 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-40/2/b/solution

Use the positive variable

Y

from part (a). Since

0 \leq (S - B)^{+} \leq S

and

(S - B)^{+} \geq S - B

0 \leq c = E [Y (S - B)^{+}] \leq E [Y S] = s, c \geq E [Y (S - B)] = s - b .

(s - b)^{+} \leq c \leq s .

For the strict assertion, both

{S > B}

and

{S < B}

have positive probability. Positivity of

Y

gives

c = E [Y (S - B)^{+}] > 0, c - (s - b) = E [Y (B - S)^{+}] > 0.

s - b \geq 0

, the second inequality is the required strict bound; if

s - b < 0

, the first is. Hence

c > (s - b)^{+}

. The weighted expectations are finite because

P

and

Y

are bounded.

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