| the PEG Ratio or Fool Ratio |
- You take the P/E Ratio example: P/E = 20
- You take the Earnings Growth Percentage example: G = 25% per year
- You divide one by t'other to get PEG: example: PEG = (P/E)/G = 20/25 = 0.8
- If it's less than "1", the stock is a good buy.
- If it's greater than "1", it ain't.
>So that's what you use to buy ... ?
No, I don't, but the idea is intriguing, no?
>No.
But the PEG ratio is some kind of time period and I'm intrigued by what it means. (But see this.)
>Huh?
The P/E Ratio is dollars_per_share divided by
dollars_per_share. That ratio has no dimension. It'd have the same
value if Price and Earnings were measured in yen_per_megashare.
It's dimensionless. On the other hand ...
>That's confusing. Dimensionless?
It's like this: you measure the circumference of a circle and you measure its diameter and you divide the two
numbers, the former by the latter, and you get a number and that number is dimensionless. It doesn't matter
whether you measured in feet or metres or light years. The ratio is π and it doesn't
depend upon any dimensions. In the case of the Growth Rate however ...
>Wait, let me do it. Growth Rate is percentage per year and ... uh ...
Percentage is dimensionless. It depends upon the ratio of new earnings to old.
However, the Growth Rate is Percentage per Year so it
has the dimensions of 1/Years. That gives PEG = (P/E)/G the dimension of
Years.
>And you're wondering "How many years?"
Well, I'm wondering what's the significance of that time. If PEG = 0.8 so
what's the significance of 0.8 years?
>Do you think this dimension stuff is important? I mean ...
Yes, it's important. Here's why:
- Volume has the dimensions of Length3 (such as cubic feet).
If R is the radius of a sphere, then
the formula V = π R2 is wrong because it has the dimensions of Length2. - Speed has the dimensions of Length/Time
(such as miles per hour). If L is distance travelled and T is the time taken, then
the formula Speed = L2/T is wrong because it has the dimensions of
Length2/Time. However,
Speed = L/T may be right because it has the correct dimensions.
>But Speed = π L/T also has the correct dimensions, so ...
Yes. This dimensional stuff can tell you when you're wrong, but ...
>So, what's your point?
I'm looking for some formula where PEG appears as a Time.
>And?
I'm still thinking about it ... but consider this:
- E(t)/P(t) is the reciprocal of the P/E Ratio, at time t (years).
- We get its rate of change by differentiating with respect to t:
[1] d/dt(E/P) = (1/P) dE/dt - (E/P2) dP/dt = (E/P) (1/E) dE/dt - (E/P) (1/P) dP/dt - Note that dE/dt is the earnings growth rate (dollars/year) and
100 (1/E) dE/dt = G is that rate expressed as a percentage of the current earnings (percentage per year)
so the first term in [1] is:
(E/P) (1/E) dE/dt = {(1/E) dE/dt}/{P/E} = {G/100}/{P/E} = 1/(100 PEG) - Hence, if x = E/P (noting that (1/P) dP/dt = d/dt (log P)), then we can rewrite the above equation [1] like so:
[2] dx/dt + x d/dt (log P) = 1/{100 PEG(t)} - Suppose we put P(t) = P(0) er t, where "r" some kind of annualized return (for the stock).
Then [2] becomes:
[3] dx/dt + r x = 1/{100 PEG(t)} - The solution to [3] is:
[4] x(t) = e-r t {x(0) + (1/100)
{1/PEG(s)} er s ds}
>So what do you conclude?
If we want a small P/E ratio
that means a big x = E/P
that means a small PEG.
>In [4], does PEG appear as a Time?
Yes (but see the correction), because x and rt are dimensionless [note that: r t = (% per year)(years) which is
dimensionless] and the other terms should also be dimensionless so,
in particular, (1/PEG) ds should be dimensionless and since ds has the dimension of
Time then (1/PEG) must have the dimension of
1/Time so PEG has the dimension ...
>I not sure I understand what ...
In any sensible formula, all the terms should have the same dimensions. You won't find a sensible
formula which involves adding a speed to a mass or an area to a volume and if there's a formula which has a term such
as ez then z must be dimensionless because ez = 1 + z + z2/2 + etc.
and it ain't sensible to add 1 and z and z2 unless z is dimensionless and even Einstein said
E = m c2 not E = m c3 because each side must have the same dimensions and ...
>Okay, so what are you going to do with [4]?
Well ... I could mention that if we change variables in the integral, in [4], from s to y = t - s,
[4] can be rewritten:
[5] x(t) = E(t)/P(t) = {E(0)/P(0)} e-r t + (1/100) {1/PEG(t-y)} e-r y dy}
>Okay, okay! What are you going to do with [5]?
I'm thinking about it.
>Yeah, I've heard that before.
However, I might point out that, if PEG changed at some "annualized" rate "a" ... like:
PEG(t) = PEG(0) ea t
then the solution to [4] (or [5]) would be:
[6] x(t) = E(t)/P(t) = {E(0)/P(0)} e-r t + {e-at - e-rt}/{100PEG(0)(r-a)}
or, expressing the growth rates as percentages R = 100r and A = 100a (so if r = 0.085 then R = 8.5%), we'd get:
[6] E(t)/P(t) = {E(0)/P(0)} e-r t + {e-at - e-rt}/{PEG(0)(R-A)}
|
>And what are you going to do with [6]?
>You realize, of course, that you don't have a SINGLE picture.
>Is that useful?
|  Figure 1 |
>I haven't the faintest idea of what ...
We have something labelled E(t), something labelled P(t) and something labelled PEG which is [P/E] / [100(1/E)dE/dt].
(The "100" is there to express it as a percentage).
We can interpret E(t) and P(t) any way we wish.
For example, E(t) might be the number of flies in my garden, at time t (in days, perhaps), and P(t) the number of tomatoes (at time t days).
Then E/P would be flies per tomatoes and 100(1/E)dE/dt would be the fly growth rate (in "percentage per day")
and PEG = [P/E] / [100(1/E)dE/dt] = [tomatoes per fly] / [percentage per day].
>And that has the dimension of time?
No. [P/E] / [100(1/E)dE/dt] always has the dimension: [dimension of Time]*[dimension of P] / [dimension of E] .
However, if P(t) were cost per tomato plant and E(t) were cost per tomato, then P and E would have the same dimensions
(even if you measured the cost in Italian lira of British pounds or whatever). Further, [100(1/E)dE/dt] would measure the rate at which our cost per
tomato changes (as more and more tomatoes grow on our plants).
In that case, our PEG = [P/E] / [100(1/E)dE/dt] would have the dimension of time:
>So I should take Figure 1 into the garden when I do my planting?
Definitely.
>Okay, now tell us about the e-mail from Jim T.
>If you don't, I will.
>So PEG is Years2.
>So where's your new formula?
>And the stuff above. It's all wrong, eh?
>So?
>So you're looking for a formula where PEG appears as years2.
An aside about dimensional analysis stuff:
When my daughter was younger, she couldn't remember which, among 2πr and
πr2, was the area of a circle.
Once upon a time I submitted a question for the Junior Math Contest, written by over 20,000 high school students
across Canada - probably many more, now. The test is multiple choice and my question was:
There was a great cry of pain from high school teachers: "Unfair! We don't teach that stuff!"
This dimensional analysis stuff can be quite interesting. For example, the "classical" example is this:
>And that's a "correct" formula? What's k?
Recently I ran across a discussion at the
No Fee Boards which mentioned the use of
a ratio such as Yield / Pricek to rank various stocks (just as the PEG ratio might rank a set of stocks).
If the value of k were "1", then it'd be just the garden-variety E/P ratio (with E interpreted as Dividends per share instead of Earnings per share).
However, other values of k may be useful, like k = 0.5 or maybe k = 1.5 or maybe ...
>So you're going to consider variations in k?
Do I have to?
Okay. Jim points out (correctly!) that Earnings have the dimensions of dollars per year per share, not dollars per share.
That gives P/E the dimensions of {dollars per share} / {dollars per year per share} or Years.
Recall that G = 100 {(1/E) dE/dt} is the annual growth rate as a percentage per year, hence has the dimension of 1/Years
{P/E} / G = PEG then has the dimensions of [Years] / [1/Years]
Uh ... yes ... Years2.
I'm working on it.
Not at all. For example:
[4] x(t) = e-r t {x(0) + (1/100) {1/PEG(s)} er s ds}
is okay. We just have to recognize that the x = E/P has the dimension of 1/Years
and 1/PEG(s) is in 1/Years2 and ds is in Years so
the terms are dimensionally the same, namely 1/Years, and ...
So, I'm thinking ...
I'm thinking ...
I gave her my "dimensional stuff" lecture (noting that if r was in feet, then r2 was square feet)
... but it did little good 
Among the answers was V = 2 π2a2b
... which
everybuddy will recognize as having the correct dimensions of Length3.
The variables "a" and "b" were lengths and there were other choices like:
V = π(a2 - b2) etc.
Well, for small oscillations one normally uses T = 2π SQRT[L/g] which (of course)
is dimensionally consistent.
Why not?
