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](https://arxiv.org/pdf/math/0209021.pdf) 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](https://www.jstor.org/stable/2237258?casa_token=k0z7kEIC5zQAAAAA%3AGUNJrIaK7tlNuZMXk0GGZgURhDXke5QdrH4Klrv1EUFHJrx38434gnUID43_RuUUBU2mFSOMbACAUyjvcyb1u_rAdoLs9aITyTW3aH6GWyHZe3v9zgb8)) 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.



Bounds on discrepancy metric of product measures
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.


