How big are the exponential moments of a truncated normal distribution?

Given a random variable $$X$$ valued on $$[-1,1]$$ with mean zero. We can use say Hoeffding's Lemma to get $$\mathbb E[e^{\lambda X}] \le e^{\lambda^2/2}$$

I believe this bound cannot be improved much if for example $$X$$ behaves like a $$\pm 1$$ coin.

Could someone provide a reference for some better exponential moments when $$X$$ is more strongtly clustered around its mean? I would be happy with the case of $$X$$ either a truncated normal or uniform distribution.

Definition: Suppose $$X$$ is normally distributed with mean $$\mu =0$$ and variance $$\sigma^2$$. We say $$Y$$ is distributed as the truncation of $$X$$ to $$[-1,1]$$ to mean $$P(Y \in A) = \frac{P(X \in A)}{P(X \in [0,1])}$$ for each measurable subset $$A \subset [-1,1]$$.

I would expect better tail bounds based on the variance. Smaller variance should give tighter bounds. The zero-variance should correspond to a point mass and as the variance gets large we should recover the uniform case.

Another way of saying clustered around the mean would be that the PDF is increasing up to the mean and then decreasing. But this only makes sense when the PDF exists. I'm not sure how you would express that condition in terms of the distribution itself.

• Why do you think Hoeffding's inequality can be improved for a uniform distribution ? – Gabriel Romon May 21 at 20:03
• Unlike the general case, computing the expectation of $e^{\lambda X}$ for $X$ uniform on $[-1, 1]$ is completely elementary. By definition this equals $\int_{x = -1}^1 e^{\lambda X} (1/2)dx$ where $1/2$ is the p.d.f of $X$. Now the integral equals $\frac{1}{2\lambda}(e^\lambda - e^{-\lambda}) = \frac{1}{\lambda}\sinh(\lambda)$. You can plot this function and $e^{\lambda^2}$ as functions of $\lambda$ in Wolfram Alpha, say and see that $(1/\lambda) \sinh(\lambda)$ is MUCH smaller for $\lambda \neq 0$ – Vincent May 21 at 21:52
• Your initial line does not look correct in general: if $X$ is concentrated at $1$ then $\mathbb E[e^{\lambda X}] = e^\lambda \gt e^{\lambda^2/2}$ for $0 \lt \lambda \lt 2$ – Henry May 21 at 22:35
• Whoops, $X$ needs to have mean zero! – Daron May 22 at 10:20