I have the dividend discount model, which is the following expression:

$$ P_{j,t} = \sum_{\tau=1}^{\infty}D_\tau(1+g)^\tau(1+r)^{-\tau}=\frac{D_{\tau+1}}{r-g} $$

Where $D_t$, is the dividend at time $t$ ,$g$ represents the constant growth over time and $r$ represents the required rate of return which is assumed to be constant over time too.

My questions are: why under constant growth and constant required rate of return can we re-write it this way. Why does the sum sign disappears?


@ZRH's solution is correct. I also found a website with more intermediate steps here: http://www.calculatinginvestor.com/2011/05/18/gordon-growth-model/

Thank you

  • $\begingroup$ It is a property of Geometric Series that allows the summation to be written in closed form this way en.wikipedia.org/wiki/Geometric_series#Formula $\endgroup$ – Alex C Mar 23 '19 at 15:25
  • $\begingroup$ Very important to note that this equation only holds if $r>g$. $\endgroup$ – Alex C Mar 23 '19 at 15:47
  • $\begingroup$ @AlexC Thank you. The assumption that $r>g$ is a general assumption of the model, otherwise, the company would grow bigger than the entire economy in perpetuity $\endgroup$ – Adrian Mar 23 '19 at 16:19
  • $\begingroup$ Yes, it can be interpreted as an economic assumption. But it is also mathematically necessary for the infinite sum to converge. $\endgroup$ – Alex C Mar 23 '19 at 17:21
  • $\begingroup$ Ah. I see. That makes sense. Thank you for the additional explanations. Been a great help $\endgroup$ – Adrian Mar 23 '19 at 17:23

Using Alex C's link, and further assuming the dividends $D_\tau$ to be constant (else you cannot really come up with a simple formula) you get:

$P=D\sum_{\tau=1}^\infty \left(\frac{1+g}{1+r}\right)^{\tau}=\frac{\frac{1+g}{1+r}}{1-\frac{1+g}{1+r}}=D\frac{1+g}{r-g}$

I presume that in your above formula, you mean $D(1+g)$ when you write $D_{\tau+1}$, as after executing the summation there should not be an index $\tau$ anymore

  • $\begingroup$ Your solution is correct. It is as you have said, the dividends are constant. I actually found in the meantime the derivations in the Gordon growth model 1959 here: calculatinginvestor.com/2011/05/18/gordon-growth-model. Thank you for your help! $\endgroup$ – Adrian Mar 23 '19 at 16:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.