Gordon Growth Model (2-Step) Approximation

Question

It is easy to prove that the sum of the infinite geometric series $$\frac{D(1+g)^n}{(1+r)^n}$$ is $$S = \frac{D(1+g)}{r-g}$$ where n = 1, 2, 3, ... and 0 < g < r < 1.

Now, if the infinite series becomes:

$$\frac{D(1+g_0)^n}{(1+r)^n}, n = 1, 2, ... , A$$

$$\frac{D(1+g_0)^A(1+g_1)^{n-A}}{(1+r)^n}, n = A+1, A+2, ...$$

Show that the sum approximates the following (assume $0< g_1 < g_0 < r < 1$):

$$S = \frac{D}{r-g_1}[(1+g_1) + A(g_0-g_1)]$$

Answer 1

I have figured out graphically and suspect that the authors did so similarly.

We are basically trying to approximate the area under the solid blue and black curves (leaving out the discounting by $(1+r)^n$ first, hence the exponential growth).

This area is approximately the same as (a bit smaller than) the area under the blue curve (solid and dashed). To calculate that area, we need to figure out the value at which the blue curve intersects the vertical axis.

Since we already know the value at which the black curve intersects the vertical axis: $D(1+g_0)$, we just need to calculate the gap as marked on the graph. The gap is approximately the difference between $a_A$ and $a_A^*$ if $a_1$ had grown at $g_1$. Then

$$Gap = \frac{D(1+g_0)^A}{(1+g_1)^A}$$

Linear approximation gives:

$$Gap = D[(1 + Ag_0) - (1 + Ag_1)] = DA(g_0 - g_1)$$

Going back to the Gordon Growth Model, all the calculation we just did were for the purpose of adjusting the numerator, which corresponds to the vertical axis intersect on our graph.

Voila! The numerator is now approximated as $D(1+g_0) + DA(g_0 - g_1)$

Finally we have the approximation of $S$:

$$\frac{D[(1 + g_0) + A(g_0 - g_1)]}{r - g_1}$$

Answer 2

For the infinite geometric series $\frac{D(1+g)^n}{(1+r)^n},0<g<r<1$, the sum is indeed: $$S=\frac{\frac{D(1+g)}{1+r}}{1-\frac{1+g}{1+r}}=\frac{D(1+g)}{r-g}.$$ Reference: Gordon growth model.

1-method: The sum of $A$ terms: $$S_A=\frac{\frac{D(1+g_0)}{1+r}\cdot \left(1-\left(\frac{1+g_0}{1+r}\right)^A\right)}{1-\frac{1+g_0}{1+r}}=\frac{D(1+g_0)[(1+r)^A-(1+g_0)^A]}{(r-g_0)(1+r)^A}$$ The sum of the remaining infinite terms (from $(A+1)$ to infinity) is: $$S_{A^*}=\frac{\frac{D(1+g_0)^A(1+g_1)}{(1+r)^{A+1}}}{1-\frac{1+g_1}{1+r}}=\frac{D(1+g_0)^A(1+g_1)}{(r-g_1)(1+r)^A}$$ Now add $S_A$ and $S_{A^*}$ and approximate to get $S$.

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Update: 2-method: Refer to the fragment from the referrenced article:

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First, note that it should be denoted by $P$ not $S$, because it is a present (not future) value of a stock. Second, note that $g_a=g_0, g_n=g_1, D_A=D_0(1+g_a)^A$. So, derivation of $6$:

$$\text{PV(Step 1)}=D_0\sum_{t=1}^A\left(\frac{1+g_a}{1+r}\right)^t=D_0\sum_{t=1}^{\infty}\left(\frac{1+g_a}{1+r}\right)^t-D_0\sum_{t=A+1}^{\infty}\left(\frac{1+g_a}{1+r}\right)^t=\\ \frac{D_0(1+g_a)}{r-g_a}-D_0\cdot \frac{\frac{(1+g_a)^{A+1}}{(1+r)^{A+1}}}{1-\frac{1+g_a}{1+r}}=\frac{D_0(1+g_a)}{r-g_a}\cdot \left[1-\frac{(1+g_a)^A}{(1+r)^A}{}\right];\\ \text{PV(Step 2)}=\frac{\frac{D_0(1+g_a)^A(1+g_n)}{(1+r)^{A+1}}}{1-\frac{1+g_n}{1+r}}=\frac{D_0(1+g_a)^A(1+g_n)}{(r-g_n)(1+r)^A};\\ \text{PV(Step 1)+PV(Step 2)}=\frac{D_0(1+g_a)}{r-g_a}\cdot \left[1-\frac{(1+g_a)^A}{(1+r)^A}+\frac{(1+g_a)^{A-1}(1+g_n)(r-g_a)}{(r-g_n)(1+r)^A}\right]=\\ \frac{D_0(1+g_a)}{r-g_a}\cdot \left[1-\frac{(1+g_a)^{A-1}[(1+g_a)(r-g_n)-(1+g_n)(r-g_a)]}{(r-g_n)(1+r)^A}\right]=\\ \frac{D_0(1+g_a)}{r-g_a}\cdot \left[1-\frac{(1+g_a)^{A-1}[\require{cancel}\cancel{r}+g_ar-g_n-\cancel{g_ag_n}-\cancel{r}-g_nr+g_a+\cancel{g_ag_n}]}{(r-g_n)(1+r)^A}\right]=\\ \frac{D_0(1+g_a)}{r-g_a}\cdot \left[1-\left(\frac{1+g_a}{1+r}\right)^{A-1}\cdot \frac{g_a-g_n}{r-g_n}\right].$$

However, next fragment shows the derivation of $7$ is concealed:

$\hspace{3cm}$

Good luck!