1
$\begingroup$

Income percentile ranks are numbers from $0$ to $100$ that indicate where you sit in the income distribution — for example, $75$ would mean you earn more income than three quarters of people, but less than the top quarter. Let's assume a child's expected income percentile is a linear function of their parents' income percentile. It turns out the intercept is enough to pin down the slope. Do you see why? If the intercept is $20$ — which means the lowest-income parents have children who end up at the $20^{th}$ percentile, on average — what is the slope?

This problem is from an app called probability puzzles.

My approach: I realised $y=ax+b$ could be the line for child but we don't have any information about the slope. We know that percentile is uniformly distributed between $0$ and $100$ so expected value is $50$ but I don't see how to proceed from here. Help!

$\endgroup$
1
  • 1
    $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here $\endgroup$ Jul 24, 2022 at 17:08

3 Answers 3

0
$\begingroup$

I agree with your linear function: $y=ax+b$, where $x$ is the ith parents' income percentile. And $y$ is the child's expected income percentile. Then the crucial information are

If the intercept is 20 — which means the lowest-income parents have children who end up at the 20th percentile, on average.

This give us the equation $y(0)=20$. A child cannot end up below the 20th percentile: $y(0)=a\cdot 0+b=b=20$.

Then the highest income parents percentile ($x=100$) can have children who end up at the highest income percentile ($y=100$), on average. It cannot be more, since the codomain of the percentile is between $0$ and $100$. Thus the equation is

$$100=a\cdot 100+20$$

Finally, calculate the value of $a$.

$\endgroup$
4
  • $\begingroup$ This answer doesn't match with the given answer. I have accepted it though since I find the explanation plausible but maybe could you try thinking from a different perspective? I am too trying the same. $\endgroup$
    – Charlie
    Jul 24, 2022 at 18:31
  • $\begingroup$ @Charlie What is the given answer? $\endgroup$ Jul 24, 2022 at 18:50
  • $\begingroup$ I don't know. The app shows correct if it is correct else it just shows incorrect! $\endgroup$
    – Charlie
    Jul 27, 2022 at 9:09
  • $\begingroup$ @Charlie Mmh, then we cannot use it as a hint. $\endgroup$ Jul 27, 2022 at 17:27
0
$\begingroup$

Hint: we know that the expected value of percentile is $50$, so $E(y)=E(x)=50$. What does that tell you about $a$?

$\endgroup$
3
  • $\begingroup$ Maybe just 0, you mean? $\endgroup$
    – Charlie
    Jul 27, 2022 at 9:10
  • $\begingroup$ @Charlie Not 0. By linearity of expectation, E(y)=aE(x)+b. Since we know b=20, we should be able to find what a is. $\endgroup$
    – st. ivan
    Jul 28, 2022 at 5:18
  • $\begingroup$ yep got it, thanks $\endgroup$
    – Charlie
    Jul 28, 2022 at 7:11
0
$\begingroup$

This question is ambiguous, and I believe for more than one reason. One source of ambiguity is the fact that income percentiles change when the population changes. I believe the only productive approach is to understand that a child of the lowest-income parents have an income in the 20th percentile relative to the new population consisting of parents and children (i.e. not relative to the old population consisting of only parents). If it were relative to the old population only, then we wouldn’t have enough information to solve the problem.

Another ambiguity is the question of whether the number of children is independent of income percentile. If wealthy parents have fewer children, then the answer changes. Let’s assume the number of children is independent of income.

This allows us to solve the problem, since the constraint that income percentiles are uniformly distributed by definition implies that half the (new) population must belong to the bottom 50th percentile, and half must belong to the top. The only way this can be is if the slope is 3/5, so the 0th percentile maps to the 20th and the 100th to the 80th.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .