Volatility Stuff
motivated by questions raised by George L.

It's "standard practice" to calculate the volatility of monthly returns, multiply by the square-root-of-12, and take that as the annual volatility.
One convenient thing about this algorithm is that there are more monthly returns in the last 30 years than there are annual returns.
I've already explained this stuff here.
>Ah, yes, because there are 12 months in a year.
Right. So one could also consider taking the volatility of weekly returns and multiplying by the square-root-of-52.
>Or daily volatility multiplied ...
Yes, yes. So it'd be interesting to see how that works ... or if it's reasonable and how that ritual compares to the real, live annual volatility.

We start with ...
>Wait! Does annual volatility really change all that much? Does it really matter if ...?
Here's a picture:

Continuing, we start with some stat-stuff, obtained here, namely:
Suppose x and y are random variables.
Let M[x], S[x], VAR[x] and r(x,y) denote the Mean, Standard Deviation, Variance and Correlation. Note: Variance = StandardDeviation2

[1]   Mxk] = ΣM[xk] ... the Mean of a sum is the sum of the Means

[2]   VARxk] = ΣVAR[xk] + 2 Σr(xk, xj) S[xk] S[xj] ... the Variance of a sum is the sum of the Variances + something that depends upon the correlation

[2A]   VAR[x+C] = VAR[x] ... for any constant "C"

[3]   VARxk] = ΣVAR[xk] ... if the variables are uncorrelated, so r = 0

[4]   VAR[xy] = M2[x]VAR[y] + M2[y]VAR[x] + VAR[x]VAR[y] ... if the variables are uncorrelated

[4A]   VAR[xy] = 2M2 S2 + S4 ... if the variables also have the same Mean and Standard Deviation

Now let's consider our random variables to be uncorrelated monthly returns with common mean M and Standard Deviation S.
If x and y are returns for 2 successive months, what's the 2-month return?

>It's ... uh, it's ...
It's (1+x)(1+y) - 1
If what's the standard deviation of this 2-month return?

>Is it SQRT(2)*S ?
We'll see.

We stare intently at the stat stuff we've noted above (especially [2A] and [4A]) and get:

VAR[(1+x)(1+y) - 1] = VAR[(1+x)(1+y)] = M2[1+x]VAR[1+y] + M2[1+y]VAR[1+x] + VAR[1+x]VAR[1+y]

or, in terms of the common Mean and Standard Deviation of 1+x and 1+y, namely (1+M) and S:

[5]   VAR[(1+x)(1+y) - 1] = 2(1+M)2S2 + S4 = 2 S2 [(1+M)2 + 0.5 S2 ].

To get some idea of the size of these terms, let's assume
M = 0.01 ... a 1% monthly return or (roughly) 12% annualized return
S = 0.06 ... a 6% volatility for monthly returns or (roughly) 21% annualized volatility.

Then (1 + M)2 + 0.5*S2 = 1.012 + 0.5*0.062 ≈ 1.03
so we'd get from [5]:
[5A]   S2[(1+x)(1+y)-1] ≈ 2.06*S2   so   S[(1+x)(1+y)-1] ≈ SQRT(2.06) S
So the volatility for the 2-month return is very nearly SQRT(2)*S

>Close, but no cigar, eh?
Yes, but even this result assumes zero correlation between monthly returns. That means there are no "trends" in the prices.

In particular, I'd expect the actual, real-live annual returns to be somewhat larger than what we'd get with that square-root-of-time stuff.

>Is that a proof?
But I can generate 5000 random monthly returns, calculate their volatility, then look at actual 2-month, 4-month, etc. returns, then calculate their volatilities, then compare with what I'd get by multiplying the 1-month volatility by SQRT(2), SQRT(4), etc., then ...

>A picture is worth a thousand ...
Here's a picture:

>Is that a proof?
Hardly ...