Inequality of expectation of product

Question

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.

Answer 1

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.

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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.