Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 221 / 4 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

By consistency of potential outcomes and conditional exchangeability,

E [Y (z) ∣ X] = E [Y (z) ∣ Z = z, X] = E [Y ∣ Z = z, X] .

Taking conditional expectations given

X_{1} = x_{1}

and subtracting the cases

z = 1

and

z = 0

proves the outcome-regression identification formula.

ii

Solution

Conditional on

X

,

E [\frac{Z Y}{e ( X )} X] = E [Y ∣ Z = 1, X], E [\frac{( 1 - Z ) Y}{1 - e ( X )} X] = E [Y ∣ Z = 0, X] .

Multiplying by

1 {X_{1} = x_{1}}

, taking expectations, dividing by

P (X_{1} = x_{1})

, and using part (i) proves the inverse probability weighting formula.

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