I came across the following maximization problem over Markov Chians in the literature:
$$\sup_{U-X-Y-\hat{U}} \log{\frac{\Pr(U=\hat{U})}{\max_{u\in\mathcal{U}}P_U(u)}}$$
and it says that we can easily rewrite the formula as follows: $$ \sup_{U-X-Y} \log{\frac{\sum_{y\in\mathcal{Y}}\max_{u\in\mathcal{U}} P_{UY}(u,y)}{\max_{u\in\mathcal{U}}P_U(u)}}$$
I think I am missing something about Markov Chains here. I tried re-writing $\Pr(U=\hat{U})=\sum_u P(U=u,\hat{U}=u)= \sum_y\sum_u P(U=u,\hat{U}=u,Y=y)$ and then, using the Markov Chain Property I get, $\sum_y\sum_u P_{UY}(u,y)P_{\hat{U}|Y}(u|y)$ but I am not sure on how I should proceed, also, where does that $\max$ come from? Shouldn't there be a $\leq $ relationship between these two?
Thanks for any help!