Question: 2.58 ( ) The result (2.226) showed that the negative gradient of ln g() for the exponential family is given by the expectation of u(x).
2.58 () The result (2.226) showed that the negative gradient of ln g(η) for the exponential family is given by the expectation of u(x). By taking the second derivatives of
(2.195), show that
−∇∇ln g(η) = E[u(x)u(x)T] − E[u(x)]E[u(x)T] = cov[u(x)]. (2.300)
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