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 ga ...
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.
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