an x while latexing it up. Let me edit the OP. Is that correct now?

Reply 3

Phichi

Original post by newblood

Ah sorry, I seem to have mislaid an x while latexing it up. Let me edit the OP. Is that correct now?

Perfect. What are your thoughts for

\theta (p)

, using the information from the question, so we can see what exactly you're stuck on. Which bit has got you stumped?

(edited 9 years ago)

Reply 4

newblood

Original post by Phichi

Perfect. What are your thoughts for

\theta (p)

, using the information from the question, so we can see what exactly you're stuck on. Which bit has got you stumped?

To be honest Im finding it quite hard to interpret what

\theta (p)

represents from the slightly long definition given, which in turn is why I am struggling to formulate it.

Reply 5

Phichi

Original post by newblood

To be honest Im finding it quite hard to interpret what

\theta (p)

represents from the slightly long definition given, which in turn is why I am struggling to formulate it.

I have to agree the wording is some what ambiguous.

So, 'Further, let θ(p) be the proportion of the total income which is earned by people with income less than x as a function of the proportion p of the population which has income less than x'.

I think you're simply finding the proportion of the total income due to the people with income less than x (p).

So lets say, 100 people have income less than x, and the sum of all the incomes for these 100 people is £10.

The total income of all N of the population is say, £30.

This would lead the proportion of the income of earners below x, to the total income of the entire population to be

\dfrac{1}{3}

Could you go from there?

The real problem solving is really just how to represent the total income of the people of income less than x.

(edited 9 years ago)

Reply 6

newblood

Original post by Phichi

\dfrac{1}{3}

Could you go from there?

But how can we work out the total income of the population using our functions?

Reply 7

Phichi

Original post by newblood

But how can we work out the total income of the population using our functions?

Well, if you had a total income of £1000, spread amongst 50 people, the average is

\dfrac{1000}{50} = £20

So if you were given the fact that a population had N people, and an average income of say, A, then the total is just NA.

p is the proportion of people with income less than x, if given a population of N people, this equates to NP people with income less than x.

(edited 9 years ago)

Reply 8

newblood

Original post by Phichi

Well, if you had a total income of £1000, spread amongst 50 people, the average is

\dfrac{1000}{50} = £20

But were not told how many people are in the population. Atleast I cant see how to work it out

Reply 9

Phichi

Original post by newblood

But were not told how many people are in the population. Atleast I cant see how to work it out

Assume the population is just N.

For the first question, finding an expression for

\mu

I'd go about it like so:

Let the total income of the population be represented by

T

T = \displaystyle N\int_0^{\infty} xf(x) \, dx

This gives you the total income.

Nf(x)dx

is the number of people with income lying between x and x+dx.

Thus

xNf(x)dx

gives you the sum of their incomes. Integrating from 0 to infinity gives you the total income.

The average is just the total divided by the number of people, N.