Exercise
Answer
Answer: A
[math]E\left[T_{0}\right]=\int_{0}^{8}{ }_{t} p_{0} d t=\int_{0}^{8}\left(1-\frac{t^{2}+t}{72}\right) d t \rightarrow \frac{1}{72}\left[72 t-\frac{t^{3}}{3}-\frac{t^{2}}{2}\right]_{0}^{8}=5.1852[/math]
[math]E\left[T_{0}{ }^{2}\right]=2 \int_{0}^{8}\left({ }_{t} p_{0} \times t\right) d t=\frac{2}{72} \int_{0}^{8}\left(72 t-t^{3}-t^{2}\right) d t=\frac{2}{72}\left[36 t^{2}-\frac{t^{4}}{4}-\frac{t^{3}}{3}\right]_{0}^{8}=30.815[/math]
[math]\operatorname{Var}\left[T_{0}\right]=E\left[T_{0}^{2}\right]-\left(E\left[T_{0}\right]\right)^{2}=3.9287[/math]
Copyright 2024 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.