# Prove that $\mathbb{E}\exp{\lambda\xi} \le \exp\left(\lambda^2 \Vert\xi\Vert_{\psi_2}^2\right)$

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

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

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 $$r\in[0,1]$$

• Thank you for the proof of the remaining case. The answer is perfectly true Commented Feb 15 at 14:41

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.

• 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. Commented Feb 13 at 10:47
• Hahahaha, this I don't know yet!
– Plop
Commented Feb 13 at 10:48
• I have just got a hint. I hope we can discuss on it Commented Feb 13 at 10:49