my notes doesn't explain these concepts well and I am very stuck at this example solution. I don't get where the numbers come from at all.

Suppose that the forced rate of interest is $\delta(t)$ and is given by he following:

$$\delta(t)=\begin{cases} 0.09 & \text{ if } 0 \leq t < 5 \\ 0.08 & \text{ if } 5 \leq t < 10 \\ 0.07 & \text{ if } 10 \leq t \end{cases} $$ Find simple expressions for $v(t)$ when $t \geq 0$. $v(t)$ is defined as $v(t) = e^{-\int_0^t \delta(s)ds}$ which is the discounted present value at time $t=0$ of one unit of money at time $t$.

The solution(which I don't understand)

According to the definition of $v(t)$,

$$v(t)=\begin{cases} e^{-0.09t} & \text{ if } 0 \leq t < 5 \\ e^{-0.05-0.08t} & \text{ if } 5 \leq t < 10 \\ e^{-0.15-0.07t} & \text{ if } 10 \leq t \end{cases}$$

QUESTION: From where do the $-0.05$ and $-0.15$ come from in the above exponents? The bit $-0.15-0.07t$ say, I undrstand getting the term $-0.07t$ from integrating $\int_0^t0.07sds$ for $t$ greater than $10$ but why $-0.15$? What is this number? So this means we're NOT integrating from $0$ to $t$, but some $x$ to $t$? I "reverse engineered" setting $x$ as some unknown to equate it to the above and solve for $x$ but got some weird number.

I just don't understand this. Can someone please explain to me? The resuorces that match my issue here is very scarce on the internet...thank you very much


The force of interest $\delta(t)$ of an amount function $a(t)$ is defined by $$ \delta(t)=\frac{a'(t)}{a(t)}\tag 1 $$ The force of interest is the fraction of the instantaneous rate of change of the accumulation function $a'(t)$ and the accumulation function $a(t)$.

The (1) can be written as $$ \delta(t)=\frac{\mathrm d \log a(t)}{\mathrm d t} $$ from which integrating we find $$ \int_0^t \delta(s)\mathrm d s=\int_0^t \mathrm d \log a(s)=\log a(t)-\underbrace{\log a(0)}_{\log 1=0}=\log a(t) $$ so that $$a(t)=\mathrm e^{\int_0^t \delta(s)\mathrm d s}.$$

For a general accumulation function $a(t)$, the discount factor for payments at time $t$ is $$ v(t)=\frac{1}{a(t)}=\mathrm e^{- \int_0^t \delta(s)\mathrm d s}. $$ So integrating your force of interest you find

  • for $0 \leq t < 5$, $$v_1(t)=\mathrm e^{- \int_0^t 0.09\mathrm d s}=\mathrm e^{- 0.09\int_0^t \mathrm d s}=\mathrm e^{- 0.09 (t-0)}=\mathrm e^{- 0.09 t}$$
  • for $5 \leq t < 10$, $$ \begin{align*} v_2(t) &=v_1(5)\mathrm e^{- \int_5^t 0.08\mathrm d s}=v_1(5)\mathrm{e}^{- 0.08\int_5^t \mathrm d s}=v_1(5)\mathrm e^{- 0.08 (t-5)}\\ &=\mathrm e^{-0.45}\,\mathrm e^{0.4-0.08t}=\mathrm e^{-0.05-0.08t}\\ \end{align*} $$
  • for $t \ge 10$, $$ \begin{align*} v_3(t) &=v_2(10)\mathrm e^{- \int_{10}^t 0.07\mathrm d s}=v_2(10)\mathrm{e}^{- 0.07\int_{10}^t \mathrm d s}=v_2(10)\mathrm e^{- 0.07 (t-10)}\\ &=\mathrm e^{-0.85}\,\mathrm e^{0.7-0.07t}=\mathrm e^{-0.15-0.07t} \end{align*} $$ And finally $$v(t)=\begin{cases} \mathrm e^{-0.09t} & \text{ if } 0 \leq t < 5 \\ \mathrm e^{-0.05-0.08t} & \text{ if } 5 \leq t < 10 \\ \mathrm e^{-0.15-0.07t} & \text{ if } 10 \leq t \end{cases}$$

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