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!
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.