Supremum over Markov Chains

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I came across the following maximization problem over Markov Chians in the literature:



$$sup_U-X-Y-hatU logfracPr(U=hatU)max_uinmathcalUP_U(u)$$



and it says that we can easily rewrite the formula as follows:
$$ sup_U-X-Y logfracsum_yinmathcalYmax_uinmathcalU P_UY(u,y)max_uinmathcalUP_U(u)$$



I think I am missing something about Markov Chains here. I tried re-writing $Pr(U=hatU)=sum_u P(U=u,hatU=u)= sum_ysum_u P(U=u,hatU=u,Y=y)$
and then, using the Markov Chain Property I get, $sum_ysum_u P_UY(u,y)P_Y(u|y)$ 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!




probability markov-chains




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  • It would help to give some context, namely what $U$, $hat U$, $X$, and $Y$ are.
    – Math1000
    Aug 2 at 2:34














up vote
1
down vote

favorite












I came across the following maximization problem over Markov Chians in the literature:



$$sup_U-X-Y-hatU logfracPr(U=hatU)max_uinmathcalUP_U(u)$$



and it says that we can easily rewrite the formula as follows:
$$ sup_U-X-Y logfracsum_yinmathcalYmax_uinmathcalU P_UY(u,y)max_uinmathcalUP_U(u)$$



I think I am missing something about Markov Chains here. I tried re-writing $Pr(U=hatU)=sum_u P(U=u,hatU=u)= sum_ysum_u P(U=u,hatU=u,Y=y)$
and then, using the Markov Chain Property I get, $sum_ysum_u P_UY(u,y)P_Y(u|y)$ 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!




probability markov-chains




share|cite|improve this question



















  • It would help to give some context, namely what $U$, $hat U$, $X$, and $Y$ are.
    – Math1000
    Aug 2 at 2:34












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I came across the following maximization problem over Markov Chians in the literature:



$$sup_U-X-Y-hatU logfracPr(U=hatU)max_uinmathcalUP_U(u)$$



and it says that we can easily rewrite the formula as follows:
$$ sup_U-X-Y logfracsum_yinmathcalYmax_uinmathcalU P_UY(u,y)max_uinmathcalUP_U(u)$$



I think I am missing something about Markov Chains here. I tried re-writing $Pr(U=hatU)=sum_u P(U=u,hatU=u)= sum_ysum_u P(U=u,hatU=u,Y=y)$
and then, using the Markov Chain Property I get, $sum_ysum_u P_UY(u,y)P_Y(u|y)$ 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!




probability markov-chains




share|cite|improve this question











I came across the following maximization problem over Markov Chians in the literature:



$$sup_U-X-Y-hatU logfracPr(U=hatU)max_uinmathcalUP_U(u)$$



and it says that we can easily rewrite the formula as follows:
$$ sup_U-X-Y logfracsum_yinmathcalYmax_uinmathcalU P_UY(u,y)max_uinmathcalUP_U(u)$$



I think I am missing something about Markov Chains here. I tried re-writing $Pr(U=hatU)=sum_u P(U=u,hatU=u)= sum_ysum_u P(U=u,hatU=u,Y=y)$
and then, using the Markov Chain Property I get, $sum_ysum_u P_UY(u,y)P_Y(u|y)$ 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!





probability markov-chains





share|cite|improve this question










share|cite|improve this question




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asked Aug 1 at 14:56









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  • It would help to give some context, namely what $U$, $hat U$, $X$, and $Y$ are.
    – Math1000
    Aug 2 at 2:34
















  • It would help to give some context, namely what $U$, $hat U$, $X$, and $Y$ are.
    – Math1000
    Aug 2 at 2:34















It would help to give some context, namely what $U$, $hat U$, $X$, and $Y$ are.
– Math1000
Aug 2 at 2:34




It would help to give some context, namely what $U$, $hat U$, $X$, and $Y$ are.
– Math1000
Aug 2 at 2:34















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