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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 $\lambda \xi \le \dfrac{1}{2}\lambda^2 + \dfrac{1}{2}\xi^2$.

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  • $\begingroup$ Have you tried replacing $\Vert \xi\Vert_{\psi_2}$ by some constant $a$, and take the infimum only at the end? $\endgroup$
    – Plop
    Commented Feb 13 at 10:20
  • $\begingroup$ And wait, I don't get your last equality... $\endgroup$
    – Plop
    Commented Feb 13 at 10:21
  • $\begingroup$ @Plop Sorry for my typo. I missed the $\exp$ term and now I have added it. $\endgroup$ Commented Feb 13 at 10:25

2 Answers 2

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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 $r\in[0,1]$

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  • $\begingroup$ Thank you for the proof of the remaining case. The answer is perfectly true $\endgroup$ Commented Feb 15 at 14:41
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This is not a full answer, since it only solves the case where $\lambda$ is small enough. It was too long to be posted as a comment.

We have, for all $a$, $\mathbb{E}[\exp(\lambda^2 \xi^2)] = \mathbb{E}\left[\left(\exp\left(\frac{\xi^2}{a^2}\right)\right)^{a^2\lambda^2}\right]$. If $\lambda$ is small enough, $x\mapsto x^{a^2\lambda^2}$ is concave, so, by Jensen's inequality, we have $\mathbb{E}\left[\left(\exp\left(\frac{\xi^2}{a^2}\right)\right)^{a^2\lambda^2}\right] \leq \mathbb{E}\left[\exp\left(\frac{\xi^2}{a^2}\right)\right]^{a^2\lambda^2}$, and when $a$ converges to $\Vert \xi \Vert_{\psi_2}$, the right-hand side converges to $\exp(\lambda^2\Vert \xi \Vert^2_{\psi_2})$. So, assuming you made no mistakes in the rest, this settles the case where $\lambda$ is small enough.

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  • $\begingroup$ Actually, I did as you did before asking this question. However, how can we prove the problem holds for all $\lambda>0$ if it just only holds for $\lambda$ is small enough. $\endgroup$ Commented Feb 13 at 10:47
  • $\begingroup$ Hahahaha, this I don't know yet! $\endgroup$
    – Plop
    Commented Feb 13 at 10:48
  • $\begingroup$ I have just got a hint. I hope we can discuss on it $\endgroup$ Commented Feb 13 at 10:49

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