p(j)和p'(j)的区别和联系？（详见图片）

在多元风险模型中遇到的问题：


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Lets say they give you a double decrement table with q(i) (no prime). This means that the rate of mortality given for decrement i takes into account the possibility of failure due to other decrements. For example:

qx(1) = 0.10
qx(2) = 0.15
qx(3) = 0.20

Then the probability of surviving to all decrements is simply
px(tau) = 1 - 0.10 - 0.15 - 0.20 = 0.55.

On the other hand, say they gave you q'(i) (with a prime). This means that the rates given were obtained by supposing that decrement i was to only way to ''fail'' within the year. Therefore, to calculate the probability of survival, you have to survive to each decrement independently. For example:

qx'(1) = 0.12
qx'(2) = 0.17
qx'(3) = 0.22

Then px(tau) = (1-0.12)*(1-0.17)*(1-0.22) = 0.57

Notice that in the above examples, qx(i) < qx'(i). This is true in general.

随手转一贴过来~~~~

公式不一样。

会在特定已知条件下相互转换就行，记得是有一个公式的。

你可以把它想成電池串聯跟並聯的關係

應該不難

q(i)代表考虑所有因素的情况下，因为i 死亡的概率
q`(i)代表只考虑因素i 的情况下，死亡的概率

q(i)和q'(i) 你可以这样考虑   q(i)是{q(1),q(2),...,q(n)}->death    而q'(i)是  {q'(1)->death}*{q'(2)->death}*...*{q'(n)->death}           所以在q'(i)假设下  一个人要想生存 则需要p'(1)*p'(2)*...p'(n) 即在各个中括号下都同时生存    而在q(i)假设下 一个人要想生存 则需要这个中括号内的所有因素都不发生 所以是1-q(1)-q(2)-...-q(n)

sang747 发表于 2014-8-29 21:35
q(i)和q'(i) 你可以这样考虑   q(i)是{q(1),q(2),...,q(n)}->death    而q'(i)是  {q'(1)->death}*{q'(2)-> ...

嗯嗯，比较明白了，但是你这句我有点不懂^_^!  能不能解释一下？
“ q(i)是{q(1),q(2),...,q(n)}->death    而q'(i)是  {q'(1)->death}*{q'(2)->death}*...*{q'(n)->death}  ”

sang747 发表于 2014-8-29 21:35
q(i)和q'(i) 你可以这样考虑   q(i)是{q(1),q(2),...,q(n)}->death    而q'(i)是  {q'(1)->death}*{q'(2)-> ...

忘记写了，谢谢你的解答

tasty2012 发表于 2014-8-26 05:17
Lets say they give you a double decrement table with q(i) (no prime). This means that the rate of m ...

wow，how nuibi it is！Thx!
请问哪里贴来的？