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As a reference, this is Problem 6.2 in Albiac and Kalton's Topics in Banach Space Theory

The question involves a direct proof of Khintchine's Inequality. In part (1), we are to prove that $\cosh(t)\leq e^{t^2/2}$ for all $t\in\mathbb{R}$. Part (2) (which I don't think I need yet for the step I am stuck on) is to show that for $p\geq1$, $t^p\leq p^pe^{-p}e^t$ for I assume $t>0$. These steps I have done.

The part I am stuck on is this: Let $(\varepsilon_n)_{n=1}^\infty$ be a sequence of Rademachers (i.e. independent random variables taking on values $\pm1$ with equal probability), and let $(a_k)_{k=1}^\infty\subset\mathbb{R}$ such that $$\underset{k=1}{\overset{n}{\sum}}a_k^2=1$$ and define $$f=\underset{k=1}{\overset{n}{\sum}}a_k\varepsilon_k$$

We want to show that $$\mathbb{E}(e^f)\leq e$$ (If you have any doubts, this is expectation over the epsilons).

Here has been my attempt so far: $$\mathbb{E}(e^f)=\mathbb{E}\left(\exp\left(\underset{k=1}{\overset{n}{\sum}}a_k\varepsilon_k\right)\right)=\mathbb{E}\left(\underset{k=1}{\overset{n}{\prod}}\exp(a_k\varepsilon_k)\right)=\underset{k=1}{\overset{n}{\prod}}\left(\mathbb{E}\exp(a_k\varepsilon_k)\right)\leq\underset{k=1}{\overset{n}{\prod}}\left(\mathbb{E}(2\cosh(a_k\varepsilon_k))\right)$$ the second to last equality is by independence, and the last inequality here follows from the definition of $\cosh$. Then by part (1), we have $$\leq\underset{k=1}{\overset{n}{\prod}}\left(\mathbb{E}\left(2\exp\left(\frac{a_k^2\varepsilon_k^2}{2}\right)\right)\right)$$ But since $\varepsilon_k^2=1$ for all $k$, and we can move the expectation back outside, and put the product back into the exponent as a sum, we will get something like $$2^n\exp\left(\underset{k=1}{\overset{n}{\sum}}\frac{a_k^2}{2}\right)=2^n\exp\left(\frac{1}{2}\right)$$ by assumption on the sum of $a_n^2$. So this does not give the inequality we want.

Any suggestions or hints would be appreciated (be gentle, I am far from knowledgeable of probability theory)

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up vote 5 down vote accepted

I think you should just evaluate $\mathbb{E}\exp(a_k\varepsilon_k)$ instead of bounding it. By definition, we have $$\mathbb{E}\exp(a_k\varepsilon_k)=\exp(-a_k)\,\mathbb{P}(\varepsilon_k=-1)+ \exp(a_k)\,\mathbb{P}(\varepsilon_k=1)={\exp(-a_k)+\exp(a_k)\over 2}.$$

This gives $\cosh(a_k)$, and your expectation becomes $\prod\limits_{k=1}^n\cosh(a_k)$. Now use part (1) of the exercise to show that $\mathbb{E}(e^f)\leq e^{1/2}$.

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Probably a thick question, but why is $\mathbb{E}\exp(a_k\varepsilon_k)=\cosh(a_k)$? – Keaton Apr 13 '12 at 18:34
@Keaton I've added some explanation to my answer. – Byron Schmuland Apr 13 '12 at 18:37
Thanks for your help. As I said, I am no probabilist, I know just enough to do some interchanging of expectations and averaging. – Keaton Apr 13 '12 at 18:43
@Keaton Glad to help. – Byron Schmuland Apr 13 '12 at 18:43

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