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If $X$ and $Y$ are (not necessarily independent) random variables taking values in $\Omega=\{1,\ldots,n\}$. then:

$\sum_{i=1}^nP(X=i,Y=i)\leq1-\frac12\sum_{i=1}^n\mid P(X=i)-P(Y=i)\mid$

I am only 99.9% sure this inequality is true. I hope someone can prove it. Thanks in advance!

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1 Answer

up vote 2 down vote accepted

This inequality follows from a standard coupling argument.

Let $A$ be the set of $i\in \Omega$ with $\mathbb{P}(X=i)>\mathbb{P}(Y=i)$ and note that $${1\over 2}\sum_i |\mathbb{P}(X=i) -\mathbb{P}(Y=i)| =\mathbb{P}(X\in A)-\mathbb{P}(Y\in A).$$ The right hand side can be rewritten, then bounded
$$ \mathbb{P}(X\in A, X\neq Y) -\mathbb{P}(Y\in A, X\neq Y) \leq \mathbb{P}(X\neq Y)$$ which gives your result.

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