Show that for $p>1,$ $\mathbb{E}[Y^p] \leq \big(\frac{p}{p-1}\big)^p\mathbb{E}[X^p]$

Question

Let $X$ and $Y$ be non-negative r.v. defined on the same probability space $(\Omega,\mathcal{F},\mathbb{P})$ such that

$$\mathbb{P}(Y>y)\leq\frac1y \int_{Y\geq y}^{} X d\mathbb{P} $$

Show that for $p>1$,

$$\mathbb{E}[Y^p]\leq\Big(\frac{p}{p-1}\Big)^p \mathbb{E}[X^p]$$

Suggestion:

Recall that for a non-negative random variable $X$,

$$||X||_p^p=\mathbb{E}[X^p]=\int_{0}^{\infty}px^{p-1}\mathbb{P}(X>x)$$

and Holder's inequality states that for $p,q$ such that $\frac1p + \frac1q = 1$,

$$\mathbb{E}[ZW]\leq||Z||_p||W||_q=\mathbb{E}[Z^p]^{1/p}\mathbb{E}[W^q]^{1/q}$$

Answer 1

Big hints:

• Start with $E[Y^p]$ and use the given formula which will have a $P(Y > y)$ in the integrand.
• Use the given inequality to bound $P(Y > y)$ in the integrand.
• Exchange the order of integration and evaluate the inner integral.
• Use Hölder's inequality on the remaining integral.

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Response to comment:

$$\int_0^\infty p y^{p-2} \int_\Omega X 1_{Y \ge y} \,dP \, dy = p \int_\Omega X \int_0^Y y^{p-2} \, dy \, dP$$