Compute expectation given probability

Question

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?

Answer 1

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

Answer 2

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

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