# Why is $b^{-1}(\cdot)$ used in $\sum_{n=1}^\infty \Pr[X\geq b_n]\leq \mathrm Eb^{-1}(X)\leq\sum_{n=0}^\infty \Pr[X\geq b_n]$ rather than $b(\cdot)$?

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 $$Y:=\sum_{j=1}^\infty j\mathrm I_{\{j\leq \varphi(X) and showed $$\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.

• This simply follows from $E(X) = \int_0^\infty P(X\geq t)\,dt$ for any $X\geq 0$. In particular, $E(|X|^r) = \int_0^\infty P(|X|\geq t^{1/r})\,dt = \int_0^\infty t^{r-1}P(|X|\geq t)\,dt$ Sep 15, 2023 at 3:16
• @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 $(1)$. Sep 15, 2023 at 3:19
• 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. Sep 15, 2023 at 3:22
• 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. Sep 15, 2023 at 3:26