Supremum over Markov Chains of probability of union

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I found the following proof in the literature, but I am struggling to understand why certain steps are true:



I have the following chain of equalities:



$sup_X-Y-(V_i)_i=1^k Pr(V=hatV_1 lor cdots lor V=hatV_k) = \ sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) $



Is this equality true because the supremum is achieved when the $hatV_j$ are disjoint and thus I can decompose $P_X(V=hatV_1 lor cdots lor V=hatV_k|X=x) = sum_i=1^k P_X(v_i|x)$? If yes, why?
Also, afterwards it says:



$sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) = \ sum_yinmathcalYsum_i=1^k maxlimits_x_ineq v_1,ldots,v_i-1 sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x)
$



Why is this? Why should the maximum be achieved selecting these $v_i$ in a Greedy fashion? Is this always true?



Thank you for any help!




probability optimization markov-chains supremum-and-infimum




share|cite|improve this question



















  • It would help to give some context, namely what $X$, $Y$ and $V_i$ are.
    – Math1000
    Aug 2 at 2:25














up vote
0
down vote

favorite












I found the following proof in the literature, but I am struggling to understand why certain steps are true:



I have the following chain of equalities:



$sup_X-Y-(V_i)_i=1^k Pr(V=hatV_1 lor cdots lor V=hatV_k) = \ sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) $



Is this equality true because the supremum is achieved when the $hatV_j$ are disjoint and thus I can decompose $P_X(V=hatV_1 lor cdots lor V=hatV_k|X=x) = sum_i=1^k P_X(v_i|x)$? If yes, why?
Also, afterwards it says:



$sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) = \ sum_yinmathcalYsum_i=1^k maxlimits_x_ineq v_1,ldots,v_i-1 sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x)
$



Why is this? Why should the maximum be achieved selecting these $v_i$ in a Greedy fashion? Is this always true?



Thank you for any help!




probability optimization markov-chains supremum-and-infimum




share|cite|improve this question



















  • It would help to give some context, namely what $X$, $Y$ and $V_i$ are.
    – Math1000
    Aug 2 at 2:25












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I found the following proof in the literature, but I am struggling to understand why certain steps are true:



I have the following chain of equalities:



$sup_X-Y-(V_i)_i=1^k Pr(V=hatV_1 lor cdots lor V=hatV_k) = \ sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) $



Is this equality true because the supremum is achieved when the $hatV_j$ are disjoint and thus I can decompose $P_X(V=hatV_1 lor cdots lor V=hatV_k|X=x) = sum_i=1^k P_X(v_i|x)$? If yes, why?
Also, afterwards it says:



$sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) = \ sum_yinmathcalYsum_i=1^k maxlimits_x_ineq v_1,ldots,v_i-1 sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x)
$



Why is this? Why should the maximum be achieved selecting these $v_i$ in a Greedy fashion? Is this always true?



Thank you for any help!




probability optimization markov-chains supremum-and-infimum




share|cite|improve this question











I found the following proof in the literature, but I am struggling to understand why certain steps are true:



I have the following chain of equalities:



$sup_X-Y-(V_i)_i=1^k Pr(V=hatV_1 lor cdots lor V=hatV_k) = \ sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) $



Is this equality true because the supremum is achieved when the $hatV_j$ are disjoint and thus I can decompose $P_X(V=hatV_1 lor cdots lor V=hatV_k|X=x) = sum_i=1^k P_X(v_i|x)$? If yes, why?
Also, afterwards it says:



$sum_yinmathcalY maxlimits_substackv1,ldots,v_k\v_ineq v_j, ineq j sum_i=1^k sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x) = \ sum_yinmathcalYsum_i=1^k maxlimits_x_ineq v_1,ldots,v_i-1 sum_xinmathcalX P_X(x)P_X(v_i|x)P_Y(y|x)
$



Why is this? Why should the maximum be achieved selecting these $v_i$ in a Greedy fashion? Is this always true?



Thank you for any help!





probability optimization markov-chains supremum-and-infimum





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share|cite|improve this question




share|cite|improve this question









asked Aug 1 at 15:08









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  • It would help to give some context, namely what $X$, $Y$ and $V_i$ are.
    – Math1000
    Aug 2 at 2:25
















  • It would help to give some context, namely what $X$, $Y$ and $V_i$ are.
    – Math1000
    Aug 2 at 2:25















It would help to give some context, namely what $X$, $Y$ and $V_i$ are.
– Math1000
Aug 2 at 2:25




It would help to give some context, namely what $X$, $Y$ and $V_i$ are.
– Math1000
Aug 2 at 2:25















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