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The foce of interest at time $t$ is given by $\delta_t=.01t$. $P$ is the present value of a 12 yr annuity due of $100$ payable annually. $Q$ is the present value of a 12 yr annuity immediate of $100$ payable annually. Calculate $P-Q$.

The solution to this problem states that we do not need to calculate $P$ or $Q$ and I honestly do not see that at all.

All I know is $e^{\int_0^t{\delta_s}ds}$ is the effective accumulation. Can I get some help?

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The difference between an annuity-due and annuity-immediate is that the former pays at the beginning of the year and the latter pays at the end. So the difference in the two cash flows is simply the difference in present values of the first payment of the annuity-due and the last payment of the annuity-immediate. This is exactly analogous to a telescoping sum: all the intermediate cash flows cancel out exactly. More concretely, consider the following table of payments: $$\begin{array}{c|c|c|c|c|c} t & 0 & 1 & 2 & \ldots & 11 & 12 \\ \hline P & p_0 & p_1 & p_2 & \ldots & p_{11} & \\ Q & & q_1 & q_2 & \ldots & q_{11} & q_{12} \end{array}$$ but note that because the force of interest is the same and the level payments are also the same ($100$ for each annuity), it is clear that no matter what the actual values of $p_i$ and $q_j$ are, that $p_i = q_i$ for each $i = 1, 2, \ldots, 11$. So $P - Q = p_0 - q_{12}$.

So, the only real computation you need to do is calculate the present value of the final payment at $t = 12$. This is given by the present value discount factor $$ \exp \left( - \int_{s=0}^{12} \delta(s) \, ds \right) = e^{-18/25} \approx 0.486752,$$ hence $$P - Q = 100(1 - e^{-18/25}) \approx 100(1 - 0.486752) = 51.3248.$$

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  • $\begingroup$ The table made it very clear to understand. Thank you very much. $\endgroup$
    – hyg17
    Feb 13, 2015 at 5:59

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