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Suppose I have to random vectors: $$\mathbf{z} = (z_1, \dots, z_n)^T, \quad \mathbf{v} = (v_1, \dots, v_n)^T$$ and set $A \subset \mathbb{R}^n$.

I want to find an upper bound $B$ for the following difference:

$$|\mathbb{P}(\mathbf{z} \in A) - \mathbb{P}(\mathbf{v} \in A)| \le B(\mathbf{z}, \mathbf{v}).$$

I had in mind the following bound $B$: $$B(\mathbf{z}, \mathbf{v}) = \text{max}_i \; (\mathbb{P}(z_i \neq v_i)).$$

But I cannot prove this bound... I would be very grateful if anyone would helpmeet either to prove my bound or construct some other bound (ideally "summarizing" vectors coordinates information, e.g., max, min, sum of some coordinate functions).

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  • $\begingroup$ Do not use $\Bbb P$ to denote the probability. Use $P$ or $\mathsf P$ or $\text{P}$ instead. The blackboard-bold font is reserved to denote sets such as $\Bbb R$, $\Bbb C$, etc. $\endgroup$
    – Iosif Pinelis
    Commented Mar 21 at 12:40
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    $\begingroup$ @IosifPinelis, I think it's pretty common to use blackboard bold in this context... In fact, most of the books I've seen write it like this. $\endgroup$
    – tsnao
    Commented Mar 21 at 12:52
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    $\begingroup$ @Grigori, you will probably find what you're looking for if you Google total variation distance and its bounds. Moreover, there's some notation abuse going on in your suggested bound: the left-hand side doesn't depend on $\mathbf{z}$ or $\mathbf{v}$, but rather on their laws. $\endgroup$
    – tsnao
    Commented Mar 21 at 12:54
  • $\begingroup$ @tsnao : Of course, this notation is common. Still, it should not be used. $\endgroup$
    – Iosif Pinelis
    Commented Mar 21 at 16:31
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    $\begingroup$ @tsnao : I think this is a matter, not of a clash of notations, but of esthetics and of esthetical awareness. Isn't is just nicer when objects of the same kind are denoted similarly while objects of different kinds are denoted differently? $\endgroup$
    – Iosif Pinelis
    Commented Mar 21 at 17:18

1 Answer 1

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$\newcommand\v{\mathbf v}\newcommand\z{\mathbf z}\newcommand\R{\mathbb R}$This bound does not hold in such generality.

E.g., suppose that $n=2$, $A=\R^2\setminus\{(0,0)\}$, $P(\v=(0,0))=1$, $P(\z=(1,0))=1/2$, and $P(\z=(0,1))=1/2$. Then the inequality in question becomes $1\le1/2$, which is false.

What is true is that $$|P(\z\in A)-P(\v\in A)|\le P\Big(\bigcup_{i=1}^n\{z_i\ne v_i\}\Big) \le\sum_{i=1}^n P(z_i\ne v_i).$$

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