Consider the canonical $d$-dimensional exponential family with densities $$p(x)=\exp\left(\langle\theta,T(x)\rangle-A(\theta)\right)h(x),\quad\theta\in\Omega$$
with $\Omega\subset\Omega_0=\{\theta:A(\theta)<\infty\}$. $\Omega_0$ is the natural parameter space. The family is called full if $\Omega=\Omega_0$.
Let $Z\in\mathbb R^{n\times p}$ be a fixed deterministic matrix and $Y\sim N(Z\beta,I_n),\beta\in\mathbb R^p$.
Show that the distribution of $Y$ can be represented as an $n$-dimensional canonical exponential family and specify $A, T$, $\Omega,\Omega_0$. Is this a full family?
Under what condition on $Z$ is the above family full rank?
Do the 1 and 2 again except write it as a d-dimensional canonical family. In particular, assuming the condition on $Z$ making the d-dimensional canonical exponential family full rank holds, derive $\nabla A$ and $\nabla A^{-1}$, the inverse of the gradient map.
Current work.
We want to write the density of $Y$ in the canonical form. I have:
$$\begin{split}(2\pi)^{-n/2}|I|e^{-\frac 1 2(Y-Z\beta)^TI(Y-Z\beta)} &\propto e^{-\frac 1 2 (Y^T-\beta^TZ^T)(Y-Z\beta)}\\ &=e^{-\frac 1 2(Y^TY-Y^TZ\beta-\beta^TZ^TY+\beta^TZ^TZ\beta)}\\ &=e^{-\frac 1 2 \|Y\|^2}e^{\langle Y,Z\beta\rangle-\frac 1 2\|Z\beta\|^2}\end{split}$$
thus $h(y)=(2\pi)^{-n/2}e^{-\frac 1 2\|Y\|^2}$, $\langle \beta,T(y)\rangle=\langle Z^TY,\beta\rangle$, and $A(\beta)=\frac 1 2 \|Z\beta\|^2$. I think the natural parameter space is $\Omega_0=\{\beta:A(\beta)<\infty\}=\mathbb R^p$. I believe $\Omega=\mathbb R^p$ so this is a fully family. However, I do not know what the requirement on Z to be "full rank" is - appreciate if someone can explain it to me. Also I think this pertains to part 3, the d-dimensional canonical family? Because $\langle \beta,T(y)\rangle=\langle Z^TY,\beta\rangle$ has d terms. So how do you get the n dimensional family, is it simply writing that term as $\langle Y,Z\beta\rangle$? But then $\beta$ is not “by itself.”