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Suppose that $X$ is a positive random variable. And then we can compute that $EX^p=\int_0^\infty P\{X>t\}dt^p$. so that $EX^p<\infty$ is equivalent to $\int_0^\infty P(X>t)dt^p<\infty$.

Then I want to show that $\int_0^\infty P(X>t)dt^p<\infty$ is equivalent to $\sum_{n=1}^\infty n^{p-1}P(X\ge n)<\infty$. I just partition $[0,\infty)$ into [n,n+1). But this ends up with $\sum_{n=0}^\infty n^{p-1}P(X>n+1)<\int_0^\infty t^{p-1} P\{X>t\}dt<\sum_{n=0}^\infty (n+1)^{p-1}P(X>n)$. I am wondering how can I get that form $\sum_{n=1}^\infty n^{p-1}P(X\ge n)$?

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up vote 0 down vote accepted

For the right bound. Note that $(n+1)<2n$. Then we can show the equivalence.

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