It's my first question here, hi. In fact, it derives from my probability theory homework, which appears to be unusually difficult (or I just don't see something):

Suppose $X$ is a real valued random variable and $\varphi$ its characteristic function. Show that for any $a > 0$: $$\mathbb{P}(|X| \leq a^{-1}) \leq \frac{2}{a} \int_{|t| \leq a} |\varphi(t)|dt$$ So at first I obtained that without loss of generality we can assume $a = 1$. Thus, I tried to evaluate $\int_{-1}^{1} |\varphi(t)|dt$ in terms of $\int|f| \geq |\int f|$ inequality, used Fubini's theorem to swap the integrals and obtained inequality $$2\int_{-1}^{1} |\varphi(t)|dt \geq 4\left|\mathbb{E}\frac{\sin(X)}{X}\right|$$ but I don't even know whether $\,4\left|\mathbb{E}\frac{\sin(X)}{X}\right| \geq \mathbb{P}(|X| \leq 1)$ holds or not.

EDIT: my detailed attempts: $$2\int_{-1}^{1}|\varphi(t)|dt \geq 2\left|\int_{-1}^{1}\varphi(t)dt\right| = 2\left|\int_{-1}^{1}\mathbb{E}\cos(tX) + i\mathbb{E}\sin(tX)dt\right| = 2\left|\int_{-1}^{1}\mathbb{E}\cos(tX)dt\right| = 2\left|\int_{-1}^{1}\int_{-\infty}^{\infty}\cos(tx)d\mu_{X}(x)dt\right| = 2\left|\int_{-\infty}^{\infty}\int_{-1}^{1}\cos(tx)dtd\mu_{X}(x)\right| = 2\left|\int_{-\infty}^{\infty}2\frac{\sin(x)}{x}d\mu_{X}(x)\right| = 4\left|\mathbb{E}\frac{\sin(X)}{X}\right|.$$

  • $\begingroup$ $2\int_{-1}^{1}|\varphi(t)|dt \geq 2|\int_{-1}^{1}\varphi(t)dt| = 2|\int_{-1}^{1}\mathbb{E}cos(tX) + i\mathbb{E}cos(tX)dt| = 2|\int_{-1}^{1}\mathbb{E}cos(tX)dt| = 2|\int_{-1}^{1}\int_{-\infty}^{\infty}cos(tx)d\mu_{X}(x)dt| = 2|\int_{-\infty}^{\infty}\int_{-1}^{1}cos(tx)dtd\mu_{X}(x)| = 2|\int_{-\infty}^{\infty}2\frac{\sin(x)}{x}d\mu_{X}(x)| = 4|\mathbb{E}\frac{sin(X)}{X}|$ $\endgroup$ – tosi3k Nov 17 '14 at 0:50
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    $\begingroup$ @tosi3k I deleted my answer as it does not give what we want. I will think on the problem. $\endgroup$ – Davide Giraudo Dec 5 '14 at 12:39

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