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: R1, R2, R3, etc. (where, for a gain of 12.3% we'd put R = 0.123).

A buy-and-hold investment will then grow by a factor G1 = 1 + R1 in the first year, then by a factor G2 = 1 + R2 in the second year etc. ... and eventually by a factor GN = 1 + RN (during the last of N years).

The growth over all N years is just G1G2G3...GN and if our Annualized Gain Factor is called simply G, then

 (1)           GN = G1G2G3...GN

>A "buy-and-hold" 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 ...

Okay. The Average (or Mean) Gain Factor is just:
 (2)           A = { G1+G2+G3+...+GN }/N = (1/N)Σ Gn = (1/N)Σ (1+Rn) = 1 + (1/N)ΣRn

For convenience (later!), we'll put:

(2A)          Gn = 1+Rn = A(1+rn)
so rn = Gn/A-1 measures the deviation of the Annual Gain Factors from their Mean. That'll change Equation (1) into:

(1A)          GN = AN (1+r1)(1+r2)...(1+rN)

Further, the Standard Deviation is given by {the Mean-of-the-Squares} - {the Square-of-the-Mean}:
 (3)           SD2 = (1/N){ G12+G22+G32+ ...+GN2} - A2 = (1/N)ΣGn2 - A2 = A2{ (1/N)Σ(1+rn)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+rn) Now we use a magic formula, namely: (5)           log(1+x) = x - x2/2 + x3/3 - + ... 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 { rn - rn2/2 + rn3/3 - + } = N log(A) + Σ rn - (1/2) Σ rn2 + error1

Now we recognize some of the stuff here. In fact, from Equation (2):

• A = (1/N)Σ Gn = A(1/N)Σ (1+rn) = A + A(1/N)Σ rn     'cause Σ1=1+1+1+... = N
Hence:   Σ rn = 0
and, from equation (3), we get:
• SD2 = A2{ (1/N)Σ(1+rn)2 - 1 } = A2{ (1/N)Σ(1+2rn+rn2) - 1} = A2{ 1 + (2/N)Σrn + (1/N)Σrn2 - 1}
Hence   SD2 = A2(1/N)Σrn2   'cause Σ 1 = N and we already know that Σrn = 0.
So we get   Σrn2 = N SD2/A2

>zzzZZZ
We can now substitute for Σrn and Σrn2 in Equation (6), and get (after some fiddling):

(7)           log({G/A}2) = - SD2/A2 + (2/N) error1

Now we use the magic formula (5) again, in the form:

log(z) = (z-1) - (z-1)2/2 + - ... = (z-1) - error2     with z = {G/A}2

which gives (finally!):

{G/A}2 - 1 = - SD2/A2 + (2/N) error1 + error2

and (finally!) ...

... and finally, our approximation:

 A2 = G2 + SD2 + ...
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 + SD2 + ...
 >zzzZZZ Since we have an exact result, from Equation (3), namely: SD2 + A2 = (1/N)ΣGn2 where A = average(Gain Factors) = 1 + Average(Gain Fractions) and the approximate result, above, namely: G2 + SD2 = A2 where A = average(Gn) = 1 + Average(Rn) 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 + SD2/2

Well, let's see. Our formula, above, is:
(1+Average)2 = (1+Annualized)2 + SD2
and, if the returns are small, we'd get:
1+2(Average) = 1+2(Annualized) + SD2     'cause (1+x)2 is approximately 1+2x
and that'd give
Average = Annualized + SD2/2

>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%
SD2 = {(50-0)2+(-50-0)2}/2 so SD = 50%
Annualized = {(1+0.5)*(1-0.5)}1/2 - 1 = -0.134 or -13.4%

so we compare:
Approximation#1 = Annualized + SD2/2 = - 0.9%
and
Approximation#2 = SQRT{(1+Annualized)2 + SD2} - 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_Deviation2

 Average Return (example 10) =% Standard Deviation (example 18) =% Approximate Annualized Return = %

Try smaller and smaller Standard Deviations!

One last thingy: Here's a chart comparing Canadian and US stocks and bonds. Notice that, in 1926-1956, the TSE average return was smaller but the annualized return was higher ... because of the smaller TSE volatility (or standard deviation):