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Using the Tower Property With Nested Sigma-Algebras

Original question

I'm stuck on a conditional expectation exercise involving sigma-algebras. Can you explain how to use the tower property?
When G⊆H\mathcal G\subseteq\mathcal H and XX is integrable, first conditioning on H\mathcal H and then on G\mathcal G gives E[X∣G]E[X\mid\mathcal G]. Treat the first conditional expectation as a random variable and average it over the coarser information. Verify the sigma-algebras are nested before simplifying.

Why conditioning twice on nested information collapses

Let Y=E[X∣H]Y=E[X\mid\mathcal H] and Z=E[Y∣G]Z=E[Y\mid\mathcal G]. The variable ZZ is G\mathcal G-measurable. For any A∈GA\in\mathcal G, we also have A∈HA\in\mathcal H, so the defining property of conditional expectation gives

E[1AZ]=E[1AY]=E[1AX].E[\mathbf 1_A Z]=E[\mathbf 1_A Y]=E[\mathbf 1_A X].
Thus ZZ satisfies the definition of E[X∣G]E[X\mid\mathcal G] and agrees with it almost surely. In the reverse nesting order, E[E[X∣G]∣H]=E[X∣G]E[E[X\mid\mathcal G]\mid\mathcal H]=E[X\mid\mathcal G] because the inner variable is already measurable with respect to the larger sigma-algebra. In either order the result retains the coarser information. Taking G\mathcal G to be the trivial sigma-algebra gives the law of total expectation.

An option expires in one of four states. A signal reveals the regime, calm or turbulent. Full information distinguishes every state.

StateProbabilityRegimePayoff
10.15Calm0
20.45Calm4
30.30Turbulent2
40.10Turbulent12

Let H\mathcal H be full information, G\mathcal G the regime information, and XX the payoff. Compute the conditional expectation before the exact state is revealed, then average over the regimes.

First, E[X∣H]=XE[X\mid\mathcal H]=X. Conditional on calm, state 2 has probability 0.45/0.60=0.750.45/0.60=0.75, so the payoff expectation is 33. Conditional on turbulent, the expectation is (0.30×2+0.10×12)/0.40=4.5(0.30\times2+0.10\times12)/0.40=4.5. Thus E[X∣G]E[X\mid\mathcal G] is the random variable that equals 33 in calm states and 4.54.5 in turbulent states. Finally,

E[E[X∣G]]=0.60(3)+0.40(4.5)=3.60.E[E[X\mid\mathcal G]]=0.60(3)+0.40(4.5)=3.60.

Directly weighting the original payoffs by their four state probabilities gives the same 3.603.60. If you observe the calm regime, the relevant conditional answer is 33. If you know only the initial probabilities, the relevant unconditional answer is 3.603.60.

Within the calm regime, divide each calm state's original probability by the total calm probability before multiplying by its payoff. The weights must sum to one inside the regime. To recover the unconditional expectation, multiply the two regime means by their respective regime probabilities. Each regime probability cancels the normalisation within that regime, returning the original state weights. This is the tower identity as arithmetic. If the two information sets are not nested, that cancellation does not in general describe the successive conditional expectations.

References and next steps