Exercise


May 07'23

Answer

Solution: C

[[math]] \begin{align*} E[100(0.5)^X] = 100 E[(0.5)^X] = 100E[e^{(\ln(0.5)}X] &= 100M_X(\ln(0.5))\\ &= 100 \frac{1}{1-2\ln(0.5)}\\ &= 41.9. \end{align*} [[/math]]

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