Moment inequality via a Markov-type bound

Question

This is the original question that contains some errors.

Let $X,Y:\Omega \to \mathbb{R}$ be two non negative random variables. Assume that $Y$ is integrable and the following Markov-type inequality holds for $\lambda > 0$: $$\mathbb{P}(X \geq \lambda) \leq \frac{\mathbb{E}\left[ Y : X \geq \lambda \right]}{\lambda}.$$ Show that if $Y \in L^p(\Omega)$ with $p > 1$, then $X \in L^p(\Omega)$ and $$ \mathbb{E}[X^p]^{1/p} \leq \frac{p}{p-1}\mathbb{E}[Y^p]^{1/p}. $$

A hint is given that the solution is a "clever application" of the formula $$\mathbb{E}[X^r] = r \int_0^\infty \lambda^{r-1}\mathbb{P}(X \geq \lambda)\, \mathrm{d}\lambda.$$

I have tried, in several ways, to plug this given Markov-type inequality to the formula above, but all the bounds I seem to be able to obtain are too rough to yield the desired bound. Specifically, the Markov-type inequality is equivalent to (this is not correct!) $$\lambda\mathbb{P}(X \geq \lambda)^2 \leq \mathbb{E}[\mathbb{1}(X \geq \lambda)Y],$$ then one may attempt to iterate the integrals in different ways. I don't seem to get anywhere with that approach.

Any hints or ideas are appreciated!

EDIT: I had misinterpreted some notation.

Answer 1

By definition, $E[Y:X\ge \lambda]=\int_\Omega Y1_{(X\ge \lambda)}dP$, hence we have

\begin{align} E[X^p]&=p\int_0^\infty \lambda^{p-1}P(X\ge \lambda)d\lambda\\ &\le p\int_0^\infty \lambda^{p-1}\frac{1}{\lambda}\int_\Omega Y1_{(X\ge \lambda)}dP d\lambda=p\int_\Omega Y(\omega)\int_0^{X(\omega)}\lambda^{p-2}d\lambda dP \\ &=\frac{p}{p-1}\int_\Omega X(\omega)^{p-1}Y(\omega)dP, \end{align} then using Holder inequality with $p$ and dividing $\int_\Omega X(\omega)^{p}dP$ on both sides, you can show the inequality, which implies $X\in L^p$ by the assumption.