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I have a question about the proof of the inequality. The well known result stats

Let $Z$ be a RV and let $0<s<t$. Then

$$E(|Z|^s)^{1/s} \leq E(|Z|^t)^{1/t}$$

The proof follows almost immediately from the Holder Inequality

$$E(|XY|)\leq E(|X|^p)^{1/p} E(|Y|^q)^{1/q}$$ for $1<p,q$ such that $1/p+1/q =1$. Taking $Z=|X|^s, Y=1, p= t/s $ we can derive the Lyapunov's inequality.

even though it's not mentioned , I think one must assume that $E(Z) =E(|X|^s)<\infty$ to apply the inequality . However, I'm not assuming that in my hypothesis.

My question is, what happens if $E(Z)=\infty$? Does the inequality still holds? In other words, $E(|X|^t)=\infty$ as well? (If $1<s<t $ it follows easyly but I don't know how to argue in other cases)

C. Zhihao
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  • Holder's inequality as stated by you holds always! Integrals of non-negative measurable functions are always defined, but the values may be $\infty$. – Kavi Rama Murthy Nov 09 '17 at 09:04
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    You should not be afraid to write $\text{something} \leq \infty$. It's true. – Gabriel Romon Nov 09 '17 at 19:47
  • What I mean, if $E(Z) =E(|X^p|)= \infty$, I get $\infty \leq$ something , following your notation. But working on @KaviRamaMurthy comment I think that the proof still remains valid under the assumption of $E(Z) = \infty$ – C. Zhihao Nov 09 '17 at 21:39

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I believe you could argue thus under the above assumptions: First $ |X|^s > 1 \Rightarrow |X| > 1 \Rightarrow |X|^s < |X|^t $. So one could write: $$ \infty = \int |X|^s = \int_{X>1} |X|^s + \int_{X \leq 1} |X|^s \leq \int_{X>1} |X|^s + 1 \leq \int_{X>1} |X|^t + 1 \leq \int |X|^t + 1, $$ from which it follows that $ E(|X|^t) = \infty $.

Radu T
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