Chapter 24 · The ocean's role: interannual-to-decadal prediction

Part VI — Predictability of the second kind Knob: initialisation lead live notebook

Overview

A decadal forecast claims to say something about the next ten years. Chapter 23 showed where such a claim can come from when the forcing is prescribed. But nobody prescribes the ocean: it is part of the system, driven by the very weather that cannot be predicted past a week, and barely observed. What exactly is being initialised, and how would you know it helped?

The model

A fast Lorenz 63 “atmosphere” coupled to a single slow “ocean” variable that integrates it, $\dot S = [-S + \lambda(z - z_{\mathrm{ref}})]/T$ , with the ocean feeding back on the effective Rayleigh number. This is Hasselmann’s picture [citation needed]: the ocean’s long memory comes from integrating fast weather, not from slow internal dynamics.

Two limits are exact and are asserted as tests rather than assumed — with no forcing the fast part is bitwise Lorenz 63, and an undriven ocean relaxes as $e^{-t/T}$ to $10^{-12}$ .

Prefer to run it yourself? Download as a Jupyter notebook — it installs its own dependencies in the first cell. The sliders are static in Jupyter; edit value= and re-run to change a parameter.

Where the memory is — and a constraint that shapes the chapter

The atmosphere forgets in 0.14 time units; the ocean in 17.9 — 128 times longer, and of order $T$ . Nothing in the ocean equation is slow except that one constant.

Then a constraint that is measured, not chosen. Turning up the coupling lets the ocean move the atmosphere, but the same loop damps the ocean and its memory goes with it:

$\kappa$	$\tau_S$ (TU)	how much it moves $\rho$
0	27.6	0
−0.02	19.1	0.05
−0.10	6.3	0.19
−0.25	1.8	0.38

Loop gain and feedback amplitude both scale as $\lambda\kappa$ , so they cannot be separated: any coupling strong enough to modulate the atmosphere appreciably has already destroyed the ocean’s memory. A positive $\kappa$ is not an option at all — it is a runaway that collapses the system onto the origin, asserted as a test.

This chapter therefore runs near the memory end, and that is a real limitation worth naming: it can show what an ocean’s memory buys, and cannot show what a strongly coupled ocean like ENSO does to the atmosphere above it.

What initialising the ocean buys

Two sets of 300 forecasts identical in every respect except whether the slow variable was initialised from observation or drawn from climatology — the fast variables get the same analysis error in both.

For the ocean, initialising is worth about 8 time units. For the weather it is worth almost nothing — but not quite nothing: the initialised run is better by 0.22 of a climatological spread at lead 2, about 18 standard errors and so real, and it is gone by lead 4. The honest summary is a small brief benefit for the weather against a lasting one for the ocean.

That brief benefit exists only because the coupling is non-zero. Had the ocean been purely passive the weather panel would show exactly nothing by construction, and would have proved nothing.

Drift is not bias

Break the model gently — relax the ocean towards $z = 23.7$ where the truth uses 23.6, a parameter error under half a per cent. That displaces the model’s climatology by about one standard deviation, so an initialised forecast slides towards the model’s preferred state. The mean error grows to −1.79, correlating with lead at −0.97.

The perfect-model control is not zero either, and saying so matters: it reaches −0.27 and wanders to 0.59, correlating with lead at −0.54, because 300 cases from one trajectory is not a large independent sample. What separates drift from that scatter is not that one vanishes — it is that drift is 7× larger at the longest lead and far more strongly trended. A drift curve quoted without its control would make scatter look like signal.

And drift is a function of lead, near zero at initialisation and large at ten years. A single subtracted number cannot remove it.

How many re-forecasts does it take?

The correction is simple — measure the mean error at each lead over past cases and subtract it — and the question that decides whether it works is how many cases you need. Estimated from the first $n$ and applied to 150 held-out ones:

With enough re-forecasts it is worth 9 %. With too few it is worse than not correcting at all — at 5 and 10 re-forecasts the “corrected” score sits above the uncorrected one, because the drift estimate is then mostly sampling noise and subtracting noise adds noise. It first pays at about 20.

That is the operational answer to a question that sounds bureaucratic and is not: a decadal system’s re-forecast archive is not documentation, it is a component of the forecast, and one that must be large enough or it actively harms the product.

And it must be cross-validated. Fitting the drift on the cases it is then scored on gives 2.766 against 2.828 — 2.2 % better than anything achievable. That gap is small here because the held-out set is large; where re-forecasts number in the tens it is far larger, and a skill claim built on an in-sample correction is measuring its own fitting procedure.

Exercises

Section 1’s trade-off bends sharply. Estimate the coupling at which the ocean’s memory drops below ten time units, and say what that implies for a model where the ocean genuinely does drive the atmosphere.
The uninitialised ocean forecast starts above one climatological spread. Why is that the correct starting value, and what would it mean if it started below?
The drift comes from a parameter error under half a per cent. Estimate the drift from an error ten times larger, and say whether the correction would still work.
The correction first pays at around twenty re-forecasts. What property of the system sets that number, and would a noisier ocean need more or fewer?