 Solution

For $\theta +t\in \mathcal{N}$, integrating the exponential tilt gives the moment-generating functionAt an interior natural parameter this is finite in a neighbourhood of $t=0$, so differentiation under the integral sign is justified by the nearby exponential moments. Differentiating its logarithm once and twice at zero givesAlternatively the score function is $Y-\nabla \kappa (\theta )$; its mean zero and its covariance matrix give the same exponential-family derivative identities. For any vector $v$, $v^T{\nabla }^2\kappa (\theta )v={\mathrm{Var}}_{\theta }(v^{T}Y)\ge 0$. In a minimal exponential family this variance is strictly positive whenever $v\ne 0$, proving strict convexity of the cumulant function. For a general exponential family whose sufficient statistic is $T(X)$, these identities describe the mean and covariance matrix of $T(X)$, rather than necessarily those of $X$ itself.