Prove $\mathbb{E}|X|^\alpha<\infty \Longleftrightarrow \mathbb{E}|X-a|^\alpha<\infty$

Asked May 19 '21 at 12:08

Active May 19 '21 at 12:08

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I want to prove that if $\alpha >0, a\in\mathbb{R}$ then $$\mathbb{E}|X|^\alpha<\infty \Longleftrightarrow \mathbb{E}|X-a|^\alpha<\infty$$ $\Longleftarrow$ is trivial taking $a=0$. However I am having trouble with $\Longrightarrow$. One of the strategies I have tried is proving for $X=\textbf{1}_A$ which guarantees the result for simple functions, then for positive ones and then for any real $X$. However I don't think I can use this here since I need to use the linearity of the integral and the absolute value and the power don't allow me to do this.

I have also tried Jensen's inequality since $|\cdot|^\alpha$ is convex if $\alpha \geq 1$, but this has been a failure as well. Finally, I attempted to write the random variable $X$ as an integral of an indicator function as done here, but again the power is bothering me.

I am sure there is a very easy way to prove this but I am very stuck. Can someone help me?

asked May 19 '21 at 12:08

kubo

2,198

5

$|x-a|^y<||x|+|a||^y<(2\max{|x|,|a|})^y=2^y\max{|x|^y, |a|^y}<2^y|x|^y+2^y|a|^y$ – Bananach May 19 '21 at 12:13
Replace $<$ by $\leq$ – Bananach May 19 '21 at 12:14
You cannot take $a=0$. You are asked to prove the equivalence for any fixed $a$ that is given to you. – Kavi Rama Murthy May 19 '21 at 12:14

Prove $\mathbb{E}|X|^\alpha<\infty \Longleftrightarrow \mathbb{E}|X-a|^\alpha<\infty$

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