Proof for $\sup\limits_{i_1,i_2,...i_j}|E(X_{i_1}X_{i_2}...X_{i_j})| = E(|X_{1}|^j)$

Question

Let $X_{1},X_{2},...,X_{n} $ be i.i.d such that $E(X_{1})< ∞$.

Let $i_{j}∈\{1,2,...,n\}$ with $j<n$.

Suppose $E(|X_{1}|^j)$ exists.

Could someone give me a proof that

$$\sup_{i_1,i_2,...i_j}|E(X_{i_1}X_{i_2}...X_{i_j})| = E(|X_{1}|^j)\ ?$$

Answer 1

Let $c_i \in \{0, 1, \ldots, j\}$, $i \in \{1, 2, \ldots, n\}$ such that $\displaystyle \sum_{i=1}^n c_i = j$. Now the LHS can be expressed as $$ \sup_{c_1, c_2, \ldots, c_n}\mathbb{E}\left[\prod_{i=1}^n |X_i|^{c_i}\right]$$

Note that $$ \mathbb{E}\left[\prod_{i=1}^n |X_i|^{c_i}\right] = \prod_{i=1}^n\mathbb{E}\left[ |X_i|^{c_i}\right] \leq \prod_{i=1}^n\mathbb{E}\left[ |X_i|^{j}\right]^{\frac {c_i} {j}} =\mathbb{E}\left[ |X_1|^{j}\right]^{\frac {1} {j}\sum_{i=1}^nc_i} = \mathbb{E}\left[ |X_1|^{j}\right] $$

where the first equality follows by independence, the inequality is the Jensen's inequality (we apply on those $c_i$ are non-zero, and for those $c_i = 0$, equality holds as both equal to $1$ anyway). The next equality follows by the identical distribution and collecting the like terms, and the last equality follows by the assumption.

Equality holds if and only if there is exactly one of the $c_i = j$ and all others vanish. Since the above statement holds for any arbitrary $c_i$ satisfying the assumption and it is attainable, it is indeed the supremum.