Bound on expectation using the Markov's inequality I'd like to show $$E\Big(|X|^k1_{\{|X|^k\ge M\}}\Big) \le M^{1-\frac{p}{k}}E|X|^p,$$
where $X$ is a r.v., $k<p$, and $M$ is a positive real number. It is suggested to use the Markov's inequality (Asymptotic Statistics, Van der Vaart, example 2.21). It seems to me that Markov's inequality alone is not enough to obtain the rhs bound. I am wondering what other result can be used along with the Markov's inequality.
 A: $$
|X|^k1_{\{|X|^k\geqslant M\}}=|X|^p\color{red}{\left\lvert X\right\rvert^{k-p}1_{\{M^{1/k}\leqslant |X|\}}}=|X|^p\color{red}{\left\lvert X\right\rvert^{k-p}1_{\{|X|^{k-p}\leqslant M^{\left(k-p\right) /k}\}}}
\leqslant |X|^p\color{red}{ M ^{\left(k-p\right)/k} },
$$
then integrate.
A: It turns out that one way around this problem is using the Holder's inequality: $$E|XY| \le (E|X^r|)^{1/r}(E|X^s|)^{1/s},$$ where $r, s>1$ and $\frac{1}{r} + \frac{1}{s} = 1$.
Setting $r=p/k$ and $s=p/(p-k)$, Holder inequality gives:
$$E\Big(|X^k|1_{\{|X|^k\ge M\}}\Big) \le \big(E|(X^k)^{p/k}|\big)^{\frac{k}{p}}\big(E1^{p/(p-k)}_{\{|X|^k\ge M\}}\big)^{\frac{p-k}{p}}=$$
$$\big(E|X^p|\big)^{\frac{k}{p}}\big(E1_{\{|X|^k\ge M\}}\big)^{\frac{p-k}{p}} = \big(E|X^p|\big)^{\frac{k}{p}}P(|X|^k\ge M)^{\frac{p-k}{p}} = \big(E|X^p|\big)^{\frac{k}{p}}P(|X|\ge M^{1/k})^{\frac{p-k}{p}}\le$$
$$\big(E|X^p|\big)^{\frac{k}{p}}\Big(\frac{E|X^p|}{M^{p/k}}\Big)^{\frac{p-k}{p}}= M^{1-\frac{p}{k}}E|X|^p,$$
where the Markov' inequality is used to get the last inequality.
A: Partial answer: a standard argument using Fubini's Theorem gives $E|X|^{k}I_{\{|X|^{k}\geq M \}} =\int_M^{\infty} P\{|X|^{k} >t\} \, dt$. Hence $E|X|^{k}I_{\{|X|^{k}\geq M \}} = \int_M^{\infty} P\{|X|^{p} >t^{p/q}\} \, dt \leq \int_M^{\infty} t^{-p/q}E|X|^{p}$. Evaluating the integral we get $E|X|^{k}I_{\{|X|^{k}\geq M \}} \leq cM^{1-\frac p k} E|X|^{p}$ where $c=\frac 1 {\frac p k -1}$ I believe that the constant $c$ cannot be avoided but I don't have a counter-example to the original statement at this moment. 
