online-autocovariance
v0.0.1
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Autocovariance (online algorithm)
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online-autocovariance
Usage
const Autocov = require('.')
const ac1 = Autocov(3)
const ac2 = Autocov(3)
;[1, 2, 3, 4, 5, 6, 7].forEach(v => { ac1(v) })
;[5, 1, 4, 7, 8, 1, -4, 0, 3, 7].forEach(v => { ac2(v) })
console.log(ac1.values, ac1.n) // ~> [ 4, 2.286, 0.714, -0.571 ]
console.log(ac2.values, ac2.n)
// Also works when calling with array argument:
// Keep in mind, it updates already defined ac1 values
console.log(ac1([8, 9, 10])) // ~> [8.25, 5.775, 3.4, 1.225]
How it works
Autocovariance online algorithm
Start with a formula for auto-covariance approximation:
covk = (1 / n) Σi=1..n-k(xi+k - μ)(xi - μ)
Then multiply values in brackets:
covk = (1 / n) Σi=1..n-k(xi+kxi + μ2 - μxi - μxi+k)
Multiply the result with (n-k)/(n-k) to get rid of Σ:
covk = α(β + μ2 - μμi - μμi+k)) = α(β + μ(μ - μi - μi+k))
Where:
α = (n - k) / n β = (Σi=1..n-k(xi+kxi)) / (n - k) = avg(xi+kxi) μi = (Σi=1..n-k xi) / (n - k) μi+k = (Σi=1..n-k xi+k) / (n - k)
μ is constant here so (Σi=1..n-k μ) / (n - k) = μ
Using the resulting formula, for each lag k from 0 to K we only need to accumulate:
- μ - average of full (0, N) data interval; don't depend on k
- μi - average of (0, N-k)
- μi+k - average of (k, N) interval
- β - average of the product of xi+k and xi
To update μ iteratively we can just use a simple algorithm: online-mean
μi is just a lagged (t-k) value of μ: ◆◆◆◆◆ μi(0, N-k), k=0 ◆◆◆◆◇ μi(0, N-k), k=1 ◆◆◆◇◇ μi(0, N-k), k=2 To update μi we just calculate μ and push it to μi and shift all its values to the right.
μi+k and β are "delayed": ◆◆◆◆◆ μi+k(k, N), k=0 ◇◆◆◆◆ μi+k(k, N), k=1 ◇◇◆◆◆ μi+k(k, N), k=2
To update their values we need to track last k observations of x (xlag) and their weights (wlag) using online-lag module xlag: [x[t], x[t-1], x[t-2], ...] wlag: [w[t], w[t-1], w[t-2], ...]
Until we get non-zero weight wlag[k], μi+k and β are zero. Each observation adds its weight to the wlag object shifting its values. So after, let's say 3 observation, 3rd weight in wlag will be 1, that gives us non-zero value for corresponding lag. I definitely should rewrite this to make everything more clear :)