Inequality of expectation of product Suppose non-negative random variables $X_1,X_2,\ldots,X_N$, and they are maybe dependent. Is the following inequality correct?
\begin{align}
E\left[\prod_{n=1}^{N} X_n^{k_n}\right] \le \max_n E\left[X_n^t\right],
\end{align}
where $t={\sum_nk_n}$ and $k_n\ge0$. I have this question because I want to bound the LHS with single variable $X_n$. Any idea is appreciated.
 A: Firstly, we prove the case of $N=2$.
I mainly use the Hölder's inequality. For positive random variables $X$ and $Y$, if $1/p+1/q=1$, then Hölder's inequality shows that
\begin{align}
E[X Y] \le (E[X^q])^{1/p} (E[Y^q])^{1/q}.
\end{align}
For the posted problem, we let $X=X_1^{k_1}$ and $Y=X_2^{k_2}$, then we have
\begin{align}
E[X_1^{k_1} X_2^{k_2}] \le (E[X_1^{k_1 p}])^{1/p} (E[X_2^{k_2 q}])^{1/q}.
\end{align}
Then, letting $p=\frac{k_1+k_2}{k_1}$ and $q=\frac{k_1+k_2}{k_2}$, obviously, we have $1/p+1/q=1$.
\begin{align}
E[X_1^{k_1} X_2^{k_2}] &\le (E[X_1^{k_1+k_2}])^{\frac{k_1}{k_1+k_2}} (E[X_2^{k_1+k_2 }])^{\frac{k_2}{k_1+k_2}}\\
&\le \max\{ E[X_1^{k_1+k_2}], E[X_2^{k_1+k_2}] \}.
\end{align}
The conclusion, $E[X_1^{k_1} X_2^{k_2}] \le (E[X_1^{k_1 p}])^{1/p} (E[X_2^{k_2 q}])^{1/q}$, will be ultized in the following.

For the general $N\ge2$, we use this conclusion and obtain
\begin{align}
E\left[X_1^{k_1} \cdots X_{N-1}^{k_{N-1}}X_{N}^{k_{N}}\right]\le \left(E\left[\left(X_1^{k_1} \cdots X_{N-1}^{k_{N-1}}\right)^{t/(t-k_N)}~\right] \right)^{(t-k_N)/t}\left(E\left[X_{N}^{k_{N} \times t/k_N}\right]\right)^{k_N/t},
\end{align}
where $p=t/(t-k_N)$ and $q=t/k_N$. The RHS above is
\begin{align}
\left(E\left[\left(X_1^{k_1} \cdots X_{N-1}^{k_{N-1}}\right)^{t/(t-k_N)}~\right] \right)^{(t-k_N)/t}\left(E\left[X_{N}^{t}\right]\right)^{k_N/t}.
\end{align}
For the first part, we can iteratively use the derived conclusion and obtain the following
\begin{align}
E\left[\prod_{n=1}^{N} X_n^{k_n}\right] &\le \prod_{n=1}^N\left(E\left[X_{n}^{t}\right]\right)^{k_n/t}\\
& \le \max_n E\left[X_n^t\right].
\end{align}
Note that when some $k_n=0$, the problem can be reduced to smaller $N$. Thus, without loss of generality, we assume $k_n\neq 0$ in the proof.
