There is a result stated in Chow, Teicher's text on probability:

Let $\{b_n, n\geq 0\}$ be a strictly increasing sequence with $0\leq b_n\uparrow \infty$ and let $b(\cdot)$ be a strictly monotone extension of $\{b_n\}$ to $[0,\infty).$ Then for any r.v. $X\geq 0, $ a.e. $$\sum_{n=1}^\infty \Pr[X\geq b_n]\leq \mathrm Eb^{-1}(X)\leq\sum_{n=0}^\infty \Pr[X\geq b_n] .\tag 1$$ In particular, for any $r>0,$ $$\sum_{n=1}^\infty \Pr[|X|\geq n^{1/r}]\leq \mathrm E|X|^r\leq\sum_{n=0}^\infty \Pr[|X|\geq n^{1/r}] .\tag 2$$

In proving the former result, the authors took $\varphi(x):= b^{-1}(X)$ and defined $Y:=\sum_{j=1}^\infty j\mathrm I_{\{j\leq \varphi(X)<j+1\}}; Z = \sum_{j=0}^\infty(j+1)\mathrm I_{\{j\leq \varphi(X)<j+1\}} $ and showed $\mathrm EY\leq \mathrm E\varphi\leq \mathrm EZ$ using $\Pr[X\geq b_n]\leq \Pr[\varphi(X)\geq n].$ Then they claimed the second result follows by taking $b(x) = x^r.$

I couldn't understand one thing: why did they take $b^{-1}$ at the first place instead of working with $b(\cdot);$ what was the point of it? If we take $b(x) = x^r,$ how could $(1) $ imply $(2)$ for the choice would mean we would be getting $\mathrm Eb^{-1}(X) = \mathrm E X^{-r}?$ Also what is the argument behind $\Pr[X\geq b_n]\leq \Pr[\varphi(X)\geq n]?$

Finally there is also another result $$\sum_{n=1}^\infty n^{r-1} \Pr[X\geq n]\leq \mathrm EX^r\leq \sum_{n=0}^\infty n^{r-1} \Pr[X\geq n];$$ (ref. Gut's book on Probability) could we be able to conclude this result from $(1)$ above? They seem to be looking alike. Is there any connection between them?

Any hints/insight would be appreciated.

  • $\begingroup$ This simply follows from $E(X) = \int_0^\infty P(X\geq t)\,dt$ for any $X\geq 0$. In particular, $E(|X|^r) = \int_0^\infty P(|X|\geq t^{1/r})\,dt = \int_0^\infty t^{r-1}P(|X|\geq t)\,dt$ $\endgroup$
    – Andrew
    Sep 15, 2023 at 3:16
  • $\begingroup$ @Andrew, Thanks for commenting and yes I know the trick of variable change but I am particularly concerned with why the authors took $b^{-1}$ and how they used $\Pr[X\geq b_n]\leq \Pr[\varphi(X)\geq n]$ and how they got $(2)$ from $(1)$. $\endgroup$ Sep 15, 2023 at 3:19
  • $\begingroup$ Every result stated in your post follows from the representation $E(X) = \int_0^\infty P(X\geq t)\,dt$ for $X\geq 0$. If you are averse to using Fubini-Tonelli's theorem, then you can use a more bare bones argument as presented by the authors. $\endgroup$
    – Andrew
    Sep 15, 2023 at 3:22
  • $\begingroup$ Again I agree with you @Andrew but as I reiterated above, I want to understand the approach of the authors; what was the point of working with $b^{-1}$ not $b()$ alone and taking $b(x) = X^r $ apparently would provide $\mathrm EX^{-r}$ from $(1)$. I wanted to understand the reasoning behind this. $\endgroup$ Sep 15, 2023 at 3:26


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