Write the partitioned mean as and assume is nonsingular, as required by the displayed inverse. For the conditional multivariate normal distribution, form the residual
A linear transformation of a multivariate normal distribution is again normal. Direct calculation gives
The jointly normal blocks and are therefore independent: their joint characteristic function has no cross term in its quadratic exponent and factors into the two marginal characteristic functions. Conditioning leaves the distribution of unchanged. Reconstructing proves the full conditional law, including the requested Schur complement covariance:
This describes a regular conditional law; it does not require the equality event to have positive probability.
Put , and
The difference vector is . Its mean is zero because the two sons have the same mean vector. Its covariance matrix is
For example, . Thus
The positive definiteness assumption supplies and : these are eigenvalues of the original matrix on within-son contrast and between-son difference directions. In particular the resulting difference covariance is positive definite.
Taking the first marginal of the normal vector in part (i) gives
Equivalently its variance is ; the covariance between the two sons must be retained when taking their difference.
Applying the conditional multivariate normal distribution to the two differences gives, more generally,
At the result is
Both this law and the marginal law have mean zero, but the conditional variance is smaller. Specifically, with , the conditional variance is the marginal variance times . The assumptions imply , so the decrease is strict. Matching the breadth differences to zero gives information about the correlated length differences; it does not force the length differences themselves to be zero. The residual uncertainty remains positive. The original PDF has the between-son differences used here; the self-subtractions in the converted TeX are transcription errors.

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