Curtate Expectation of life aged x
So I did @k=0, P(60)= 0.9
@k=1, P(61)= 0.9*[0.9*(1-(0.1*1)) = 0.729
@k=2, P(62)= 0.90.729 [0.9*(1-(0.1*2)) = 0.47239
Not sure if that's right, or where to go from there. Can anyone help with a detailed working for this question?
I assume $e_x$ denotes the complete expectation of life at age $x$. And $p_x $ denotes the probability that the person of age $x$ will survive til the next year.
$$\begin{array} {r c l } e_{60} &=& \displaystyle \sum_{k\geqslant1} \phantom0_k p_{60} \\ &=& \displaystyle p_{60} + \phantom0_2 p_{60} + \phantom0_3 p_{60} + \phantom0_3 p_{60} \cdot e_{63} \\ &=& 0.9 + 0.9 \cdot 0.81 + 0.9 \cdot 0.81 \cdot 0.72 (1 + 13.5) \\ &=& \boxed{9.23976 } \end{array}$$