c) The force of interest at time $$ t $$ is given by: $$ \delta(t)=\left\{\begin{array}{cl}0.01+0.01 t, & 0 \leq t

Question

c) The force of interest at time $$ t $$ is given by: $$ \delta(t)=\left\{\begin{array}{cl}0.01+0.01 t, & 0 \leq t<4 \ 0.15-0.003 t, & 4 \leq t<6 \ 0.06 & t \geq 6\end{array}\right. $$ 1. Find an expression for the value at time $$ t=0 $$ of a payment of 100 at time $$ t $$

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To find the present value of a payment of 100 at time $$ t $$ given the force of interest $$ \delta(t) $$, we need to use the relationship between the force of interest and the accumulated value. The present value $$ V(0) $$ of a payment $$ P $$ at time $$ t $$ can be expressed as:

$$
V(0) = P e^{-\int_0^t \delta(s) \, ds}
$$

In this case, $$ P = 100 $$. We will need to evaluate the integral $$ \int_0^t \delta(s) \, ds $$ for different intervals of $$ t $$ based on the piecewise definition of $$ \delta(t) $$.

### Step 1: Determine the intervals for $$ t $$

1. **For $$ 0 \leq t < 4 $$**:
   $$
   \delta(t) = 0.01 + 0.01 t
   $$

2. **For $$ 4 \leq t < 6 $$**:
   $$
   \delta(t) = 0.15 - 0.003 t
   $$

3. **For $$ t \geq 6 $$**:
   $$
   \delta(t) = 0.06
   $$

### Step 2: Calculate the integral for each interval

#### Case 1: $$ 0 \leq t < 4 $$

We need to calculate:
$$
\int_0^t (0.01 + 0.01 s) \, ds
$$

Calculating the integral:
$$
\int_0^t (0.01 + 0.01 s) \, ds = \int_0^t 0.01 \, ds + \int_0^t 0.01 s \, ds
$$

Calculating each part:
1. $$\int_0^t 0.01 \, ds = 0.01 t$$
2. $$\int_0^t 0.01 s \, ds = 0.01 \cdot \frac{s^2}{2} \bigg|_0^t = 0.01 \cdot \frac{t^2}{2} = 0.005 t^2$$

Combining these results:
$$
\int_0^t (0.01 + 0.01 s) \, ds = 0.01 t + 0.005 t^2
$$

Thus, the present value is:
$$
V(0) = 100 e^{-(0.01 t + 0.005 t^2)}
$$

#### Case 2: $$ 4 \leq t < 6 $$

We need to calculate:
$$
\int_0^4 (0.01 + 0.01 s) \, ds + \int_4^t (0.15 - 0.003 s) \, ds
$$

The first integral (from 0 to 4) is:
$$
\int_0^4 (0.01 + 0.01 s) \, ds = 0.01 \cdot 4 + 0.005 \cdot 4^2 = 0.04 + 0.08 = 0.12
$$

Now for the second integral (from 4 to $$ t $$):
$$
\int_4^t (0.15 - 0.003 s) \, ds = \left(0.15 s - 0.0015 s^2\right) \bigg|_4^t
$$

Calculating this:
$$
= \left(0.15 t - 0.0015 t^2\right) - \left(0.15 \cdot 4 - 0.0015 \cdot 16\right)
$$
$$
= 0.15 t - 0.0015 t^2 - (0.6 - 0.024) = 0.15 t - 0.0015 t^2 - 0.576
$$

Combining both parts:
$$
\int_0^t \delta(s) \, ds = 0.12 + 0.15 t - 0.0015 t^2 - 0.576 = 0.15 t - 0.0015 t^2 - 0.456
$$

Thus, the present value is:
$$
V(0) = 100 e^{-(0.15 t - 0.0015 t^2 - 0.456)}
$$

#### Case 3: $$ t \geq 6 $$

We need to calculate:
$$
\int_0^4 (0.01 + 0.01 s) \, ds + \int_4^6 (0.15 - 0.003 s) \, ds + \int_6^t 0.06 \, ds
$$

The first integral (from 0 to 4) is $$ 0.12 $$ as calculated before.

The second integral (from 4 to 6):
$$
\int_4^6 (0.15 - 0.003 s) \, ds = \left(0.15 s - 0.0015 s^2\right) \bigg|_4^6
$$
Calculating this:
$$
= \left(0.15 \cdot 6 - 0.0015 \cdot 36\right) - \left(0.15 \cdot 4 - 0.0015 \cdot 16\right)
$$
$$
= (0.9 - 0.054) - (0.6 - 0.024) = 0.846 - 0.576 = 0.27
$$

Now for the third integral (from 6 to $$ t $$):
$$
\int_6^t 0.06 \, ds = 0.06(t - 6)
$$

Combining all parts:
$$
\int_0^t \delta(s) \, ds = 0.12 + 0.27 + 0.06(t - 6) = 0.39 + 0.06t - 0.36 = 0.06t + 0.03
$$

Thus, the present value is:
$$
V(0) = 100 e^{-(0.06t + 0.03)}
$$

### Summary of Results

1. For $$ 0 \leq t < 4 $$:
   $$
   V(0) = 100 e^{-(0.01 t + 0.005 t^2)}
   $$

2. For $$ 4 \leq t < 6 $$:
   $$
   V(0) = 100 e^{-(0.15 t - 0.0015 t^2 - 
For $$ 0 \leq t < 4 $$: \[ V(0) = 100 e^{-(0.01 t + 0.005 t^2)} $$ For $$ 4 \leq t < 6 $$: $$ V(0) = 100 e^{-(0.15 t - 0.0015 t^2 - 0.456)} $$ For $$ t \geq 6 $$: $$ V(0) = 100 e^{-(0.06t + 0.03)} $$

c) The force of interest at time is given by:

Find an expression for the value at time of a payment of 100 at time

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Solución

Step 1: Determine the intervals for

Step 2: Calculate the integral for each interval

Case 1:

Case 2:

Case 3:

Summary of Results

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