REPLY: Well if you’ve proven it to be a non proof by those methods, then show you work. - AnthonyHere you are.
The following is in response to the conclusions drawn from data presented in Figure 3 and Figure 6 of Dr. Spencer's guest article "Spencer Part 2: More CO2 Peculiarities - The C13/C12 Isotope Ratio. By the time I had written this up, the thread was closed for comments. Please note, that since that posting I have registered at wordpress and had to change my moniker from "blue" to "bluegrue" in the process. You can find this entire post in my livejournal entry, too. I recommend to read it there, because of the better layout.
Notation
Some of the expressions in the following proof can get quite long, so I am using a very compact notation, suitable for a text blog:
- Variables are one letter only and case-sensitive.
- x(t) denotes a timeseries, i.e. a collection of n tuples (ti,xi) where i = 1 .. n
- I'll need a lot of sums over all elements of a timeseries and elementwise sums or products of these timeseries. Multiplication signs are dropped. I'll use square brackets to denote them. Let [x] be the sum over all terms of the indicated timeseries. Examples:
- [x] = x1 + x2 + ... + xn
- [xy] = (x1*y1) + (x2*y2) + ... + (xn*yn)
- The mean value of a timeseries in this notation is [x]/n
- Caveat:
[1] = 1 + 1 + .... + 1 = n
We will need the slope m of the linear regression line of y(x), where x(t) and y(t) are series with n members:
m = (n[xy]-[x][y]) / (n[xx]-[x][x])You can verify that in literature or see e.g. here at MathWorld equation 15. If requested, I can give you the derivation for this from least square principles.
Now on to the proof
Let u(t) and w(t) be two time series with n elements each, sampled at equidistant timesteps (i.e. daily, monthly or similar data).
Create two new time series, by linearly detrending the above time series u(t) and w(t), where e,f,g and h are constants (Dr. Spencer used the regression lines for this, I'm proving the more general case) :
U(t) := u(t) - et - f
W(t) := w(t) - gt - hNow take the derivatives with respect to time
x(t) := du(t)/dt
y(t) := dw(t)/dt
X(t) := dU(t)/dt = d(u(t) - et - f)/dt = x(t) - e
Y(t) := dW(t)/dt = d(w(t) - gt - h)/dt = y(t) - g We can observe, that only the constants g and e survive, when taking the derivatives.x(t) and y(t) are what Dr. Spencer called the "Trend Signal".
X(t) and Y(t) are, what Dr. Spencer called the "Interannual Signal".
The correlation coefficient of the "Trend Signal" is just
m = (n[xy]-[x][y]) / (n[xx]-[x][x])
Now let us calculate the correlation coefficient of the "Interannual Signal":
M = (n[XY]-[X][Y]) / (n[XX]-[X][X])
= (n[(x-e)(y-g)]-[(x-e)][(y-g)]) / (n[(x-e)(x-e)]-[(x-e)][(x-e)])
= (n[xy-gx-ey+eg]-([x]-ne)([y]-ng)) / (n[xx-2ex+ee]-([x]-ne)([x]-ne))
= (n[xy]-ng[x]-ne[y]+nneg - [x][y]+ng[x]+ne[y]-nneg)
/ (n[xx]-ne[x]-ne[x]+nnee - [x][x]+ne[x]+ne[x]-nnee)
= (n[xy]-[x][y])/(n[xx]-[x][x])
= m So the correlation coefficient of Dr. Spencers "Trend Signal" and "Interannual Signal" of any equidistantly spaced time series are identical by mathematical neccessity. Arguing "m equals M, therefore A holds true" is equivalent to saying "4 = 4, therefore A holds true"
Anthony, feel free to copy this post to where ever you feel appropriate.

