$X$ is in the form of exponential family i.e.
$$\mathbb{P\_\theta}x = h(x)e^{\langle \theta,T(x)\rangle-\phi(\theta)}$$
where $\theta\in \mathbb{R}^d$. If $\nabla\phi(\theta)$ is L-Lipschitz i.e.
$$\vert\nabla\phi(\theta\_1) - \nabla\phi(\theta\_2)\vert \leq L\vert\theta\_1-\theta\_2\vert,$$
how can we prove $Z = \langle v, T(X)\rangle$ for fixed $v$ satisfying $\Vert v\Vert\_2=1$ is sub-Gaussian?


I thought $Z$ should be like the linear combination of $X$, but I failed to reach what the Lipschitz condition can control here.



Is the main part of certain exponential family sub-Gaussian?
$\newcommand\th\theta\newcommand\la\lambda\newcommand\R{\mathbb R}$Note that for $t\in\R^d$ we have  

$$M\_\th(t):=E\_\th e^{t\cdot T(X)}
=\int\_{\R^d}dx\,h(x)e^{(t+\th)\cdot T(x)-\phi(\th)}
=e^{\phi(t+\th)-\phi(\th)},$$
where $\cdot$ denotes the dot product. So,
$$E\_\th T(X)=\nabla M\_\th(0)=\nabla\phi(\th)$$
and, for any real $\la$,
$$E\_\th e^{\la(Z-E\_\th Z)}
=E\_\th\exp\{\la v\cdot T(X)-\la v\cdot E\_\th T(X))\} \\ 
=\exp\{\phi(\la v+\th)-\phi(\th)-\la v\cdot \nabla\phi(\th)\}. 
\tag{1}\label{1}$$
By the mean-value theorem, for some $a\in(0,1)$ depending on $\la,v,\th$, we have
$$\phi(\la v+\th)-\phi(\th)=\la v\cdot\nabla\phi(a\la v+\th).$$
Also, by the Lipschitz condition on $\nabla\phi$,
$$|\nabla\phi(a\la v+\th)-\nabla\phi(\th)|\le La\la|v|\le L\la.$$
Thus,
$$\phi(\la v+\th)-\phi(\th)-\la v\cdot \nabla\phi(\th)
=\la v\cdot(\nabla\phi(a\la v+\th)-\nabla\phi(\th))
\le L\la^2$$
and hence, by \eqref{1},
$$E\_\th e^{\la(Z-E\_\th Z)}
\le e^{L\la^2},$$
so that $Z$ is (uniformly) [sub-Gaussian](https://en.wikipedia.org/wiki/Sub-Gaussian_distribution#More_equivalent_definitions) for all $\th$.


