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.

We’ll do Exercises 1 and 2 as a class. You’ll work on Exercise 3 and 4 in groups.

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 \(Y^{a=1}\) and \(Y^{a=0}\), and fill them in using the conditional exchangeability assumption.
Within \(Y^{a=1}\) and \(Y^{a=0}\), 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 (\(P(A\mid Z)\))?
Write out the calculation for the ACE (\(P(Y^{a=1}=\text{high}) - P(Y^{a=0}=\text{high})\)) 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 \(Y^{a=1}\) and \(Y^{a=0}\) 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)?

Assume that we have conditional exchangeability given \(Z\).

\(n\)	\(Z\)	\(A\)	\(E[Y\mid A, Z]\)
80	A	1	40
20	A	0	20
20	B	1	30
80	B	0	20

Create columns \(E[Y^{a=1} \mid Z]\) and \(E[Y^{a=0} \mid Z]\), and fill them in using the conditional exchangeability assumption.
Within \(Y^{a=1}\) and \(Y^{a=0}\), 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 (\(P(A\mid Z)\))?
Write out the calculation for the ACE (\(E[Y^{a=1}] - E[Y^{a=0}]\)) 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)?

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 (\(P(Y^{a=1}=\text{high}) - P(Y^{a=0}=\text{high})\)) in two ways:

Assume that we have conditional exchangeability given \(Z\).

\(n\)	\(Z\)	\(A\)	\(E[Y\mid A, Z]\)
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 (\(E[Y^{a=1}] - E[Y^{a=0}]\)) in two ways: