$l_1$ distance between $n$th fold product distributions $p^n$ and $q^n$

Question

Let $p$ and $q$ be two probability distributions on a finite alphabet $\mathcal{X}$. The $l_1$ distance between them is defined as $$l_1(p, q) = \sum_{x \in \mathcal{X}} |p(x) - q(x)|$$

Consider $n$th fold product distribution of $p$, namely $p^n$ given by $$p^n(x^n) = \prod_{i=1}^n p(x_i)$$ for all $x^n \in \mathcal{X}^n$.

Can we express $l_1(p^n, q^n)$ in terms of $l_1(p, q)$ or at least provide an upper or lower bound on $l_1(p^n, q^n)$ in terms of $l_1(p, q)$?

\begin{align} l_1(p^n, q^n) &= \sum_{x^n \in \mathcal{X}^n} |p^n(x^n) - q^n(x^n)|\\ &= \sum_{x^n \in \mathcal{X}^n} \Bigg | \prod_{i=1}^n p(x_i) - \prod_{i=1}^n q(x_i) \Bigg | \end{align}

Answer 1

I figured it out.

Consider first the case for $n=2$ and arbitrary product distributions $p_1 \times p_2$ and $q_1 \times q_2$

\begin{align*} l_1(p_1 \times p_2, q_1 \times q_2) &= \sum_{x_1} \sum_{x_2} |p_1(x_1) p_2(x_2) - q_1(x_1) q_2(x_2)|\\ &= \sum_{x_1} \sum_{x_2} |p_1(x_1) p_2(x_2) - p_1(x_1) q_2(x_2) + p_1(x_1) q_2(x_2) - q_1(x_1) q_2(x_2)|\\ &= \sum_{x_1} \sum_{x_2} |p_1(x_1)(p_2(x_2) - q_2(x_2)) + q_2(x_2) (p_1(x_1) - q_1(x_1)) |\\ &\leq \sum_{x_1} \sum_{x_2} p_1(x_1) |p_2(x_2) - q_2(x_2)| + \sum_{x_1} \sum_{x_2} q_2(x_2) |p_1(x_1) - q_1(x_1)|\\ &= \sum_{x_2} |p_2(x_2) - q_2(x_2)| + \sum_{x_1} |p_1(x_1) - q_1(x_1)|\\ &= l_1(p_1, q_1) + l_1(p_2, q_2) \end{align*}

This implies $$l_1(p^2, q^2) \leq 2 l_1(p, q)$$

Now use induction with similar arguments as before.