# Bounds on discrepancy metric of product measures

Consider two measurable spaces $$X_1 = (\mathbb{R}^m,\mathcal{B}(\mathbb{R}^m),\mu_1)$$ and $$X_2 = (\mathbb{R}^m,\mathcal{B}(\mathbb{R}^m),\mu_2)$$ and the product spaces $$X_1^{q} = (\times_{i=1}^q\mathbb{R}^m,\otimes_{i=1}^q\mathcal{B}(\mathbb{R}^{m}),\mu_1^{\otimes q})\ \ \ \text{ and }\ \ \ X_2^{q} = (\times_{i=1}^q\mathbb{R}^m,\otimes_{i=1}^q\mathcal{B}(\mathbb{R}^{m}),\mu_2^{\otimes q})$$ where $$\mu_1^{\otimes q} = \underbrace{\mu_1\otimes \cdots \otimes\mu_1}_{q}$$ and $$\mu_2^{\otimes q} = \underbrace{\mu_2\otimes \cdots \otimes\mu_2}_{q}$$, respectively.

Let $$D(\mu,\nu) = \sup_{\substack{\text{all closed balls } B \text{ of }\mathbb{R}^m\\ \text{in the Euclidean norm} }} \left|\mu(B)-\nu(B)\right|$$ be the discrepancy metric between probability measures $$\mu$$ and $$\nu$$ (see e.g. this paper for more details on this metric).

My question. For probability measures $$\mu_1$$ and $$\mu_2$$, does there exist $$k>0$$ independent of $$\mu_1$$ and $$\mu_2$$ such that $$D(\mu_1^{\otimes q},\mu_2^{\otimes q})\le k D(\mu_1,\mu_2)\ \ \ ?$$

Note. For other probability metrics the answer is in the affirmative (e.g., for the total variation metric, see Eq. (4.5) of this paper) but I couldn't find anything about the discrepancy metric above. I suspect that, if true, this should be a rather known result. Any suggestion or comment is very welcome.

• So, you want $k$ to depend only on $q$, right? Commented Aug 8, 2022 at 22:01
• @IosifPinelis: Yes. Commented Aug 9, 2022 at 6:23
• I’m guessing you want the Euclidean distance to define the closed balls in $\mathbb R^d$? If you are allowed to use $\ell_\infty$ distance instead, this seems as though it becomes fairly straightforward? Commented Aug 10, 2022 at 6:54
• @AnthonyQuas: Yes, the balls are in the Euclidean norm. I edited the question to clarify this. Thanks for your comment. Commented Aug 10, 2022 at 7:09
• Something is not clear to me. The two sides of the inequality are not homogeneous. So the best constant for $2\mu_1$ and $2\mu_2$ is $2^{q-1}$ times the best constant for $\mu_1$ and $\mu_2$, so there exists no $k$ independent from $\mu_1$ and $\mu_2$. Maybe you are talking of probability measures? Commented Aug 19, 2022 at 14:10

Analogous to the TV metric, the requested upper bound holds for the discrepancy metric with $$k=q$$. The result given below can also be easily extended to general product probability measures $$\mu=\otimes_{i=1}^q \mu_{i}$$ and $$\nu=\otimes_{i=1}^q \nu_i$$ to obtain $$D(\mu,\nu)\le \sum_{i=1}^q D(\mu_i, \nu_i)\;.$$
Suppose that $$\mu_1$$ and $$\mu_2$$ are probability measures. Let $$B = \{ (x_1, \dots, x_q) : \sum_i (x_i - c_i)^2 \le R^2 \}$$ be a closed ball in $$\mathbb{R}^q$$ with radius $$R>0$$ centered at $$(c_1, \dots, c_q) \in \mathbb{R}^q$$. Let $$\mathbf{1}_A$$ denote the indicator function of the set $$A$$.
By telescoping, and invoking Tonelli’s theorem to write $$(\mu_1^{\otimes q}-\mu_2^{\otimes q})(B)$$ as an iterated integral, note that \begin{align*} & |(\mu_1^{\otimes q}-\mu_2^{\otimes q})(B)|= \left| \int_{\mathbb{R}^q} \mathbf{1}_B(x_1, \dots, x_q) \biggl( d\mu_1(x_1) \cdots d\mu_1(x_q) - d\mu_2(x_1) \cdots d\mu_2(x_q) \biggr) \right| \\ &= \left| \int_{\mathbb{R}^q} \mathbf{1}_B(x_1, \dots, x_q) \sum_{ i=1}^q d\mu_1(x_1) \cdots d\mu_1(x_{i-1}) \biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i})\biggr) d\mu_2(x_{i+1}) \dots d\mu_2(x_q) \right| \\ &= \left| \sum_{ i=1}^q \int_{\mathbb{R}^{q-1}} \left(\int_{\mathbb{R}} \mathbf{1}_B(x_1, \dots, x_q)\biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i})\biggr) \right) d\mu_1(x_1) \cdots d\mu_1(x_{i-1}) d\mu_2(x_{i+1}) \dots d\mu_2(x_q) \right| \\ &\le \sum_{ i=1}^q \int_{\mathbb{R}^{q-1}} \left| \int_{\mathbb{R}} \mathbf{1}_B(x_1, \dots, x_q)\biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i})\biggr) \right| d\mu_1(x_1) \cdots d\mu_1(x_{i-1}) d\mu_2(x_{i+1}) \dots d\mu_2(x_q) \\ &\le \sum_{i=1}^q \sup_{\gamma_i, \rho_i} \left| \int_{\mathbb{R}}\mathbf{1}_{(x_i-\gamma_i)^2 \le \rho_i^2} \biggl(d\mu_1(x_{ i}) - d\mu_2(x_{ i}) \biggr) \right| \\ & \qquad \int_{\mathbb{R}^{q-1}} d\mu_1(x_1) \cdots d\mu_1(x_{i-1}) d\mu_2(x_{i+1}) \cdots d\mu_2(x_q) \\ &\le q D(\mu_1, \mu_2) \;. \end{align*} Since $$B$$ is arbitrary, the conjectured upper bound holds with $$k=q$$.
Remark. One can do much better than this linear upper bound in some special cases. For instance, suppose that $$\mu_1 = \mathcal{N}(0,1)$$ and $$\mu_2 = \mathcal{N}(0, \sigma^2)$$. Then, by moving to hyperspherical coordinates, it's not too hard to show that $$D(\mu_1^{\otimes q},\mu_2^{\otimes q}) = \left| \frac{\Gamma(\frac{q}{2}, \frac{q \log{\sigma}}{\sigma^2-1}) - \Gamma(\frac{q}{2}, \frac{q \sigma^2 \log(\sigma)}{\sigma^2-1})}{ \Gamma(\frac{q}{2}) } \right| \;$$ which converges to one with $$q$$ and $$k(q):= D(\mu_1^{\otimes q},\mu_2^{\otimes q}) /D(\mu_1,\mu_2)$$ grows sublinearly. This Gaussian case is rather exceptional because the corresponding product measure can be directly written in terms of the Euclidean distance.