Topic 11 Inverse Probability Weighting
Learning Goals
- Connect inverse probability weighting to the exchangeability computations we have done previously
Slides from today are available here.
Exercises
We’ll do Exercises 1 and 2 as a class. You’ll work on Exercise 3 and 4 in groups.
Exercise 1
Assume that we have conditional exchangeability given Z. For the outcome, let high = 1, low = 0.
n | Z | A | Y |
---|---|---|---|
30 | 1 | 1 | 90% high (27), 10% low (3) |
30 | 1 | 0 | 40% high (12), 60% low (18) |
30 | 0 | 1 | 70% high (21), 30% low (9) |
10 | 0 | 0 | 20% high (2), 80% low (8) |
Create columns Ya=1 and Ya=0, and fill them in using the conditional exchangeability assumption.
Within Ya=1 and Ya=0, total the number of highs and lows within Z=1 and within Z=0.
Verify that you would get the same totals by…
- Scaling the 30 with Z=1, A=1 up to 60 by giving the original 30 a weight of 2.
- Scaling the 30 with Z=0, A=1 up to 40 by giving the original 30 a weight of 4/3.
- Scaling the 30 with Z=1, A=0 up to 60 by giving the original 30 a weight of 2.
- Scaling the 10 with Z=0, A=0 up to 40 by giving the original 10 a weight of 4.
- How are all of these weights related to the fraction of those who receive their particular value of treatment among the Z subgroup (P(A∣Z))?
Write out the calculation for the ACE (P(Ya=1=high)−P(Ya=0=high)) in two ways:
- Directly using the total number of highs in the two columns
- As a weighted mean
We can view the data in columns Ya=1 and Ya=0 as a pseudopopulation in which all 100 units exist twice: once as their treated version and once as their untreated version. Verify that within Z=1 half of the “pseudounits” recive treatment and the other half don’t. Do the same for Z=0. What does this tell us about the relationship between Z and A in the weighted sample (the pseudopopulation)?
Exercise 2
Assume that we have conditional exchangeability given Z.
n | Z | A | E[Y∣A,Z] |
---|---|---|---|
80 | A | 1 | 40 |
20 | A | 0 | 20 |
20 | B | 1 | 30 |
80 | B | 0 | 20 |
Create columns E[Ya=1∣Z] and E[Ya=0∣Z], and fill them in using the conditional exchangeability assumption.
Within Ya=1 and Ya=0, find the total (sum) outcome within Z=1 and within Z=0.
Verify that you would get the same totals by…
- Scaling the 80 with Z=A, A=1 up to 100 by giving the original 80 a weight of 100/80.
- Scaling the 20 with Z=B, A=1 up to 100 by giving the original 30 a weight of 5.
- Scaling the 20 with Z=A, A=0 up to 100 by giving the original 30 a weight of 5.
- Scaling the 80 with Z=B, A=0 up to 100 by giving the original 10 a weight of 100/80.
- How are all of these weights related to the fraction of those who receive their particular value of treatment among the Z subgroup (P(A∣Z))?
Write out the calculation for the ACE (E[Ya=1]−E[Ya=0]) in two ways:
- Directly using the total in the two columns
- As a weighted mean
Verify that within Z=A half of the “pseudounits” recive treatment and the other half don’t. Do the same for Z=B. What does this tell us about the relationship between Z and A in the weighted sample (the pseudopopulation)?
Exercise 3
Assume that we have conditional exchangeability given Z.
n | Z | A | Y |
---|---|---|---|
10 | 1 | 1 | 60% high (6), 40% low (4) |
40 | 1 | 0 | 50% high (20), 50% low (20) |
10 | 0 | 1 | 50% high (5), 50% low (5) |
50 | 0 | 0 | 40% high (20), 60% low (30) |
Use the same process we went through in Exercises 1 and 2 to compute the ACE (P(Ya=1=high)−P(Ya=0=high)) in two ways:
- Directly using the total number of highs in the two columns (the “old” way)
- As a weighted mean (inverse probability weighting (IPW))
Exercise 4
Assume that we have conditional exchangeability given Z.
n | Z | A | E[Y∣A,Z] |
---|---|---|---|
80 | A | 1 | 30 |
20 | A | 0 | 20 |
40 | B | 1 | 60 |
60 | B | 0 | 10 |
Use the same process we went through in Exercises 1 and 2 to compute the ACE (E[Ya=1]−E[Ya=0]) in two ways:
- Directly using the total number of highs in the two columns (the “old” way)
- As a weighted mean (IPW)