Question: 1.15. If p > 0, e'(IXIP) < 00, then xP~{IXI > x} = 0(1) as x --+ 00. Conversely, if xP3l{IXI > x} = 0(1),

1.15. If p > 0, e'(IXIP) < 00, then xP~{IXI > x} = 0(1) as x --+ 00.

Conversely, if xP3l{IXI > x} = 0(1), then (f'(IXIP-E) < 00 for 0 < E < p.

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