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Problem

This question arises from a paper I am reading now. The original reasoning could be translated into following

$$ \Pr[X>\alpha+\beta+t] \leq f(t) \Rightarrow \mathbb{E}[X] \leq \alpha+\beta + \int_{t}f(t)dt $$ where $x$ is a random variable and $\alpha, \beta$ are constants.

It seems that the author tries to do integration on both sides and somehow remove the probability sign and replace it with expectation. But to my understanding, I need pdf of $X$ to compute its expectation and $\Pr[X>\alpha+\beta+t] \leq f(t)$ is not enough.

So could someone help me with the reasoning the author uses here?

Mr.Robot
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  • Are you familiar with the equality $\mathbb EX=\int_0^{\infty}P(X>t)dt$ for nonnegative random variable $X$? See here for instance. No PDF or PMF needed. – drhab Mar 05 '19 at 08:33
  • @drhab First time I have seen something like this. Quite useful result! Thank you so much. – Mr.Robot Mar 05 '19 at 08:47
  • You are welcome. Also note that in the answer of Kavi it is not used actually that $X$ is nonnegative. – drhab Mar 05 '19 at 08:48

2 Answers2

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Indeed, you can derive this using the formula given by drhab above: If $X$ is nonnegative, you have \begin{eqnarray}\mathbb{E}[X]&=&\int_0^\infty\mathbb{P}(X>t)~\mathrm{d}t\\&=&\int_{-\alpha-\beta}^\infty\mathbb{P}(X>\alpha+\beta+t)~\mathrm{d}t\\&\leq& \int_{-\alpha-\beta}^0 1~\mathrm{d}t+\int_{0}^\infty\mathbb{P}(X>\alpha+\beta+t)~\mathrm{d}t \\&\leq& \alpha+\beta+\int_{0}^\infty f(t)~\mathrm{d}t. \end{eqnarray}

Edit: Based on the other contributors' comments, I add a solution without the nonnegativity assumption: Let $Y:=X-\alpha-\beta$. Then $$\mathbb{E}[Y]\leq \mathbb{E}[Y^+]=\int_0^\infty\mathbb{P}(Y^+>t)~\mathrm{d}t= \int_0^\infty\mathbb{P}(Y>t)~\mathrm{d}t\leq \int_0^\infty f(t)~\mathrm{d}t.$$ Plugging in the definition of $Y$ immediately tells us $$\mathbb{E}[X]\leq\alpha+\beta+\int_0^\infty f(t)~\mathrm{d}t.$$

Mau314
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  • Not necessary to assume that $X$ is non-negative. – Kavi Rama Murthy Mar 05 '19 at 08:47
  • @KaviRamaMurthy I'm not sure: If $\alpha+\beta<0$ we don't necessarily know that $P(X^+>\alpha+\beta+t)\leq f(t)$. More precisely: If $\alpha+\beta+t<0$, clearly $P(X^+>\alpha+\beta+t)=1$ while we may have $P(X>\alpha+\beta+t)\leq f(t)<1$. – Mau314 Mar 05 '19 at 09:00
  • BY considering $X-\alpha -\beta$ the proof reduces to the case $\alpha =\beta=0$ so we don't have to worry about $\alpha +\beta <0$. – Kavi Rama Murthy Mar 05 '19 at 09:05
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$EX \leq EX^{+} = \int_0^{\alpha +\beta} P(X^{+} >t)\, dt +\int_{\alpha +\beta}^ {\infty} P(X^{+} >t)\, dt \leq \alpha +\beta +\int_0^ {\infty} f (s)\, ds$.

Actually, by considering $Y=X-\alpha -\beta$ we can reduce to the proof to the case $\alpha =\beta =0$.

Kavi Rama Murthy
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