For $r′ < r,$ show that $|X|^{r′} ≤ 1 + |X|^r $ and conclude that if $E|X|^r < \infty$, then $E|X|^{r′} < E|X|^r$ for all $0< r′ < r.$

Question

For $r′ < r$, show that $|X|^{r′} \leq 1 + |X|^r$ and conclude that if $E[|X|^r] < \infty$, then $E[|X|^{r′}]<\infty$ for all 0 < r′ < r.

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I tried to solve this by the Basic Properties of the Expectation of an R.V we have

If E|X|r < ∞ for some r > 0, where X is an r.v., then E|X|r′< ∞ for all 0 < r′ < r

This is a consequence of the obvious inequality |X|r′ ≤ 1 + |X|r

could any one help me please

Answer 1

Hint for the inequality $|X|^{r'} \le 1+ |X|^r$ for all $1 \le r' \le r$:
Consider $|X|^{r'} = |X|^{r'} \mathbf{1}_{\{|X|^{r'} \le 1\}} + |X|^{r'} \mathbf{1}_{\{|X|^{r'}>1\}}$.

And consider the function $y \mapsto x^y$ for $x \in [0,\infty]$.

Answer 2

to prove the inequality |X|r′ ≤ 1 + |X|r

considering |X|r′=|X|r′1{|X|r′≤0}+|X|r′1{|X|r′>0}

as r′< r then |X|r′≤ |X|r1{|X|r′≤0}+|X|r1{|X|r′>0}

And consider the function y↦x^y for x∈[0,∞], also 0< r′< r

thus |X|r′≤ 1 + |X|r

is this correct