MHB Can You Tackle Holder's Inequality with Finite Moments in Statistics?

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Here is this week's POTW:

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Let $X$ and $Y$ be random variables. Suppose $-\infty < q < 0$ and $p > 0$ such that $\frac{1}{p} + \frac{1}{q} = 1$. Show that if $X$ and $Y$ have finite $p$-th and $q$-th absolute moments, respectively, then

$$\left(\Bbb E[|X|^p]\right)^{1/p} \cdot \left(\Bbb E[|Y|^q] \right)^{1/q}\le \Bbb E|XY|.$$

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No one answered this week's problem. You can read my solution below.

Spoiler

Since $\frac{1}{p} + \frac{1}{q} = 1$, then $p + q = pq$. Using that fact, we find that the numbers $-q/p$ and $1/p$ are conjugate exponents; they are positive numbers such that

$$\frac{1}{-q/p} + \frac{1}{1/p} = -\frac{p}{q} + p = \frac{-p + pq}{q} = \frac{-p + (p + q)}{q} = \frac{q}{q} = 1.$$

Hence, by Holder's inequality,

$$\Bbb E[|X|^p] = \Bbb E[|XY|^p|Y|^{-p}] \le \left(\Bbb E[(|XY|^p)^{1/p}]\right)^{p} \left(\Bbb E[(|Y|^{-p})^{-q/p}]\right)^{-p/q} = (\Bbb E[|XY|])^p\, (\Bbb E[|Y|^q])^{-p/q}.$$

Taking the $p$th root results in

$$(\Bbb E[|X|^p])^{1/p} \le \Bbb E[|XY|]\, (\Bbb E[|Y|^q])^{-1/q}.$$

Multiplying this inequality by $(\Bbb E[|Y|^q])^{1/q}$ gives the result.