Q&A

# "the people who move have a percentage of Democrats which is between these two values" — How does this furnish intuition for Simpson's Paradox?

+0
−0

Why does the phrase colored in red below ($\color{red}{\text{the people who move have a percentage of Democrats which is between these two values}}$) matter? How does it assist with intuiting the peaceful coexistence of $P_{new}(D|B) > P_{old}(D|B)$ and $P_{new}(D|B^C) > P_{old}(D|B^C)$?

1. The book Red State, Blue State, Rich State, Poor State by Andrew Gelman  discusses the following election phenomenon: within any U.S. state, a wealthy voter is more likely to vote for a Republican than a poor voter, yet the wealthier states tend to favor Democratic candidates!

(a) Assume for simplicity that there are only 2 states (called Red and Blue), each of which has 100 people, and that each person is either rich or poor, and either a Democrat or a Republican. Make up numbers consistent with the above, showing how this phenomenon is possible, by giving a 2 2 table for each state (listing how many people in each state are rich Democrats, etc.). So within each state, a rich voter is more likely to vote for a Republican than a poor voter, but the percentage of Democrats is higher in the state with the higher percentage of rich people than in the state with the lower percentage of rich people.

(b) In the setup of (a) (not necessarily with the numbers you made up there), let D be the event that a randomly chosen person is a Democrat (with all 200 people equally likely), and B be the event that the person lives in the Blue State. Suppose that 10 people move from the Blue State to the Red State. Write $P_{old}$ and $P_{new}$ for probabilities before and after they move. Assume that people do not change parties, so we have $P_{new}(D) = P_{old}(D)$. Is it possible that both $P_{new}(D|B) > P_{old}(D|B)$ and $P_{new}(D|B^C) > P_{old}(D|B^C)$ are true? If so, explain how it is possible and why it does not contradict the law of total probability $P(D) = P(D|B)P(B) + P(D|B^C)P(B^C)$; if not, show that it is impossible.

## Solution:

(b) Yes, it is possible. Suppose with the numbers from (a) that 10 people move from the Blue State to the Red State, of whom 5 are Democrats and 5 are Republicans. Then $P_{new}(D|B) = 75/90 > 80/100 = P_{old}(D|B)$ and $P_{new}(D|B^C) = 30/110 > 25/100 = P_{old}(D|B^C)$. Intuitively, this makes sense since the Blue State has a higher percentage of Democrats initially than the Red State, and $\color{red}{\text{the people who move have a percentage of Democrats which is between these two values}}$.

This result does not contradict the law of total probability since the weights $P(B), P(B^C)$ also change: $P_{new}(B) = 90/200$, while $P_{old}(B) = 1/2$. The phenomenon could not occur if an equal number of people also move from the Red State to the Blue State (so that P(B) is kept constant).

Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 59, p 96.

Why does this post require moderator attention?
Why should this post be closed? 