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Am solving some exercises in probability theory in order to prepare myself for the exams and i came across the following inequality which i cant prove:

Prove that for every $\alpha>0$ the following inequality holds $$\mathsf{P}(|X| \geq 2/\alpha) \leq \frac{1}{\alpha}\int_{-\alpha}^{\alpha}(1-\phi(t))dt$$ where $\phi(t) = \mathbb{E}(e^{itX})$ , there characteristic function of a random variable $X$.

I tried to use Fubini in order to evaluate the right hand side and i derived the expression

$$\begin{align*} \int_{-\alpha}^{\alpha}(1-\phi(t))dt &= \int_{-\alpha}^{\alpha}(1-\mathbb{E}(\cos tX))\, dt \\ &=\int_{-\alpha}^{\alpha}\mathbb{E}(1-\cos tX)\, dt \\ &\overset{1-\cos tX\geq 0}{\geq} \int_{-\alpha}^{\alpha}\mathbb{E}\biggl((1-\cos tX)I_{\{|X|\geq 2/\alpha\}}\biggr) \, dt \\ &=\int_{-\alpha}^{\alpha}\biggl(\int_{\mathbb{R}}(1-\cos tx)I_{\{|x|\geq 2/\alpha\}} \, dF_X(x)\biggr) \, dt \\ &= \int_{\mathbb{R}}\biggl(\int_{-\alpha}^{-\alpha}(1-\cos tx)I_{\{|x|\geq 2/\alpha\}}\, dt\biggr) \, dF_X(x)\end{align*} $$

But now I cant figure out what to do with integrable function inside.

If you have any idea on how to continue or if there is other way , let me know!

Thanks in advance!

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  • $\begingroup$ Either you have to assume that $X$ is symmetric or you have to replace $\phi(t)$ by $\text{Re} \, \phi(t)$... note that the relation "$w \leq z$" is not well-defined for complex numbers $w,z$. $\endgroup$ – saz Jun 19 '18 at 15:10
  • $\begingroup$ isnt the $\int_{-\alpha}^{\alpha}\phi(t)dt$ a real number since $\int_{-\alpha}^{\alpha}\mathbb{E}(sintX)dt = 0?$ $\endgroup$ – dem0nakos Jun 19 '18 at 15:17
  • $\begingroup$ Yes, that's correct... the statement would be, however, somewhat clearer if you would use the real part. Right now you leave it to the reader to make sense of your equation which is, in personal oppinion, not a good approach. $\endgroup$ – saz Jun 19 '18 at 15:44
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Your calculation shows

$$\int_{-\alpha}^{\alpha} (1-\text{Re} \, \phi(t)) \, dt \geq \int_{\mathbb{R}} \int_{-\alpha}^{\alpha} (1-\cos(tx)) 1_{|x|>2/\alpha} dt \, dF_X(x) \tag{1}$$

(note that there are two typos in your computations; instead of $\alpha/2$ it should read $2/\alpha$). Clearly,

$$\int_{-\alpha}^{\alpha} (1-\cos(tx)) \, dt = 2\alpha - 2\alpha \frac{\sin(\alpha x)}{\alpha x}.$$

Since

$$\forall |r| \geq 2: \qquad \left| \frac{\sin r}{r} \right| \leq \frac{1}{2} $$

we find that

$$\int_{-\alpha}^{\alpha} (1-\cos(tx)) \, dt \geq 2\alpha - 2\alpha \frac{1}{2} = \alpha$$

for any $x$ such that $|\alpha x| \geq 2$. Plugging this estimate into $(1)$ proves the assertion.

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  • $\begingroup$ Nice , thanks , i missed the inequality $|\frac{sinr}{r}|\leq sin(1/2)$. $\endgroup$ – dem0nakos Jun 19 '18 at 15:59
  • $\begingroup$ @dem0nakos Actually the estimate $|\frac{\sin r}{r}| \leq \frac{1}{2}$ is good enough, see my edited answer. $\endgroup$ – saz Jun 20 '18 at 4:45

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