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Let F be the distribution function of random variable X, that is $F(t)=Prob[X<t]$ and know that $E[X]=0$ and $Var[X]=1$. I am trying to understand how to show the following inequality
$\int_{1/u}^{\infty} e^{ut}dF(t) \leq\sum_{k=1}^\infty e^{k+1} Prob[X\geq k/u]$

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note that $ut \leq [ut] + 1$ therefore

$$ \int_{1/u}^{\infty} e^{ut}dF(t) \leq \int_{1/u}^{\infty} e^{[ut] + 1}dF(t) = \int_{1/u}^{\infty} \sum_{k=1}^\infty e^{k+1} 1_{\frac{k}{u}<t\leq \frac{k+1}{u}}dF(t)= \sum_{k=1}^\infty e^{k+1} \bigg(F\bigg(\frac{k+1}{u} \bigg) - F\bigg(\frac{k}{u} \bigg)\bigg) = \sum_{k=1}^\infty e^{k+1} \Bbb{P}\bigg(\frac{k}{u} <X\leq\frac{k+1}{u} \bigg) \leq \sum_{k=1}^\infty e^{k+1} \Bbb{P}\bigg(\frac{k}{u} <X \bigg) $$

Note that we didn't use that $\Bbb{E}(X) = 0$ nor that $Var(X)=1$

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  • $\begingroup$ why is $$ \int_{1/u}^{\infty} e^{[ut] + 1}dF(t) =\sum_{k=1}^\infty e^{k+1} [F\bigg(\frac{k+1}{u} \bigg) - F\bigg(\frac{k}{u} \bigg)] $$ ? (I am assuming there should an extra parenthesis there ) Is it an approximation or an exact result? $\endgroup$
    – user126540
    Sep 3, 2015 at 14:20
  • $\begingroup$ I edited the question @SlugPue $\endgroup$ Sep 3, 2015 at 14:33

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