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Exercise


ABy Admin
May 14'23

Answer

Key: A

[[math]] \begin{aligned} &\operatorname{E}[( S − 3)_+ ] = \operatorname{E}[ S ) − 3 + 3 f_S (0) + 2 f_S (1) + 1 f_ S (2) \\ &\operatorname{E}[ S ) = 2[0.6 + 2(0.4)] = 2.8 \\ &f_S (0) = e^{−2} , f_S (1) = e^{−2} (2)(0.6) = 1.2e^{−2} \\ &f_S(2) = e^{-2}(2)(0.4) + \frac{e^{-2}2^2}{2!}(0.6)^2 = 1.52e^{-2} \\ &\operatorname{E}[( S − 3)_+ ] = 2.8 − 3 + 3e^{−2} + 2(1.2e^{−2} ) + 1(1.52e^{−2} ) = −0.2 + 6.92e^{−2} = 0.7365 \end{aligned} [[/math]]

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