Average vs Annualized Gains
a continuation of Part I

Once upon a time, while roaming
Morningstar, I larned something neat:
We suppose that the annual gains for a sequence of years are:
R_{1}, R_{2}, R_{3}, etc. (where, for a gain of 12.3% we'd put
R = 0.123).
A buyandhold investment will then grow by a factor G_{1} = 1 + R_{1}
in the first year, then by a factor
G_{2} = 1 + R_{2} in the second year etc.
... and eventually by a factor
G_{N} = 1 + R_{N} (during the last of N years).
The growth over all N years is just
G_{1}G_{2}G_{3}...G_{N}
and if our Annualized Gain Factor is called simply G, then
(1)
G^{N} = G_{1}G_{2}G_{3}...G_{N}

>A "buyandhold" investment? Why do you ...?
The annualized gain isn't that easy to calculate if you're putting money in and/or taking money
out, like Dollar Cost Averaging, for example. So, to make things simpler, we'll
just invest $1.00 and leave it there  and watch it grow for N years. Otherwise, you'd have to
take a look at the Rate of Return stuff ...
>Please continue.
Okay. The Average (or Mean) Gain Factor is just:
(2)
A = {
G_{1}+G_{2}+G_{3}+...+G_{N}
}/N = (1/N)Σ G_{n}
= (1/N)Σ (1+R_{n})
= 1 + (1/N)ΣR_{n}

For convenience (later!), we'll put:
(2A)
G_{n} = 1+R_{n} = A(1+r_{n})
so
r_{n} = G_{n}/A1 measures the deviation of the Annual Gain Factors
from their Mean. That'll change Equation (1) into:
(1A)
G^{N} = A^{N}
(1+r_{1})(1+r_{2})...(1+r_{N})
Further, the Standard Deviation is given by
{the MeanoftheSquares} 
{the SquareoftheMean}:
(3)
SD^{2} = (1/N){
G_{1}^{2}+G_{2}^{2}+G_{3}^{2}+
...+G_{N}^{2}} 
A^{2} =
(1/N)ΣG_{n}^{2}  A^{2}
= A^{2}{
(1/N)Σ(1+r_{n})^{2}  1
}

Our objective is to see which of G and A is larger and ...
>But you already said that A is larger than G, in Part I.
Yes, but now I'd like to estimate how much larger ... with a wee bit o' math.
>Wake me when you're done.
Let's first take the logarithm of each side, in Equation (1A). That'll give:
(4)
N log(G) = N log(A) +
Σlog(1+r_{n})
Now we use a magic formula, namely:
(5)
log(1+x) = x  x^{2}/2 + x^{3}/3  + ...
which, as you can see, is pretty good when x is small
>That's assuming you're using natural logs, to the base e, so why don't you say LN instead of ....
Go back to sleep.


Anyway, we get:
(6)
N log(G) = N log(A) +
Σ
{ r_{n}  r_{n}^{2}/2 +
r_{n}^{3}/3  + }
= N log(A) + Σ r_{n}
 (1/2) Σ r_{n}^{2} +
error_{1}
Now we recognize some of the stuff here. In fact, from Equation (2):
 A = (1/N)Σ G_{n}
= A(1/N)Σ (1+r_{n})
= A + A(1/N)Σ r_{n}
'cause Σ1=1+1+1+... = N
Hence:
Σ r_{n} = 0
and, from equation (3), we get:
 SD^{2} = A^{2}{
(1/N)Σ(1+r_{n})^{2}  1
}
= A^{2}{
(1/N)Σ(1+2r_{n}+r_{n}^{2})
 1}
= A^{2}{
1 + (2/N)Σr_{n}
+ (1/N)Σr_{n}^{2}
 1}
Hence SD^{2} =
A^{2}(1/N)Σr_{n}^{2}
'cause Σ 1 = N
and we already know that Σr_{n} = 0.
So we get Σr_{n}^{2}
= N SD^{2}/A^{2}
>zzzZZZ
We can now substitute for Σr_{n}
and Σr_{n}^{2} in Equation (6),
and get (after some fiddling):
(7)
log({G/A}^{2})
=  SD^{2}/A^{2} + (2/N) error_{1}
Now we use the magic formula (5) again, in the form:
log(z) = (z1)  (z1)^{2}/2 +  ...
= (z1)  error_{2}
with z = {G/A}^{2}
which gives (finally!):
{G/A}^{2}
 1 =  SD^{2}/A^{2} +
(2/N) error_{1}
+ error_{2}
and (finally!) ...
>You already said finally!
... and finally, our approximation:
A^{2} = G^{2} + SD^{2} + ...

or, in terms of Average and Annualized fractional gains
(a 12.3% percentage gain means a 0.123 fractional gain and 1.123 gain FACTOR ... okay?):
(1 + Average)^{2} = (1 + Annualized)^{2} + SD^{2} + ...

>zzzZZZ
Since we have an exact result, from Equation (3), namely:
SD^{2} + A^{2} = (1/N)ΣG_{n}^{2}
where A = average(Gain Factors) = 1 + Average(Gain Fractions)
and the approximate result, above, namely:
G^{2} + SD^{2} = A^{2}
where A = average(G_{n}) = 1 + Average(R_{n})
we get a neat approximate picture of the geometric relationship between the
Geometric Mean G and the Arithmetic Mean A and the Standard Deviation SD
and ...


>Finally! A picture!
Here's another, using the approximation generated above:
>Yeah, but just how good is the approximation?
Good question. Here's a collection of annual returns, over ten years. For each we calculate the
Exact annualized return and the Approximation and note the
Error = Exact  Approximation:
>I've seen the approximation: Average = Annualized + SD^{2}/2
Well, let's see. Our formula, above, is:
(1+Average)^{2} = (1+Annualized)^{2} + SD^{2}
and, if the returns are small, we'd get:
1+2(Average) = 1+2(Annualized) + SD^{2}
'cause (1+x)^{2} is approximately 1+2x
and that'd give
Average = Annualized + SD^{2}/2
>Both are approximations, right?
Right, but I like mine better.
Suppose, for example, we have two returns: +50% and 50%. Then we'd have:
Actual Average = {(50%)+(50%)}/2
= 0.00%
SD^{2} =
{(500)^{2}+(500)^{2}}/2 so SD = 50%
Annualized =
{(1+0.5)*(10.5)}^{1/2}  1 = 0.134 or 13.4%
so we compare:
Approximation#1 = Annualized + SD^{2}/2 =  0.9%
and
Approximation#2 =
SQRT{(1+Annualized)^{2} + SD^{2}}  1
= 0.000 or 0.00% which agrees with the actual Average.
>You win!
Well ... thank you. In the meantime, I'll leave you with the Mean Annual Return for the
S&P 500 (Averaged over a time period from some time in the remote past to the present ...
well ... May, 2001) and the actual Annualized Return  again Annualized over the same time period 
and the Approximation#2 (using the Standard Deviation over the same time period):
You can play with the approximation:
(1+Annualized)^{2} = (1+Average)^{2}  Standard_Deviation^{2}
Try smaller and smaller Standard Deviations!
One last thingy:
Here's a chart comparing Canadian and US stocks and bonds. Notice that, in 19261956,
the TSE average return was smaller but the annualized return was higher
... because of the smaller TSE volatility (or standard deviation):
