Problem: Let $\xi$ be a real random variable. We say that $\xi$ is $\psi_2$ when $\exists \lambda >0$ such that $\mathbb{E}\exp(\xi^2/\lambda^2) \le e$.
We denote by $\Vert \xi \Vert_{\psi_2}$ the infimum of such $\lambda$, that is
\begin{align*}
\Vert \xi \Vert_{\psi_2} = \inf\left\{\lambda > 0: \mathbb{E}\exp(\xi^2/\lambda^2) \le e\right\}.
\end{align*}
Suppose that $\xi$ is $\psi_2$ and that $\mathbb{E}\xi = 0$. Prove that $\forall \lambda >0$
$$\mathbb{E}\exp(\lambda \xi) \le \exp\left(\lambda^2 \Vert \xi\Vert_{\psi_2}^2\right).$$
My attempt: By using the inequality $e^{y} \le y+ e^{y^2}$, we have
\begin{align*}
\mathbb{E}\exp(\lambda \xi) &= \int_{-\infty}^{+\infty}\exp(\lambda t) f_{\xi}(t) dt\\
& \le \int_{\infty}^{+\infty}(\lambda t + e^{\lambda^2 t^2})f_\xi(t) dt\\
& \le \lambda \int_{-\infty}^{+\infty}t f_\xi(t)dt + \int_{-\infty}^{+\infty}e^{\lambda^2 t^2}f_\xi(t)dt \\
& = \lambda \mathbb{E}\xi + \int_{-\infty}^{+\infty}e^{\lambda^2 t^2}f_\xi(t)dt = \int_{-\infty}^{+\infty}e^{\lambda^2 t^2}f_\xi(t)dt \tag{since $\mathbb{E}\xi = 0$}\\
& = \mathbb{E}[\exp(\lambda^2 \xi^2)]\\
& = \mathbb{E}\left[\left(\exp(\xi^2/\Vert \xi\Vert_{\psi_2}^2)\right)^{\lambda^2\Vert \xi \Vert_{\psi_2}^2}\right]
\end{align*}
Now I am stuck here. I intended to use Jensen's inequality but the function in the exponential is not concave.
Hints: I have just got a hint as follows
Consider two cases are $\lambda \in ]0,1]$ and $\lambda > 1$. When $\lambda \in ]0,1]$ use the inequality $e^y \le y + e^{y^2}$, while for $\lambda >1$, use the fact that $\lambda \xi \le \dfrac{1}{2}\lambda^2 + \dfrac{1}{2}\xi^2$.
Best Answer
If $\lambda \left\|\xi\right\|_{\psi_2} > 1$ Let $\mu=\left\|\xi\right\|_{\psi_2}$, such that $\lambda\mu > 1$
$$\lambda \xi = \mu \lambda \times \frac\xi\mu \le \frac12\mu^2 \lambda^2 + \frac12\left(\frac\xi\mu\right)^2$$
Take the expectation of the exponenent, use the fact $u\mapsto \sqrt u$ is concave, you have $$\mathbb E \left[\exp \lambda \xi\right]\le \sqrt e\times \exp \left(\frac12\lambda^2\mu^2\right) \le \exp \left(\lambda^2\mu^2\right)$$ Then you have the inequality you are looking for.
For the other case $\lambda \left\|\xi\right\|_{\psi_2}\le 1$ use the computations that you did and the fact that $u\mapsto u^r$ is concave for $r\in[0,1]$