Combinatorics - Probability

get rich

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compounded interest

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1. Get rich!

Play with P, r and n to get rich!

When we deposit money into a savings account, we lend money to the bank. This bank pays interest on our deposit. If we deposit an amount P at an interest rate r per period (year for example), after n periods (years), how much our investment will be worth ?

We deposit P now . At the end of the:
-- First period, we will have:
P + rP = P(1 + r)
-- Second period, we will have:
(P + rP) + r(P + Pr) = (P + rP)(1 + r) = P (1 + r)²
-- Third period, we will have:
P (1 + r)² + rP (1 + r)² = P (1 + r)³
....
-- Last (nth) period, we will have:
P (1 + r)ⁿ

After n periods (years), our investment will be worth
W = P (1 + r)ⁿ

Examples:

1. How much do we have to invest at the interest rate of 10% yearly to become millionaire in 10 years?
W = P (1 + r)^{- n} = 10⁶ (1 + 10%)^{- 10} =
10⁶ (1.10)^{- 10} = 385543.29 $

2. How many periods (frequencies to compound the interest) do we need to get the double of an investment at an interest rate of r? We have:
W = 2P = P (1 + 1r)ⁿ or (1 + 10%)ⁿ = 2
n ln(1 + r) = ln 2
n = ln 2 /ln (1 + r)
The Maclaurin series of ln(1 + r) is:
ln (1 + r) = r - r²/2 + r³/3 - r⁴/4 + ...
"r" is small, the terms, after the first, of the Maclaurin expansion may be neglected, then ln (1 + r) = r , therefore:
n = ln 2 /r = 0.7/r
if r = 10%, then n = 7 periods.

2. What's about a continuously compounded interest?

If the interest is compounded n times a period p (for example a year) at an interest rate of r, after t periods, an investment P grows to : G (t) = P(1 + r/n)^nt.

If G(t) is the value of the investment at t periods, the earned interest will be G(t) x r δt. After a short time of periods, δt, G(t) becomes:
G(t + δt) = G(t) + G(t) x r δt

This formula can be written as:
G(t + δt) - G(t) = G(t) x r δt, or
δ G(t) = G(t) x r δt, or
δ G(t) / G(t) = r δt.

The integration gives:
Log G(t) = r t + cst, or
G(t) = Const x exp {rt}
At t = 0, G(t) = G(0) = Const
Therefore:
G(t) = G(0) exp {rt}

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G(t) = G(0) exp {rt}
G(t): the investment at the time t
G(0) = P: the principal or initial investment
r: the interest rate
t: the number of periods

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Example:

G(0) = 1000 $
r = 10% per year
t = 5 (5 years)
Then:
G(t) = 1000 x exp{10% x 5} = 1000 exp{0.5} = 1000 x 1.65 = 1650 $

The application of the formula:
P(t) = P(1 + r)ⁿ gives:
P(t) = 1000 x (1 + 10%)⁵ = 1000 x 1.610 = 1610 $

Remark:

To double an investment, we get directly the time of period we need:
G(t) = 2G(0) = G(0) exp {rt}
gives:
rt = ln 2 and then t = ln 2/r = 0.7/r .


Other example:

The banker knows the formula:

lim (1 + 1/n)ⁿ = e = 2.718
n → ∞

We make a deposit of $1.00 with an interest of 100% per year.

If the interest is credited once a year, at the end of the year, the value of the account will be:
$1.00 + (100/100)$1.00 = $1.00[1 + (100%)] = $2.00.

If the interest is credited two times a year, at the end of the year, the value of the account will be:
$1.00[1 + (100%/2)]² = $1.00[1 + (50%)]² = $1.00 (1.5)² = $2.25

If the interest is credited three times a year, at the end of the year, the value of the account will be:
$1.00[1 + (100%/3)]³ = $1.00[1 + 1/3]³ = $1.00 (4/3)³ = $2.37

If the interest is credited monthly during a year, at the end of the year, the value of the account will be:
$1.00[1 + (100%/12)]¹² = $1.00[1 + 1/12]¹² = $1.00 (13/12)¹² = $2.61

If the interest is credited daily during a year, at the end of the year, the value of the account will be:
$1.00[1 + (100%/365)]³⁶⁵ = $1.00[1 + 1/365]³⁶⁵ = $1.00 (366/365)³⁶⁵ = $2.71

If the interest is credited indefinitely (continuous compound) during a year, at the end of the year, the value of the account will be:
$1.00[1 + (100%/∞)]^∞ = $1.00[1 + 1/∞]^∞ =
$1.00 x lim (1 + 1/n)ⁿ = $e = $2.718
n → ∞

That is: even though the interest, at the rate of 100%, is credited indefinitely during a year, the deposit, at the end of the year, won't be even tripled.