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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?

user22222969
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jul 13 '23 at 16:07
  • Whaz is the meaning of $e_x$ and $ P_{x+k}$? – callculus42 Jul 13 '23 at 16:25

1 Answers1

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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.

$$\begin{array} {r c l } e_{60} &=& \displaystyle \sum_{k\geqslant1} \phantom0_k p_{60} \\ &=& \displaystyle p_{60} + \phantom0_2 p_{60} + \phantom0_3 p_{60} + \phantom0_3 p_{60} \cdot e_{63} \\ &=& 0.9 + 0.9 \cdot 0.81 + 0.9 \cdot 0.81 \cdot 0.72 (1 + 13.5) \\ &=& \boxed{9.23976 } \end{array}$$

GohP.iHan
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