A Primer on Phylogenetic Generalised Least Squares

Abstract

Phylogenetic generalised least squares (PGLS) is one of the most commonly employed phylogenetic comparative methods. The technique, a modification of generalised least squares, uses knowledge of phylogenetic relationships to produce an estimate of expected covariance in cross-species data. Closely related species are assumed to have more similar traits because of their shared ancestry and hence produce more similar residuals from the least squares regression line. By taking into account the expected covariance structure of these residuals, modified slope and intercept estimates are generated that can account for interspecific autocorrelation due to phylogeny. Here, we provide a basic conceptual background to PGLS, for those unfamiliar with the approach. We describe the requirements for a PGLS analysis and highlight the packages that can be used to implement the method. We show how phylogeny is used to calculate the expected covariance structure in the data and how this is applied to the generalised least squares regression equation. We demonstrate how PGLS can incorporate information about phylogenetic signal, the extent to which closely related species truly are similar, and how it controls for this signal appropriately, thereby negating concerns about unnecessarily ‘correcting’ for phylogeny. In addition to discussing the appropriate way to present the results of PGLS analyses, we highlight some common misconceptions about the approach and commonly encountered problems with the method. These include misunderstandings about what phylogenetic signal refers to in the context of PGLS (residuals errors, not the traits themselves), and issues associated with unknown or uncertain phylogeny.

Appendix

1.1 Further Mathematical Details of the Calculation of OLS and PGLS Using Our Worked Example

An alternative way of expressing the ordinary least squares regression formula that is quicker and more effective for analysis with more than one predictor is using matrix algebra. Here, the equation to obtain regression estimates is given as

$$ {\varvec{\upbeta}} = \left( {{\mathbf{X}}^{{\prime }} {\mathbf{X}}} \right)^{ - 1} {\mathbf{X}}^{{\prime }} {\mathbf{y}} $$

In this case, β is the vector consisting of the parameter estimates (b₀, b₁, and so on if more than one predictor variable). X is a matrix consisting of n rows and (m + 1) columns (m is the number of predictor variables), where the first column represents a constant (given the value 1 on each row), and the subsequent columns are the X values for each predictor variable. In the matrix formulation, the term ${\mathbf{X}}^{{\prime }}$ denotes the ‘transpose’ of X—simply put, the rows become columns, and the columns become rows.

$$ {\mathbf{X}} = \left[ {\begin{array}{*{20}c} 1 & {1.02} \\ 1 & {1.06} \\ 1 & {0.96} \\ 1 & {0.92} \\ 1 & {0.89} \\ \end{array} } \right]\quad {\mathbf{X}}^{{\prime }} = \left[ {\begin{array}{*{20}c} 1 & 1 & 1 & 1 & 1 \\ {1.02} & {1.06} & {0.96} & {0.92} & {0.89} \\ \end{array} } \right] $$

When multiplied together, these become ${\mathbf{X}}^{{\prime }} {\mathbf{X}}$, calculated as follows:

$$ {\mathbf{X}}^{{\prime }} {\mathbf{X}} = \left[ {\begin{array}{*{20}c} 5 & {4.85} \\ {4.85} & {4.724} \\ \end{array} } \right] $$

Here, the value in row i, column j of ${\mathbf{X}}^{{\prime }} {\mathbf{X}}$ equals the sum total of row i elements of ${\mathbf{X}}^{{\prime }}$ multiplied by their respective column j elements of X. So for example, row 2, column 2 of ${\mathbf{X}}^{{\prime }} {\mathbf{X}}$ is (1.02 × 1.02) + (1.06 × 1.06) + (0.96 × 0.96) + (0.92 × 0.92) + (0.89 × 0.89) = 4.724.

Finally, the suffix ⁻¹ applied to ${\mathbf{X}}^{{\prime }} {\mathbf{X}}$ indicates the ‘inverse’ matrix. The way the inverse matrix is calculated is somewhat complex but it is the matrix that when multiplied by it original form (${\mathbf{X}}^{{\prime }} {\mathbf{X}}$) produces a matrix with 1s in the diagonal elements, and 0s in the off-diagonals (this is known as the identity matrix—see below).

$$ ({\mathbf{X}}^{{\prime }} {\mathbf{X}})^{ - 1} = \left[ {\begin{array}{*{20}c} {48.21} & { - 49.49} \\ { - 49.49} & {51.02} \\ \end{array} } \right] $$

y is the vector of n rows, containing the values of Y.

$$ {\mathbf{y}} = \left[ {\begin{array}{*{20}c} {1.38} \\ {1.41} \\ {1.36} \\ {1.22} \\ {1.13} \\ \end{array} } \right] $$

As with ${\mathbf{X}}^{{\prime }} {\mathbf{X}}$, for the ${\mathbf{X}}^{{\prime }} {\mathbf{y}}$ vector, the row i value is the overall total of each of the row i elements of ${\mathbf{X}}^{{\prime }}$ multiplied by their respective counterparts in the column of y (i.e. row 2 = (1.02 × 1.38) + (1.06 × 1.41) + (0.96 × 1.36) + (0.92 × 1.22) + (0.89 × 1.13) = 6.336.

$$ {\mathbf{X}}^{{\prime }} {\mathbf{y}} = \left[ {\begin{array}{*{20}c} {6.5} \\ {6.336} \\ \end{array} } \right] $$

Hence, when $({\mathbf{X}}^{{\prime }} {\mathbf{X}})^{ - 1}$is then multiplied by ${\mathbf{X}}^{{\prime }} {\mathbf{y}}$, we get the OLS solution for β

$$ {\varvec{\upbeta}} = \left[ {\begin{array}{*{20}c} {(48.21 \times 6.5)} + {( - 49.49 \times 6.336)} \\ {( - 49.49 \times 6.5)}+ {(51.02 \times 6.336)} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - 0.229} \\ {1.577} \\ \end{array} } \right] $$

where the first value (−0.229) is the intercept (b₀) and the second value is the slope estimate (b₁).

For generalised least squares, an additional element is added to the regression equation, in the form of the variance–covariance matrix, which represents the expected covariance structure of the residuals from the regression equation. In the case of OLS regression, the assumption is that there is no covariance between residuals (i.e. all species are independent of each other, and residuals from closely related species are not more similar on average than residuals from distantly related species). This (n × n) variance–covariance matrix is denoted as C, and the regression equation becomes

$$ {\varvec{\upbeta}} = \left( {{\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{X}}} \right)^{ - 1} {\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{y}} $$

Under the assumption that there is no covariance among the residuals and they are normally distributed, with mean = 0 and standard deviation σ_ε, then

$$ {\mathbf{C}} = \left[ {\begin{array}{*{20}c} {\varvec{\sigma}_{\varvec{\varepsilon}}^{2} } & 0 & 0 & 0 & 0 \\ 0 & {\varvec{\sigma}^{2}_{\varvec{\varepsilon}} } & 0 & 0 & 0 \\ 0 & 0 & {\varvec{\sigma}^{2}_{\varvec{\varepsilon}} } & 0 & 0 \\ 0 & 0 & 0 & {\varvec{\sigma}^{2}_{\varvec{\varepsilon}} } & 0 \\ 0 & 0 & 0 & 0 & {\varvec{\sigma}^{2}_{\varvec{\varepsilon}} } \\ \end{array} } \right] $$

The diagonal elements (the line of values from top left to bottom right) therefore represent the variance of the residuals, while the other off-diagonal elements = 0, meaning there is no covariation among the residuals. The inverse of this matrix, ${\mathbf{C}}^{ - 1}$, has essentially the same properties (all the off-diagonal elements remain as 0) except the diagonal elements now equal $1/\sigma_{\varepsilon }^{2}$. When this variance–covariance structure is assumed, the results of GLS are the same as those of OLS (the C part of the regression equation essentially drops out). On the other hand, if the variances are not equal, then you have a standard weighted least squares regression.

For phylogenetic generalised least squares, our expected variance–covariance matrix is C_phyl (see main text), and its inverse

$$ \begin{aligned}{\mathbf{C}}_{{{\mathbf{phyl}}}} &= \left[ {\begin{array}{*{20}c} 3 & 2 & 1 & 1 & 0 \\ 2 & 3 & 1 & 1 & 0 \\ 1 & 1 & 3 & 2 & 0 \\ 1 & 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 0 & 3 \\ \end{array} } \right] \\ {\mathbf{C}}_{{{\mathbf{phyl}}}}^{ - 1} &= \left[ {\begin{array}{*{20}c} {0.619} & { - 0.381} & { - 0.048} & { - 0.048} & 0 \\ { - 0.381} & {0.619} & { - 0.048} & { - 0.048} & 0 \\ { - 0.048} & { - 0.048} & {0.619} & { - 0.381} & 0 \\ { - 0.048} & { - 0.048} & { - 0.381} & {0.619} & 0 \\ 0 & 0 & 0 & 0 & {0.333} \\ \end{array} } \right] \end{aligned}$$

Taking apart the components of the GLS regression equation, we first calculate the product ${\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1}$whose row i and column j values are the total of the ith row of ${\mathbf{X}}^{{\prime }}$ multiplied by the jth column of ${\mathbf{C}}^{ - 1}$. So, for example, row 2, column 3 of ${\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1}$ is (1.02 × -0.048) + (1.06 × -0.048) + (0.96 × 0.619) + (0.92 × -0.381) + (0.89 × 0) = 0.144

$$ \begin{aligned} {\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} & = \left[ {\begin{array}{*{20}c} 1 & 1 & 1 & 1 & 1 \\ {1.02} & {1.06} & {0.96} & {0.92} & {0.89} \\ \end{array} } \right] \times \left[ {\begin{array}{*{20}c} {0.619} & { - 0.381} & { - 0.048} & { - 0.048} & 0 \\ { - 0.381} & {0.619} & { - 0.048} & { - 0.048} & 0 \\ { - 0.048} & { - 0.048} & {0.619} & { - 0.381} & 0 \\ { - 0.048} & { - 0.048} & { - 0.381} & {0.619} & 0 \\ 0 & 0 & 0 & 0 & {0.333} \\ \end{array} } \right] \\ & = \left[ {\begin{array}{*{20}c} {0.142} & {0.142} & {0.142} & {0.142} & {0.333} \\ {0.137} & {0.177} & {0.144} & {0.104} & {0.296} \\ \end{array} } \right] \\ \end{aligned} $$

In similar fashion ${\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{X}}$is therefore

$$ {\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{X}} = \left[ {\begin{array}{*{20}c} {0.142} & {0.142} & {0.142} & {0.142} & {0.333} \\ {0.137} & {0.177} & {0.144} & {0.104} & {0.296} \\ \end{array} } \right] \times \left[ {\begin{array}{*{20}c} 1 & {1.02} \\ 1 & {1.06} \\ 1 & {0.96} \\ 1 & {0.92} \\ 1 & {0.89} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {0.901} & {0.859} \\ {0.859} & {0.825} \\ \end{array} } \right] $$

The inverse of which is

$$ ({\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{X}})^{ - 1} = \left[ {\begin{array}{*{20}c} {130.141} & { - 135.389} \\ { - 135.389} & {142.060} \\ \end{array} } \right] $$

The second component of the GLS regression equation ${\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{y}}$ follows likewise as

$$ {\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{y}} = \left[ {\begin{array}{*{20}c} {0.142} & {0.142} & {0.142} & {0.142} & {0.142} \\ {0.137} & {0.177} & {0.144} & {0.104} & {0.296} \\ \end{array} } \right] \times \left[ {\begin{array}{*{20}c} {1.38} \\ {1.41} \\ {1.36} \\ {1.22} \\ {1.13} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {1.139} \\ {1.097} \\ \end{array} } \right] $$

where, for example, the first row value (1.139) = (0.142 × 1.38) + (0.142 × 1.41) + (0.142 × 1.36) + (0.142 × 1.22) + (0.142 × 1.13).

Finally, we can combine our two products to obtain the PGLS solution for β.

$$ {\varvec{\upbeta}}_{{{\mathbf{PGLS}}}} = \left( {{\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{X}}} \right)^{ - 1} {\mathbf{X}}^{{\prime }} {\mathbf{C}}^{ - 1} {\mathbf{y}} = \left[ {\begin{array}{*{20}c} {130.141} & { - 135.389} \\ { - 135.389} & {142.060} \\ \end{array} } \right] \times \left[ {\begin{array}{*{20}c} {1.139} \\ {1.097} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - 0.276} \\ {1.616} \\ \end{array} } \right] $$

where b₀ = −0.276 and b₁ = 1.616.