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Let $p(x)_{t}$ be the probability that a person aged $x$ survives until at least age $x+t$. Suppose we are given the following:

  • $p(x)_{1} = 0.99$
  • $p(x+1)_{1} = 0.985$
  • $p(x+1)_{3} = 0.95$
  • $q(x+3)_1 = 0.02$

Note that $q(x)_{t} = 1-p(x)_{t}$. What is $p(x+1)_{2}$?

So we want to find the probability that a person aged $x+1$ survives until at least age $x+3$. So $$p(x+1)_{2} = p(x+1)_{1} \cdot p(x+2)_{1}$$

But we don't know those values. That is, we don't know $ p(x+2)_{1}$.

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1 Answer 1

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$p(x+1)_3 = p(x+1)_1 p(x+2)_1 p(x+3)_1$, so $p(x+1)_1 p(x+2)_1 = \frac{p(x+1)_3}{p(x+3)_1} = \frac{0.95}{0.98} \approx 0.97$.

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