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1 Derivation of Expected Covariances

To understand what it is about the observed statistics that suggests sibling interactions in our twin data we must follow through a little algebra. We shall try to keep this as simple as possible by considering the path model in Figure 8.3, which depicts the influence of an arbitrary latent

Figure 8.3: Path diagram showing influence of arbitrary exogenous variable $X$ on phenotype $P$ in a pair of relatives (for univariate twin data, incorporating sibling interaction).

variable, $X$, on the phenotype $P$. As long as our latent variables -- A, C, E, etc. -- are independent of each other, their effects can be considered one at a time and then summed, even in the presence of social interactions. The linear model corresponding to this path diagram is
$\displaystyle P_1$ $\textstyle =$ $\displaystyle sP_2 + xX_1$ (47)
$\displaystyle P_2$ $\textstyle =$ $\displaystyle sP_1 + xX_2$ (48)

Or, in matrices:

\left( \begin{array}{r} P_1 \ P_2 \end{array} \right) =
\left( \begin{array}{r} X_1\ X_2 \end{array} \right)

which in turn we can write more economically as

\bf y = \bf B \bf y + \bf G \bf x \end{displaymath}

Following the rules for matrix algebra set out in Chapters 4 and [*], we can rearrange this equation, as before:
$\displaystyle \bf y - \bf B \bf y$ $\textstyle =$ $\displaystyle \bf G \bf x$ (49)
$\displaystyle {\bf I}\bf y - \bf B \bf y$ $\textstyle =$ $\displaystyle \bf G \bf x$ (50)
$\displaystyle ({\bf I} - \bf B) \bf y$ $\textstyle =$ $\displaystyle \bf G \bf x \; ,$ (51)

and then, multiplying both sides of this equation by the inverse of (I - B), we have
$\displaystyle \bf y$ $\textstyle =$ $\displaystyle ({\bf I} -\bf B)^{-1} \bf G \bf x \; .$ (52)

In this case, the matrix (I - B) is simply

\left( \begin{array}{rr} 1&-s\ -s&1 \end{array} \right) \;,

which has determinant $1-s^2$, so $({\bf I} -\bf B)^{-1}$ is

\left( \begin{array}{rr} 1&s\ s&1 \end{array} \right) \; .

The symbol $\otimes$ is used to represent the Kronecker product, which in this case simply means that each element in the matrix is to be multiplied by the constant $\frac{1}{1-s^2}$. We have a vector of phenotypes on the left hand side of equation 8.8. In the chapter on matrix algebra (p. [*]) we showed how the covariance matrix could be computed from the raw data matrix $\bf T$ by expressing the observed data as deviations from the mean to form matrix $\bf U$, and computing the matrix product $\bf UU^\prime$. The same principle is applied here to the vector of phenotypes, which has an expected mean of 0 and is thus already expressed in mean deviate form. So to find the expected variance-covariance matrix of the phenotypes $P_1$ and $P_2$, we multiply by the transpose:
$\displaystyle {\cal E} \left\{ \bf y\bf y'\right\}$ $\textstyle =$ $\displaystyle \left\{({\bf I}-\bf B)^{-1}\bf G \bf x\right\}
\left\{({\bf I}-\bf B)^{-1}\bf G \bf x\right\}^\prime$ (53)
  $\textstyle =$ $\displaystyle ({\bf I}-\bf B)^{-1}\bf G
{\cal E} \left\{ \bf x\bf x'\right\} \bf G'({\bf I}-\bf B)^{-1'}
\; .$ (54)

Now in the middle of this equation we have the matrix product ${\cal E}
\left\{ \bf x\bf x'\right\}$. This is the covariance matrix of the x variables. For our particular example, we want two standardized variables, $X_1$ and $X_2$ to have unit variance and correlation $r$ so the matrix is:

\begin{displaymath}\left( \begin{array}{rr} 1 & r\ r& 1 \end{array} \right) \; . \end{displaymath}

We now have all the pieces required to compute the covariance matrix, recalling that for this case,
$\displaystyle \bf G$ $\textstyle =$ $\displaystyle \left( \begin{array}{rr} x & 0\  0& x \end{array} \right)$ (55)
$\displaystyle ({\bf I}-\bf B)^{-1}$ $\textstyle =$ $\displaystyle \frac{1}{1-s^2}\otimes
\left( \begin{array}{rr} 1&s\  s&1 \end{array} \right)$ (56)
$\displaystyle {\cal E}\left\{ \bf x\bf x'\right\}$ $\textstyle =$ $\displaystyle \left( \begin{array}{rr} 1&r\  r&1 \end{array} \right) \; .$ (57)

The reader may wish to show as an exercise that by substituting the right hand sides of equations 8.11 to 8.13 into equation 8.10, and carrying out the multiplication, we obtain:
$\displaystyle {\cal E} \left\{ \bf y\bf y'\right\}$ $\textstyle =$ $\displaystyle \frac{x^2}{(1-s^2)^2}\otimes
\left( \begin{array}{rr} 1+2sr+s^2& r+2s+rs^2\  r+2s+rs^2 & 1+2sr+s^2
\end{array} \right)$ (58)

We can use this result to derive the effects of sibling interaction on the variance and covariance due to a variety of sources of individual differences. For example, when considering:
  1. additive genetic influences, $x^2=a^2$ and $r=\alpha$, where $\alpha$ is 1.0 for MZ twins and 0.5 for DZ twins;
  2. shared environment influences, $x^2=c^2$ and $r=1$;
  3. non-shared environmental influences, $x^2=e^2$ and $r=0$;
  4. genetic dominance, $x^2=d^2$ and $r=\delta$, where $\delta=1.0$ for MZ twins and $\delta=0.25$ for DZ twins.
These results are summarized in Table 8.3.

Table 8.3: Effects of sibling interaction(s) on variance and covariance components between pairs of relatives.
Source Variance Covariance
Additive genetic $\omega (1+2s \alpha +s^2)a^2$ $\omega (\alpha+2s+\alpha s^2)a^2$
Dominance genetic $\omega (1+2s \delta +s^2)d^2$ $\omega (\delta+2s+\delta s^2)d^2$
Shared environment $\omega (1+2s+s^2)c^2$ $\omega (1+2s+s^2)c^2$
Non-shared environment $\omega (1+s^2)e^2$ $\omega 2se^2$
$\omega$ represents the scalar $\frac{1}{(1-s^2)^2}$ obtained from equation 8.14.

next up previous index
Next: 2 Numerical Illustration Up: 4 Consequences for Variation Previous: 4 Consequences for Variation   Index
Jeff Lessem 2002-03-21