 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2023/2/b/ii/solution

B

and

U

B

gives

M_{Z} (t) = (1 - p) + p M_{U} (t) = 1 - p + p \frac{sinh t}{t},

with the continuous value

M_{Z} (0) = 1

. Indeed

Z = 0

when

B = 0

, while the conditional probability distribution of

Z

given

B = 1

Under the literal assumptions alone, the law of total expectation gives only

M_{Z} (t) = 1 - p + p E [e^{t U} ∣ B = 1] .

This is not determined by

M_{U}

and

p

. For an explicit counterexample, put

p = 1/2

and

B = 1_{{U > 0}}

. Then

B

has the required Bernoulli distribution, but

M_{Z} (t) = \frac{1}{2} + \frac{1}{2} \int_{0}^{1} e^{t u} d u = \frac{1}{2} + \frac{e ^{t} - 1}{2 t} (t \neq = 0),

which differs from the independent-case formula.

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