# 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?

• 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. Mar 5, 2019 at 8:33
• @drhab First time I have seen something like this. Quite useful result! Thank you so much. Mar 5, 2019 at 8:47
• You are welcome. Also note that in the answer of Kavi it is not used actually that $X$ is nonnegative. Mar 5, 2019 at 8:48

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 $$\mathbb{E}[X]\leq\alpha+\beta+\int_0^\infty f(t)~\mathrm{d}t.$$

• Not necessary to assume that $X$ is non-negative. Mar 5, 2019 at 8:47
• @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 $P(X>\alpha+\beta+t)\leq f(t)<1$. Mar 5, 2019 at 9:00
• BY considering $X-\alpha -\beta$ the proof reduces to the case $\alpha =\beta=0$ so we don't have to worry about $\alpha +\beta <0$. Mar 5, 2019 at 9:05

$$EX \leq EX^{+} = \int_0^{\alpha +\beta} P(X^{+} >t)\, dt +\int_{\alpha +\beta}^ {\infty} P(X^{+} >t)\, dt \leq \alpha +\beta +\int_0^ {\infty} f (s)\, ds$$.

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

• No need of "nonnegative" here. Mar 5, 2019 at 8:50