0

I ran into this post which shows $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$, $p \leq q$.

So I guess it's true $({\Bbb E|X|^p})^{\frac{1}{p}} \leq ({\Bbb E|X|^q})^{\frac{1}{q}}$, if $p \leq q$, but can't find a way to show that.

Community
  • 1
Bender
  • 57
  • 6
  • This may help you. In particular, see the first answer. – David Mitra Mar 06 '14 at 15:12
  • @DavidMitra Thanks again. The link is of great help. Hoelder's inequality quoted in the first answer is slightly more general than the one in my textbook and wikipedia. I have to figure out why it holds. – Bender Mar 06 '14 at 15:19
  • Maybe this would help: http://en.wikipedia.org/wiki/Holder%27s_inequality#Probability_measure – Cm7F7Bb Mar 06 '14 at 15:19
  • @V.C. Thank you. I think it's promising, but haven't figure out how. In the first answer of Mr David Mitra's link, there's a slightly general version of Hoelder's inequality, which can be applied directly, but I don't know why it's true. – Bender Mar 06 '14 at 15:25

1 Answers1

2

Yes, it's true and follows from the Holder's inequality:

For any for $a,b>0$ such that $\frac1a+\frac1b=1$ we have

$$\int|fg|\leq(\int |f|^a)^\frac1a(\int|g|^b)^\frac1b$$

Now set $f=|X|^p$, $g=1$, $a=\frac pq$ and $b=\frac{q-p}q$, this is possible for $p<q$: $$\int||X|^p\cdot1|\leq\left(\int\left(|X|^p\right)^\frac qp\right)^\frac pq\left(\int 1^\frac q{q-p}\right)^\frac{q-p}q$$ $$\left(\int|X|^p\right)^\frac1p\leq\left(\int|X|^q\right)^\frac1q\cdot1$$

user2345215
  • 16,422