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Equation 31 · Part 1 · How a Model Actually Gets Small Enough to Run on a Phone

Symbol E

E ⁣[(x−x^)2]≈s212\mathbb{E}\!\left[(x - \hat{x})^2\right] \approx \frac{s^2}{12}
E\mathbb{E}

What this part means

The expected value operator: the probability-weighted average of the quantity inside its brackets.

Its job in the formula

E is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

The passage around this formula

for a signed integer of b bits. An 8-bit integer has 2^8=256 representable levels; a 4-bit integer has 2^4=16 . That sixteen-fold reduction in available codes is the entire cost of quantization in one number. Treating the rounding error as approximately uniform over one quantization step s , its expected squared magnitude is the classical result E ⁣[(x−x^)2]≈s212\mathbb{E}\!\left[(x - \hat{x})^2\right] \approx \frac{s^2}{12}. so halving the number of bits, which roughly doubles s at fixed range, roughly quadruples the expected squared error per weight. That is why INT8 is usually described as close to free and INT4 is not: the error a network has to absorb does not grow gently as bits are removed, it grows quadratically in the step size, and every bit…

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Learn the underlying idea

Probability assigns a number from 0 to 1 to an event under a stated model. Zero means impossible within that model; one means certain.

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See this notation across published equations →

Sources cited in the article section

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