= Solution
For $\theta+t\in\mathcal N$, integrating the exponential tilt gives the <moment-generating function>
$$
\boxed{M_\theta(t)=\mathbb E_\theta e^{t^TY}=\exp\{\kappa(\theta+t)-\kappa(\theta)\}.}
$$
At 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 gives
$$
\boxed{\mathbb E_\theta Y=\nabla\kappa(\theta),\qquad \operatorname{Cov}_\theta(Y)=\nabla^2\kappa(\theta).}
$$
Alternatively 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 $v$, $v^T\nabla^2\kappa(\theta)v=\operatorname{Var}_\theta(v^TY)\geq0$. In a <minimal exponential family> this <variance> is strictly positive whenever $v\ne0$, 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.
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