Finding the present value of a continuously varying perpertuity

actuarial-sciencecalculusfinanceintegrationsolution-verification

A perpetuity provides for continuous payments. The annual rate of payment at time $t$ is $1$ if $0\le t<10$ and $(1.03)^{t-10}$ if $t\ge 10$. Using an annual effective interest rate of $6\%$, the present value at time $t = 0$ of this perpetuity is $x$. Calculate $x$.

I know that the present value of an $n$-period continuously increasing annuity with a constant force of interest is given by

$$\int_0^n t v^t \; dt = \frac{\overline{a}_{n\rceil i}-nv^n}{\delta}$$

Using this formula, I figured that the PV of the perpetuity in question is given by

$$\int_0^{10} 1 \cdot (1.06)^{-t} \; dt + \lim_{b \to\infty}\int_{10}^b (1.03)^{t-10} \cdot 1.06^{-t} \; dt, $$

but this expression is not correct. The correct solution is $27.03$. Where have I gone wrong?

Best Answer

Your expression $$\int_{t=0}^{10} (1.06)^{-t} \, dt + \int_{t=10}^\infty (1.03)^{t-10} (1.06)^{-t} \, dt$$ is correct as written. If you did not get $27.028160912907478246\ldots$ then you have evaluated it incorrectly. Note that this can be written $$\int_{t=0}^{10} e^{-t \log 1.06} \, dt + (1.03)^{-10} \int_{t=10}^\infty e^{-t \log (1.06/1.03)} \, dt \\ = \frac{1}{\log 1.06} \int_{u=0}^k e^{-u} \, du + \frac{1}{(1.03)^{10} \log \frac{1.06}{1.03}} \int_{u=m}^\infty e^{-u} \, du,$$ where $k = 10 \log 1.06$ and $m = 10 \log \frac{1.06}{1.03}$.

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