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I've been trying to think up different examples of functions such that $EZ^p = \infty$ (with $Z>0$) for all $p$, but each time it becomes rather messy. Can anyone suggest some interesting but simple examples to me?

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up vote 0 down vote accepted

On $[0,1]$, try $$Z:=\sum_{j=1}^{+\infty}e^{e^j}\chi_{(2^{—j},2^{-(j+1)})}.$$

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Thanks very much – user Nov 18 '12 at 21:22

$\Omega = \mathbb{R}$ and $\mu((a,b)) = b-a$. $$Z = 1 ,\,\,\,\,\,\,\, \forall x \in \Omega$$ Then $\forall p \in \mathbb{R}$, $$\int_{\Omega} Z^p d \mu = \int_{\mathbb{R}} d\mu = \mu(\mathbb{R}) = \infty$$

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On which probability space are you working? – Davide Giraudo Nov 18 '12 at 21:22
@DavideGiraudo Thanks for pointing out the mistake. For some reason, I thought the previous answer I had will be $\infty$ for $p < -1$, as well. – user17762 Nov 18 '12 at 21:38

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