4
$\begingroup$

Let $X_i, Y_i$ be i.i.d Bernoulli $0/1$ random variables with $\mathbb{E}[X_i] = p$ and $\mathbb{E}[Y_i] = q$.

Let

\begin{align*} X &= X_1 X_2 + Χ_2 Χ_3 + \ldots +X_{n-2} X_{n-1}+ X_{n-1} X_n\\ Y &= Y_1 Y_2 + Y_2 Y_3 + \ldots +Y_{n-2} Y_{n-1}+ Y_{n-1} Y_n \end{align*}

I would like to get an upper bound on the total variation distance of $X$ and $Y$.

Right now I have the following bound using data processing, and Pinsker's inequalities.

\begin{align*} d_{tv}(X,Y) &\leq d_{tv}((X_1,\ldots,X_n),(Y_1, \ldots, Y_n)) \\ &\leq \sqrt{\frac12 D((X_1,\ldots,X_n)\|(Y_1, \ldots, Y_n))} \\ &= \sqrt{\frac{n}{2} D(X_1\|Y_1)} \\ &= \sqrt{\frac{n}{2} \left(p \log \frac{p}{q} + (1-p) \log \frac{1-p}{1-q} \right)}\\ &\leq 4 \sqrt{\frac{n}{(1-p)p}} | p-q| \end{align*}

I would like to get a better upper bound, since I suspect that the correct bound should be $4 \sqrt{\frac{n}{(1-p)}} | p-q|$.

The following plot shows the exact total variation distance (blue line) for $p \in (0,1)$, $q = p+0.01$,$n=100$ and the upper bound $4 \sqrt{\frac{n}{(1-p)}} | p-q|$ (green line).

tvd

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.