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Let $c>0$ and $X$ be a real random variable such that for any $\lambda\in\mathbb R$:

$\displaystyle \mathbb E\left(e^{\lambda X}\right)\leq e^{c\frac{\lambda^2}{4}}$.

Prove that, for any $\delta > 0$,

$\displaystyle \mathbb P\left(|X|>\delta\right)\leq 2e^{-\frac{\delta^2}{c}}$

I think this may follow from Markov's Inequality but perhaps not since we haven't covered Markov's inequality yet in class. I'd appreciate any input about how to solve this one. Thanks!

Markov's Inequality: $\mathbb P(X\geq\delta)\leq\frac{\mathbb E[X]}{\delta}$

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  • $\begingroup$ $\frac{1}{e^{\lambda \cdot \delta}} \mathbb{E}(e^{\lambda \cdot X}) + \frac{1}{e^{\lambda \cdot \delta}} \mathbb{E}(e^{-\lambda \cdot X})\leq \frac{1}{e^{\lambda \cdot \delta}} \exp \left( c \cdot \frac{\lambda^2}{4} \right) + \frac{1}{e^{\lambda \cdot \delta}} \mathbb{E}(e^{-\lambda \cdot X})$ $\endgroup$ – Chris Feb 21 '13 at 10:47
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    $\begingroup$ You can apply the inequality also to the second addend. Simply note that $\mathbb{E}(e^{-\lambda \cdot X}) = \mathbb{E}(e^{\mu \cdot X})$ where $\mu := -\lambda$. Now use $\mathbb{E}(e^{\mu \cdot X}) \leq e^{c \frac{\mu^2}{4}}$. $\endgroup$ – saz Feb 21 '13 at 11:07
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    $\begingroup$ Exactly. Note that $\frac{e^x}{e^y} = e^{x-y}$. Thus you have to solve an equation of the form $p(\lambda)=0$ where $p$ is a polynomial of degree 2. $\endgroup$ – saz Feb 21 '13 at 11:36
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    $\begingroup$ So $\lambda =\frac{2\delta}{c}$? $\endgroup$ – Chris Feb 21 '13 at 12:08
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    $\begingroup$ Yes, that's it. $\endgroup$ – saz Feb 21 '13 at 12:28
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The inequality does not follow from Markov's inequality, but the idea for the proof is similar.

By the monotonicity of $x \mapsto e^x$ we have

$$\begin{align} \mathbb{P}(|X|>\delta) &= \mathbb{P}(X>\delta)+ \mathbb{P}(-X>\delta) = \int_{[X>\delta]} 1\, d\mathbb{P} + \int_{[-X>\delta]} 1 \, d\mathbb{P} \\ &\leq \int_{[X>\delta]} \frac{e^{\lambda \cdot X}}{e^{\lambda \cdot \delta}} \, d\mathbb{P} + \int_{[-X>\delta]} \frac{e^{-\lambda \cdot X}}{e^{\lambda \cdot \delta}} \, d\mathbb{P} \\ &\leq \frac{1}{e^{\lambda \cdot \delta}} \mathbb{E}(e^{\lambda \cdot X}) + \frac{1}{e^{\lambda \cdot \delta}} \mathbb{E}(e^{-\lambda \cdot X})\end{align}$$

for arbritrary $\lambda \in \mathbb{R}$. Now use the assumption $\mathbb{E}(e^{\lambda \cdot X}) \leq \exp \left( c \cdot \frac{\lambda^2}{4} \right)$. Choose $\lambda$ such that you obtain the inequality you are looking for. If you don't get along with it, don't hesitate to ask...

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  • $\begingroup$ Hello saz, I'm not seeing where to go from where you left off, although I have definitely been trying. Thank you very much for offering to explain the rest. $\endgroup$ – Chris Feb 21 '13 at 10:36

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