 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/8/a/ii/solution

X

E [X 1_{B}] = \sum_{x} x P ({X = x} \cap B) = P (B) \sum_{x} x P (X = x ∣ B) .

Dividing by the positive probability

P (B)

proves $E [X ∣ B] = E [X 1_{B}] / P (B)$ . On

B^{c}

the product vanishes, so no additional contribution is omitted.

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