Chapter 2 Chapter 2

2.1 Simple Interest

Interest $r$,

$$V(1) = (1+r)V(0)$$

$$V(2) = (1+2r)V(0)$$

$$V(n) = (1+nr)V(0)$$

$$V(m) = (1+m\frac{r}{365})V(0)$$

Assume $m$ is $km = 1r$ For instance, $365\times day = 1 year$

$$V(m) = (1+m\frac{r}{k})V(0)$$

2.2 Exercise 2.1

A sum of $9, 000 paid into a bank account for two months (61 days) to attract simple interest will produce $9, 020 at the and of the term. Find the interest rate r and the return on this investment.

Recall the formula that after $n$ days:

$$V(\frac{n}{365}) = V(0)\times(1+\frac{n}{365}r)$$

$$9020=9000×(1+\frac{61}{365}×r) \rightarrow r=0.0133\\ K(0,61)=\frac{9020−9000}{9000}=0.0022$$

2.3 Exercise 2.4

Find the principal to be deposited initially in an account attracting simple interest at a rate of 8% if $1, 000 is needed after three months (91days).

$$1000=P×(1+\frac{91}{365}×0.08) \rightarrow P=980.45$$

Or we can simply use the formula:

$$V (0) = V (t)(1 + rt)^{−1}.$$

This number is called the present or discounted value of $V (t)$ and $(1+rt)^{−1}$ is the discount factor.

2.4 Periodic Compounding

$$V(1) = (1+r)V(0)\\ V(2) = (1+r)(1+r)V(0) = (1+r)^2V(0) \\ V(n) = (1+r)V(n-1) = (1+r)^nV(0)$$

$$V(1 d) = (1+\frac{r}{365})V(0) \\ V(2d) = (1+\frac{r}{365})V(1d) = (1+\frac{r}{365})^2V(0) \\ V(md) = (1+\frac{r}{365})^mV(0)$$

Assume compound $m$ times in 1 years, after $t$ years, interest $r$

$$V(t) = (1+\frac{r}{m})^{tm}V(0)$$

$(1+\frac{r}{m})^{tm}$ is growth rate.

2.5 Proposition

The future value V (t) increases if any one of the parameters $m$, $t$, $r$ or $P$ (or $V(0)$) increases, the others remaining unchanged.

2.5.1 Proof

$$V(1)_1 = (1+r)V(0)\\ V(1)_{365} = (1+\frac{r}{365})^{365}V(0)$$

$$V(1)_m = (1+\frac{r}{m})^{tm}V(0) \\ V(1)_k = (1+\frac{r}{k})^{tk}V(0)$$

Recall that the binomal formula:

$$(1+x)^n = 1+ nx +\frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3+\dots$$

$$\begin{align} (1+\frac{r}{m})^{m} &= 1 + r + \frac{m(m-1)}{2!}(\frac{r}{m})^2 + \frac{m(m-1)(m-2)}{3!}(\frac{r}{m})^3+\dots\\ &= 1 + r + \frac{1(1-\frac{1}{m})}{2!}(r)^2 + \frac{1(1-\frac{1}{m})(1-\frac{2}{m})}{3!}(r)^3+\dots + \frac{1(1-\frac{1}{m})\dots(1-\frac{m-1}{m})}{3!}(r)^m \\ &< 1 + r + \frac{1(1-\frac{1}{m})}{2!}(r)^2 + \frac{1(1-\frac{1}{m})(1-\frac{2}{m})}{3!}(r)^3+\dots + \frac{1(1-\frac{1}{m})\dots(1-\frac{m-1}{m})}{3!}(r)^k \\ &< 1 + r + \frac{1(1-\frac{1}{m})}{2!}(r)^2 + \frac{1(1-\frac{1}{k})(1-\frac{2}{k})}{3!}(r)^3+\dots + \frac{1(1-\frac{1}{k})\dots(1-\frac{m-1}{k})}{3!}(r)^k \\ &< 1 + r + \frac{1(1-\frac{1}{m})}{2!}(r)^2 + \frac{1(1-\frac{1}{k})(1-\frac{2}{k})}{3!}(r)^3+\dots + \frac{1(1-\frac{1}{k})\dots(1-\frac{k-1}{k})}{3!}(r)^k \\ &= (1+\frac{r}{k})^{k} \end{align} \\ $$

Proof done.

2.6 Exercise 2.8

Which will deliver a higher future value after one year, a deposit of $1, 000 attracting interest at 15% compounded daily, or at 15.5% compounded semi-annually?

# Principle
V_0 = 1000
# First one
V_a = V_0*(1+0.15/365)^365
V_a

## [1] 1161.798

# Second one
V_b = V_0*(1+0.155/2)^2
V_b

## [1] 1161.006

2.7 Exercise 2.9

What initial investment subject to annual compounding at 12% is needed to produce $1, 000 after two years?

# 1000 = (1+0.12/1)^2*P
P = 1000*(1+0.12/1)^{-2}
P

## [1] 797.1939

2.8 Annuity

Annuities are insurance contracts that promise to pay you regular income immediately or in the future

Assume that dealer gives $C$ , you need $V(0)$

$$V(0) = V(0)_1 + V(0)_2 + \dots + V(0)_n$$ $$V(0)_1 = \frac{C}{(1+r)}\\ C = V(1) = (1+r)V(0)_1\\ C = (1+r)^2V(0)_2\\ C = (1+r)^nV(0)_n$$

$$V(0) = 10, V(0)_1 = 4, V(0)_2 = 3, V(0)_3 = 2, V(0)_4 = 1 \\ V(0) = \frac{C}{(1+r)} + \frac{C}{(1+r)^2}+ \frac{C}{(1+r)^3}+ \frac{C}{(1+r)^4} \\ V(0) = PA(r,n)\times C\\ PA(r,4) = \frac{1}{(1+r)} + \frac{1}{(1+r)^2}+ \frac{1}{(1+r)^3}+ \frac{1}{(1+r)^4}$$

$$1+ q + q^2 + \dots + q^n = x\\ xq = q + q^2 + \dots + q^{n+1} \\ (q-1)x = q^{n+1}-1 \longrightarrow x = \frac{q^{n+1}-1}{q-1}\\ 1+ q + q^2 + \dots + q^n =\frac{q^{n+1}-1}{q-1}$$

$$PA(r,n) = \sum_{i=1}^n\frac{1}{(1+r)^i} = \frac{1-(1+r)^{-n}}{r}$$

2.9 Exercise 2.17

Find a formula for the present value of an infinite stream of payments of the form $C, C(1+g), C(1+g)2, . . . $, growing at a constant rate $g$. By the tail-cutting procedure find a formula for the present value of $n$ such payments.

$$V(t) = \frac{C}{(1+r)} + \frac{C(1+g)}{(1+r)^2}+\frac{C(1+g)^2}{(1+r)^3}+ \dots= \frac{C}{r-g}$$

$$\begin{align} V(t) &= \frac{C}{(1+r)} + \frac{C(1+g)}{(1+r)^2}+\dots+\frac{C(1+g)^{n-1}}{(1+r)^n}+\frac{C(1+g)^{n}}{(1+r)^{n+1}}+\frac{C(1+g)^{n+1}}{(1+r)^{n+2}}+ \dots \\ &= V(t)_{1-n} + V(t)_{n+1-\infty} \end{align}\\ V(t)\frac{(1+g)^n}{(1+r)^n}=\frac{C(1+g)^{n}}{(1+r)^{n+1}}+\frac{C(1+g)^{n+1}}{(1+r)^{n+2}}+ \dots \\ \longrightarrow V(t)_{(n+1)-\infty} = V(t)\times \frac{(1+g)^n}{(1+r)^n}\\ $$

$$V(t)_{1-n} = V(t) - V(t)_{(n+1)-\infty} = \frac{C}{r-g} - \frac{C}{r-g}\times\frac{(1+g)^n}{(1+r)^n} = C\times \frac{1-(\frac{1+g}{1+r})^n}{r-g}$$