## Sunday, December 23, 2012

### Deriving Graham's Intrinsic Value Formula

In Chapter 11 of Benjamin Graham's The Intelligent Investor, Graham introduces a formula for valuing growth stocks.

$$P = E (8.5 + 2G)$$

where P is the value or price, E are current (or normal) earnings and G is the growth rate.

In a footnote, Graham remarks:
Note that we do not suggest that this formula gives the "true value" of a growth stock, but only that it approximates the results of the more elaborate calculations in vogue.
He mainly used the formula to show how absurd some implied growth expectations are. In other words:

$$G = \frac{P/E - 8.5}{2}$$

For example, a company with a P/E of 100 has an implied growth rate of about 46% which seems pretty excessive for long-term growth (over the long-term, company growth rates are more or less constrained by overall economic growth, say 2-4%.)

We can derive Graham's formula from a variation of the Gordon Growth Model.

Let's suppose that you have a company which next year will earn E in earnings, grows at a perpetual constant rate g and has a required rate of return (discount rate) R (with R > g). Then it can be shown (with a bit of calculus) that the present value of those earnings is given by:

$$P = \frac{E}{R-g}$$

Next, we'll take the Taylor expansion (with respect to g) of this and just look at the first two terms which will give us a linear approximation of the above formula.

$$P = E (\frac{1}{R}+\frac{1}{R^2}g)$$

To get the 8.5 in Graham's formula, we have to use R~11.75%:

$$P = E (8.5 + 72g)$$

And since Graham's growth rate G = 100g, we can substitute that in to obtain Graham's formula:

$$P = E (8.5 + 0.72G)$$

So it's not an exact match but it's pretty close. Considering that we eliminated a number of the terms from the Taylor expansion, some of which were non-negligible, it's reasonable to conclude that Graham's formula is a decent approximation to a future earnings discount model.

1. If you're having trouble reading the above post it's due to a latex script which seems to be working in my end. I'm curious if it works in comments as well so I'll give it a try.

Here's one way to derive the Gordon's Growth Model.

We assume continuous perpetual growth at rate g and discounted at rate r:

$$P = \int_0^\infty \! E\exp^{(r-g)} \, \mathrm{d}t.$$

$$P = \frac{E}{r-g}$$

$$P = \int_0^\infty \! E e^{(g-r)} \, \mathrm{d}t.$$
since we're applying $e^g$ growth to the earnings and discounted by a factor of $e^-r$.