# Is the main part of certain exponential family sub-Gaussian?

$$X$$ 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.

$$\newcommand\th\theta\newcommand\la\lambda\newcommand\R{\mathbb R}$$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 for all $$\th$$.