B stock investments: Derivation

Derivation

We want to find out the value of $P n$ as $n \rightarrow \infty$ , where

$P_n = \sum_{t=1}^{n} D\times \frac{(1+g)^t}{(1+k)^t} = D\times \sum_{t=1}^{n} \frac{(1+g)^t}{(1+k)^t} = D\times \sum_{t=1}^{n} \left(\frac{1+g}{1+k}\right)^t$

Let

$a = \frac{1+g}{1+k}$ .

Then

$P_n = D\times \sum_{t=1}^{n} a^t$ .

Since

$1-a^{(n+1)} = (1-a) \times (1 + a + a^2 + ... + a^n) = (1-a) \times (1 + \sum_{t=1}^{n} a^t)$ ,

we get

$\frac{1-a^{(n+1)}}{1-a} - 1 = \sum_{t=1}^{n} a^t$ .

Therefore,

$P_n = D \times \left[ \frac{1-a^{(n+1)}}{1-a} - 1 \right]$ .

If $g < k$ , then $a <> and$

$a^{(n+1)} \rightarrow 0$ as $n \rightarrow \infty$ .

Thus, we get

$P_\infty = D \times \left( \frac{1}{1-a} - 1 \right) = D \times \left( \frac{a}{1-a} \right) = D \times \left(\frac{1+g}{1+k}\right) / \left[ 1 - \left(\frac{1+g}{1+k}\right) \right] = D \times \left( \frac{1+g}{k-g} \right)$ .

B stock investments

s1

Friday, September 26, 2008

Derivation

Derivation

No comments:

yout players

Blog Archive

About Me

2