 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/7/a/ii/solution

X

(B_{i})

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

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

and using

E [X 1_{B_{i}}] = P (B_{i}) E [X ∣ B_{i}]

. Integrability justifies the countable sum. The same equality is valid with nonnegative, possibly infinite expectations when

X \geq 0

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