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If $X$ is a random variable then is it true that,

$$\liminf_{p \rightarrow 0}\ \lVert X \rVert_ p = \exp [ \mathbb{E} [ \log \lvert X \rvert]]\quad ?$$

I am unable to see this to be true even for finite random variables. Like if $X$ takes two values say $x_1$ and $x_2$ with probabilities $p_1 \in [0,1]$ and $1-p_1$ respectively then the above claim is saying that,

$\liminf _{p \rightarrow 0} (p_1 x_1 ^p + (1-p_1)x_2^p )^\frac{1}{p} = x_1^{p_1} x_2^{1-p_1} $

I don't know why the above must be true!

gradstudent
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1 Answers1

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From here you get $$\displaystyle\lim_{p\to 0} \left[ \int_{\Omega}|X(\omega)|^p\mu (d\omega ) \right]^{\frac{1}{p}}=\exp \left[ \int_{\Omega}\log|X(\omega )| \mu (d\omega) \right],$$ what can be rewritten as $$\liminf_{p \rightarrow 0}\ \lVert X \rVert_ p = \exp [ \mathbb{E} [ \log \lvert X \rvert]]\quad $$

Community
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