fips.Covariance.aggregate

Covariance.aggregate(W)

Return the covariance of aggregates, \(W S W^T\).

For \(x\) with covariance \(S\), \(W x\) has covariance \(W S W^T\): one row and one column per aggregate (a total per year, the mean over a region, a trend). Its diagonal is the variance of each aggregate, and its off-diagonal entries are what makes a combination of aggregates (the difference of two years) more or less uncertain than its parts. The product is taken in the forms of \(W\) and \(S\) (through the factors of a Kronecker covariance), so nothing of the covariance’s size is formed when \(W\) has few rows.

Parameters:

W (Operator) – An aggregation, k x n, such as fips.aggregate.sum()’s: its columns the covariance’s labels (in any order), its rows the aggregates.

Returns:

\(W S W^T\), on the rows of W, with this covariance’s name.

Return type:

Covariance

Raises:

• TypeError – If W is not an Operator.

• ValueError – If the columns of W do not have exactly the covariance’s labels.

See also

fips.Posterior.cov

The posterior covariance of aggregates of the state.

fips.Posterior.uncertainty

The prior and posterior uncertainty of aggregates.

fips.Problem.aggregate

The problem of aggregated observations.

Examples

The total of two cells, each with variance 1, correlated by 0.5:

>>> import numpy as np
>>> import pandas as pd
>>> import fips
>>> space = fips.Space({"flux": pd.Index(["a", "b"], name="cell")})
>>> S = fips.Covariance(np.array([[1.0, 0.5], [0.5, 1.0]]), space)
>>> total = fips.Operator(
...     np.ones((1, 2)),
...     fips.Space({"total": pd.Index(["all"], name="region")}),
...     space,
... )
>>> S.aggregate(total).dense()
array([[3.]])