Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics III / 4 / a / ii

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- Solution ii

Solution

For an integrable random variable

X

, the law of total expectation for the partition of a sample space is

E [X] = i : P (B_{i}) > 0 \sum E [X ∣ B_{i}] P (B_{i}) .

Parts of zero probability contribute zero and need no definition of

E [X ∣ B_{i}]

. To see the identity, write

X = \sum_{i} X 1_{B_{i}}

and use linearity of expectation; integrability justifies the countable sum since

\sum_{i} E [∣ X ∣ 1_{B_{i}}] = E [∣ X ∣] < \infty

.

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