Martingale in probability and Statistics [duplicate]












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  • Supermartingale with constant Expectation is a martingale

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Let $(X_n)$ be a supermartingale such that $EX_n$ is constant. show that $X_n$ is a martingale. My problem here in this question is how can I use the constant expectation concept given to prove the statement.






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marked as duplicate by Scientifica, Davide Giraudo, NCh, jgon, Paul Frost Dec 1 '18 at 0:10


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    0















    This question already has an answer here:




    • Supermartingale with constant Expectation is a martingale

      1 answer




    Let $(X_n)$ be a supermartingale such that $EX_n$ is constant. show that $X_n$ is a martingale. My problem here in this question is how can I use the constant expectation concept given to prove the statement.






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    share|cite|improve this question













    marked as duplicate by Scientifica, Davide Giraudo, NCh, jgon, Paul Frost Dec 1 '18 at 0:10


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















      0












      0








      0








      This question already has an answer here:




      • Supermartingale with constant Expectation is a martingale

        1 answer




      Let $(X_n)$ be a supermartingale such that $EX_n$ is constant. show that $X_n$ is a martingale. My problem here in this question is how can I use the constant expectation concept given to prove the statement.






      probability







      share|cite|improve this question














      This question already has an answer here:




      • Supermartingale with constant Expectation is a martingale

        1 answer




      Let $(X_n)$ be a supermartingale such that $EX_n$ is constant. show that $X_n$ is a martingale. My problem here in this question is how can I use the constant expectation concept given to prove the statement.





      This question already has an answer here:




      • Supermartingale with constant Expectation is a martingale

        1 answer







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      asked Nov 25 '18 at 17:01









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      marked as duplicate by Scientifica, Davide Giraudo, NCh, jgon, Paul Frost Dec 1 '18 at 0:10


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Scientifica, Davide Giraudo, NCh, jgon, Paul Frost Dec 1 '18 at 0:10


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























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          We have $mathbb{E}(X_{n+1}| mathcal{F}_n) le X_n $, because $(X_n)_{n in mathbb{N}}$ is a supermartingale. Let $$A_varepsilon = { mathbb{E}(X_{n+1}| mathcal{F}_n)+varepsilon le X_n},$$
          then
          $$ int_{A_varepsilon} X_{n+1} , d mathbb{P}+ varepsilon mathbb{P}(A_varepsilon) = int_{A_varepsilon} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} + varepsilon mathbb{P}(A_varepsilon) le int_{A_varepsilon} X_n , d mathbb{P}.$$
          On the other hand
          $$int_{A_varepsilon^c} X_{n+1} , d mathbb{P} = int_{A_varepsilon^c} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} le int_{A_varepsilon} X_n , d mathbb{P}.$$
          Adding both, we get
          $$mathbb{E}[X_n]+varepsilon mathbb{P}(A_varepsilon) =mathbb{E}[X_{n+1}]+varepsilon mathbb{P}(A_varepsilon) le mathbb{E}[X_n].$$
          Thus $mathbb{P}(A_varepsilon) =0$ for any $varepsilon >0$. Finally, by $sigma$-additivity we find that
          $$mathbb{P}(mathbb{E}(X_{n+1}| mathcal{F}_n) neq X_n) = mathbb{P}(bigcup_{n=1}^infty A_{1/n}) le sum_{n=1}^infty mathbb{P}(A_{1/n}) =0.$$






          share|cite|improve this answer























          • Already in the first line you have mistook the property of a submartingale for a supermartingale.
            – LoveTooNap29
            Nov 25 '18 at 21:30






          • 1




            I always mix up 'submartingale' and 'supermartingale'. However, it does not matter, since we can replace $(X_n)_n$ by $(-X_n)_n$ and vica versa. I changed the proof for the case of supermartingales. (The requiered changes are more than obvious!)
            – p4sch
            Nov 25 '18 at 22:06










          • you might like this little mnemonic from D. Williams book, “supermartingales decrease on average, submartingales increase on average—the ‘p’ in supermartingale points down, and the ‘b’ in submartingale points up!”
            – LoveTooNap29
            Nov 25 '18 at 22:36


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          We have $mathbb{E}(X_{n+1}| mathcal{F}_n) le X_n $, because $(X_n)_{n in mathbb{N}}$ is a supermartingale. Let $$A_varepsilon = { mathbb{E}(X_{n+1}| mathcal{F}_n)+varepsilon le X_n},$$
          then
          $$ int_{A_varepsilon} X_{n+1} , d mathbb{P}+ varepsilon mathbb{P}(A_varepsilon) = int_{A_varepsilon} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} + varepsilon mathbb{P}(A_varepsilon) le int_{A_varepsilon} X_n , d mathbb{P}.$$
          On the other hand
          $$int_{A_varepsilon^c} X_{n+1} , d mathbb{P} = int_{A_varepsilon^c} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} le int_{A_varepsilon} X_n , d mathbb{P}.$$
          Adding both, we get
          $$mathbb{E}[X_n]+varepsilon mathbb{P}(A_varepsilon) =mathbb{E}[X_{n+1}]+varepsilon mathbb{P}(A_varepsilon) le mathbb{E}[X_n].$$
          Thus $mathbb{P}(A_varepsilon) =0$ for any $varepsilon >0$. Finally, by $sigma$-additivity we find that
          $$mathbb{P}(mathbb{E}(X_{n+1}| mathcal{F}_n) neq X_n) = mathbb{P}(bigcup_{n=1}^infty A_{1/n}) le sum_{n=1}^infty mathbb{P}(A_{1/n}) =0.$$






          share|cite|improve this answer























          • Already in the first line you have mistook the property of a submartingale for a supermartingale.
            – LoveTooNap29
            Nov 25 '18 at 21:30






          • 1




            I always mix up 'submartingale' and 'supermartingale'. However, it does not matter, since we can replace $(X_n)_n$ by $(-X_n)_n$ and vica versa. I changed the proof for the case of supermartingales. (The requiered changes are more than obvious!)
            – p4sch
            Nov 25 '18 at 22:06










          • you might like this little mnemonic from D. Williams book, “supermartingales decrease on average, submartingales increase on average—the ‘p’ in supermartingale points down, and the ‘b’ in submartingale points up!”
            – LoveTooNap29
            Nov 25 '18 at 22:36
















          1














          We have $mathbb{E}(X_{n+1}| mathcal{F}_n) le X_n $, because $(X_n)_{n in mathbb{N}}$ is a supermartingale. Let $$A_varepsilon = { mathbb{E}(X_{n+1}| mathcal{F}_n)+varepsilon le X_n},$$
          then
          $$ int_{A_varepsilon} X_{n+1} , d mathbb{P}+ varepsilon mathbb{P}(A_varepsilon) = int_{A_varepsilon} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} + varepsilon mathbb{P}(A_varepsilon) le int_{A_varepsilon} X_n , d mathbb{P}.$$
          On the other hand
          $$int_{A_varepsilon^c} X_{n+1} , d mathbb{P} = int_{A_varepsilon^c} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} le int_{A_varepsilon} X_n , d mathbb{P}.$$
          Adding both, we get
          $$mathbb{E}[X_n]+varepsilon mathbb{P}(A_varepsilon) =mathbb{E}[X_{n+1}]+varepsilon mathbb{P}(A_varepsilon) le mathbb{E}[X_n].$$
          Thus $mathbb{P}(A_varepsilon) =0$ for any $varepsilon >0$. Finally, by $sigma$-additivity we find that
          $$mathbb{P}(mathbb{E}(X_{n+1}| mathcal{F}_n) neq X_n) = mathbb{P}(bigcup_{n=1}^infty A_{1/n}) le sum_{n=1}^infty mathbb{P}(A_{1/n}) =0.$$






          share|cite|improve this answer























          • Already in the first line you have mistook the property of a submartingale for a supermartingale.
            – LoveTooNap29
            Nov 25 '18 at 21:30






          • 1




            I always mix up 'submartingale' and 'supermartingale'. However, it does not matter, since we can replace $(X_n)_n$ by $(-X_n)_n$ and vica versa. I changed the proof for the case of supermartingales. (The requiered changes are more than obvious!)
            – p4sch
            Nov 25 '18 at 22:06










          • you might like this little mnemonic from D. Williams book, “supermartingales decrease on average, submartingales increase on average—the ‘p’ in supermartingale points down, and the ‘b’ in submartingale points up!”
            – LoveTooNap29
            Nov 25 '18 at 22:36














          1












          1








          1






          We have $mathbb{E}(X_{n+1}| mathcal{F}_n) le X_n $, because $(X_n)_{n in mathbb{N}}$ is a supermartingale. Let $$A_varepsilon = { mathbb{E}(X_{n+1}| mathcal{F}_n)+varepsilon le X_n},$$
          then
          $$ int_{A_varepsilon} X_{n+1} , d mathbb{P}+ varepsilon mathbb{P}(A_varepsilon) = int_{A_varepsilon} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} + varepsilon mathbb{P}(A_varepsilon) le int_{A_varepsilon} X_n , d mathbb{P}.$$
          On the other hand
          $$int_{A_varepsilon^c} X_{n+1} , d mathbb{P} = int_{A_varepsilon^c} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} le int_{A_varepsilon} X_n , d mathbb{P}.$$
          Adding both, we get
          $$mathbb{E}[X_n]+varepsilon mathbb{P}(A_varepsilon) =mathbb{E}[X_{n+1}]+varepsilon mathbb{P}(A_varepsilon) le mathbb{E}[X_n].$$
          Thus $mathbb{P}(A_varepsilon) =0$ for any $varepsilon >0$. Finally, by $sigma$-additivity we find that
          $$mathbb{P}(mathbb{E}(X_{n+1}| mathcal{F}_n) neq X_n) = mathbb{P}(bigcup_{n=1}^infty A_{1/n}) le sum_{n=1}^infty mathbb{P}(A_{1/n}) =0.$$






          share|cite|improve this answer














          We have $mathbb{E}(X_{n+1}| mathcal{F}_n) le X_n $, because $(X_n)_{n in mathbb{N}}$ is a supermartingale. Let $$A_varepsilon = { mathbb{E}(X_{n+1}| mathcal{F}_n)+varepsilon le X_n},$$
          then
          $$ int_{A_varepsilon} X_{n+1} , d mathbb{P}+ varepsilon mathbb{P}(A_varepsilon) = int_{A_varepsilon} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} + varepsilon mathbb{P}(A_varepsilon) le int_{A_varepsilon} X_n , d mathbb{P}.$$
          On the other hand
          $$int_{A_varepsilon^c} X_{n+1} , d mathbb{P} = int_{A_varepsilon^c} mathbb{E}(X_{n+1}| mathcal{F}_n) , d mathbb{P} le int_{A_varepsilon} X_n , d mathbb{P}.$$
          Adding both, we get
          $$mathbb{E}[X_n]+varepsilon mathbb{P}(A_varepsilon) =mathbb{E}[X_{n+1}]+varepsilon mathbb{P}(A_varepsilon) le mathbb{E}[X_n].$$
          Thus $mathbb{P}(A_varepsilon) =0$ for any $varepsilon >0$. Finally, by $sigma$-additivity we find that
          $$mathbb{P}(mathbb{E}(X_{n+1}| mathcal{F}_n) neq X_n) = mathbb{P}(bigcup_{n=1}^infty A_{1/n}) le sum_{n=1}^infty mathbb{P}(A_{1/n}) =0.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 '18 at 22:04

























          answered Nov 25 '18 at 17:13









          p4sch

          4,760217




          4,760217












          • Already in the first line you have mistook the property of a submartingale for a supermartingale.
            – LoveTooNap29
            Nov 25 '18 at 21:30






          • 1




            I always mix up 'submartingale' and 'supermartingale'. However, it does not matter, since we can replace $(X_n)_n$ by $(-X_n)_n$ and vica versa. I changed the proof for the case of supermartingales. (The requiered changes are more than obvious!)
            – p4sch
            Nov 25 '18 at 22:06










          • you might like this little mnemonic from D. Williams book, “supermartingales decrease on average, submartingales increase on average—the ‘p’ in supermartingale points down, and the ‘b’ in submartingale points up!”
            – LoveTooNap29
            Nov 25 '18 at 22:36


















          • Already in the first line you have mistook the property of a submartingale for a supermartingale.
            – LoveTooNap29
            Nov 25 '18 at 21:30






          • 1




            I always mix up 'submartingale' and 'supermartingale'. However, it does not matter, since we can replace $(X_n)_n$ by $(-X_n)_n$ and vica versa. I changed the proof for the case of supermartingales. (The requiered changes are more than obvious!)
            – p4sch
            Nov 25 '18 at 22:06










          • you might like this little mnemonic from D. Williams book, “supermartingales decrease on average, submartingales increase on average—the ‘p’ in supermartingale points down, and the ‘b’ in submartingale points up!”
            – LoveTooNap29
            Nov 25 '18 at 22:36
















          Already in the first line you have mistook the property of a submartingale for a supermartingale.
          – LoveTooNap29
          Nov 25 '18 at 21:30




          Already in the first line you have mistook the property of a submartingale for a supermartingale.
          – LoveTooNap29
          Nov 25 '18 at 21:30




          1




          1




          I always mix up 'submartingale' and 'supermartingale'. However, it does not matter, since we can replace $(X_n)_n$ by $(-X_n)_n$ and vica versa. I changed the proof for the case of supermartingales. (The requiered changes are more than obvious!)
          – p4sch
          Nov 25 '18 at 22:06




          I always mix up 'submartingale' and 'supermartingale'. However, it does not matter, since we can replace $(X_n)_n$ by $(-X_n)_n$ and vica versa. I changed the proof for the case of supermartingales. (The requiered changes are more than obvious!)
          – p4sch
          Nov 25 '18 at 22:06












          you might like this little mnemonic from D. Williams book, “supermartingales decrease on average, submartingales increase on average—the ‘p’ in supermartingale points down, and the ‘b’ in submartingale points up!”
          – LoveTooNap29
          Nov 25 '18 at 22:36




          you might like this little mnemonic from D. Williams book, “supermartingales decrease on average, submartingales increase on average—the ‘p’ in supermartingale points down, and the ‘b’ in submartingale points up!”
          – LoveTooNap29
          Nov 25 '18 at 22:36



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