# Bounding the total variation distance of two specific random variables?

Let $X$ and $Y$ be two independent discrete random variables, and $Z$ be a function of $X$ and $Y$, i.e., $Z=f(X,Y)$. Suppose that $\Gamma$ is a set such that $$\mathrm{Pr}[(X,Y)\in\Gamma]\geq 1-\epsilon$$ Now, we constitute $\tilde{Z}$ from $Z$ as follows \begin{align} \tilde{Z}=\begin{cases} Z&(x,y)\in\Gamma\\ \star&(x,y)\in\Gamma^{c} \end{cases} \end{align} where $\star$ is a realization of $\tilde{Z}$ which does not belong to sample space of $Z$.

Question: Does the total variation distance between $p(z,x)$ and $q(\tilde{z},x)$ (joint pmfs) is small enough. In other words, is there any $\delta(\epsilon)$ goes to zero as $\epsilon\rightarrow 0$ such that \begin{align} \lVert p(x,z)-q(x,\tilde{z})\rVert_1=\sum_{(x,z\cup\star)}\lvert p(x,z)-q(x,\tilde{z}) \rvert\leq\delta(\epsilon) \end{align} where $p(x,\star)=0$. If there is not such a $\delta$, what is the best upper-bound for total variation term?

• The sentence "$\star$ is a realization of $\tilde Z$ which does not belong to the sample space of $Z$" seems a bit circular since $\star$ is used to define $\tilde Z$. Also the joint PMFs are for which random variables? – Nawaf Bou-Rabee Dec 18 '16 at 0:42
• I mean $\tilde{Z}$ is equal to $Z$ with probability $1-\epsilon$ and equal to $\star$ with probability $\epsilon$. $p(x,z)$ is joint pmf of $X$ and $Z$, and $q(x,\tilde{z})$ is joint pmf of $X$ and $\tilde{Z}$. – Math_Y Dec 18 '16 at 0:52
• Is $\star$ a constant? – Nawaf Bou-Rabee Dec 18 '16 at 0:53
• Yes, it is constant. – Math_Y Dec 18 '16 at 0:57
• Doesn't the bound follow from the coupling interpretation of the TV distance, i.e., $d_{TV}(q,p) \le Pr( (X,Z) \ne (X, \tilde Z)) = \epsilon$? – Nawaf Bou-Rabee Dec 18 '16 at 1:06