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

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

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This equation gives an approximation: it relates the quantities while allowing an approximation. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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E\mathbb{E}

Symbol E

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

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xx

Symbol x

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

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x^\hat{x}

Symbol hatx

the dequantised approximation actually used in arithmetic.

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s2s^2

Symbol s^2

the square of s; the scale.

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fraction

fraction

Divide the expression above the line by the one below it.

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≈

≈

Approximately equal to; the equality is not exact.

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subtraction

subtraction

Subtract the following term or group from the preceding one. A leading minus marks a negative quantity.

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superscript

superscript

A raised number can be a power. When it is a label or bound, it selects a case or the upper limit of a sum; the formula’s structure distinguishes these uses.

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1212

Denominator: 12

The complete quantity below the fraction bar; it must be nonzero for this division.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Its accuracy depends on the assumptions and range of use described in the article.

What the article says around this equation

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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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 below eight is removed from an already-narrow budget.

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