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For a zero mean sub-Gaussian R.V. we know that: $$ \mathbb{E}[e^{\lambda X}]\le e^{\frac{\lambda^2\sigma^2}{2}},\qquad\forall\lambda\in \mathbb{R}$$ From Taylor series expansion and equating the terms of the same power for $\lambda$ it can be easily obtained that: $$\mathbb{E}[X^2]\le \sigma^2$$

Is it possible to prove that $\mathbb{E}[X^2]=\sigma^2$ for the minimum $\sigma^2$ that the inequality $ \mathbb{E}[e^{\lambda X}]\le e^{\frac{\lambda^2\sigma^2}{2}}$ holds?

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  • $\begingroup$ Can you explain how comparing terms of the Taylor expansion works pls? I don't follow. AFAIK the following is tight $$\mathbb{E}[X^2] = \int_0^\infty \mathbb{P}(|X| > \sqrt{t}\,) \mathrm{d}t \le 2 \int_0^\infty e^{-\frac{t}{2\sigma^2}} \mathrm{d}t = 4 \sigma^2.$$ Am I wrong? $\endgroup$
    – user27182
    May 18 '20 at 13:32
  • $\begingroup$ Ok thanks, but even given what you say, how do I get $\mathbb{E}[X^2] \le \sigma^2$? $\endgroup$
    – user27182
    May 18 '20 at 19:57
  • $\begingroup$ Also, why does your $t$ in the integrand go? $$\int_0^\infty 2t \mathbb{P}(|X| > t) \, \mathrm{d}t \le 2 \int_0^\infty 2t e^{-\frac{t^2}{2{\sigma}^2}} \,\mathrm{d}t \le 4 \int_0^\infty( - \sigma^2) \frac{\mathrm{d}}{\mathrm{d}t}e^{-\frac{t^2}{2{\sigma}^2}} \, \mathrm{d}t = 4\sigma^2$$ $\endgroup$
    – user27182
    May 18 '20 at 20:09
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    $\begingroup$ @user27182 You're correct. That was a typo. You can get it from Taylor expansion of both sides. After some calculations: $$ \frac{\lambda^2}{2}(\sigma^2-E[X^2])\geq \lambda^3(P(\lambda)),$$ where $P(\lambda)$ is a polynomial. Dividing both sides by $\lambda^2$ and letting $\lambda\to0$ should give the result. $\endgroup$
    – SMA.D
    May 18 '20 at 20:51
  • $\begingroup$ ok I see now. thanks. I did not know the tail bound would be so loose. $\endgroup$
    – user27182
    May 18 '20 at 21:47
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$\newcommand{\EE}{\mathbb{E}}$ Here's a counterexample that has expectation zero:

Let $X$ take the value $1-p$ with probability $p$ and the value $-p$ with probability $1-p$. Then $\EE X = 0$ and $\EE X^2 = p(1-p)$. The moment generating function is $$\EE e^{\lambda X} = e^{\lambda(1-p)} p + e^{-\lambda p} (1-p)$$ so the smallest variance proxy we can take is $$\sigma^2 = \sup_{\lambda \in \mathbb{R}} \frac{2}{\lambda^2} \log \left( e^{\lambda(1-p)} p + e^{-\lambda p} (1-p) \right) . $$

If we take $p=1/4$ then this gives $\sigma^2 \approx 0.2276$ while $p(1-p)=0.1875$.

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No you can't. For example, a Bernoulli $X$ is s.t. $\mathbb{E}[X^2]=p$, but $$\mathbb{E}[e^{\lambda X}] = pe^{\lambda}+(1-p)\geq e^{\frac{p\lambda^2}{2}}\quad \text{for }\lambda=2.$$

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  • $\begingroup$ Thank you for answering. But in your example X is not zero mean as mentioned in the question. $\endgroup$
    – SMA.D
    Apr 5 '19 at 15:48

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