How can I show that the sum of two martingales is a martingale? [closed]











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If I have X and Y, two martingales in the same filtered environment, and a and b two constants, how can I show that :



{aXt + bYt} is also a martingale?






probability stochastic-calculus martingales







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closed as off-topic by Did, John B, KReiser, user10354138, Rebellos Dec 5 at 7:14


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  • By the linearity of conditional expectation?
    – Federico
    Nov 22 at 18:05















up vote
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down vote

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If I have X and Y, two martingales in the same filtered environment, and a and b two constants, how can I show that :



{aXt + bYt} is also a martingale?






probability stochastic-calculus martingales







share|cite|improve this question













closed as off-topic by Did, John B, KReiser, user10354138, Rebellos Dec 5 at 7:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, John B, KReiser, user10354138, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.













  • By the linearity of conditional expectation?
    – Federico
    Nov 22 at 18:05













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











If I have X and Y, two martingales in the same filtered environment, and a and b two constants, how can I show that :



{aXt + bYt} is also a martingale?






probability stochastic-calculus martingales







share|cite|improve this question













If I have X and Y, two martingales in the same filtered environment, and a and b two constants, how can I show that :



{aXt + bYt} is also a martingale?






probability stochastic-calculus martingales




probability stochastic-calculus martingales






share|cite|improve this question













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asked Nov 22 at 17:59









Chuck

1




1




closed as off-topic by Did, John B, KReiser, user10354138, Rebellos Dec 5 at 7:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, John B, KReiser, user10354138, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Did, John B, KReiser, user10354138, Rebellos Dec 5 at 7:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, John B, KReiser, user10354138, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • By the linearity of conditional expectation?
    – Federico
    Nov 22 at 18:05


















  • By the linearity of conditional expectation?
    – Federico
    Nov 22 at 18:05
















By the linearity of conditional expectation?
– Federico
Nov 22 at 18:05




By the linearity of conditional expectation?
– Federico
Nov 22 at 18:05










1 Answer
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$E(aX_{n+1}+bY_{n+1}|mathcal{F}_n)=aE(X_{n+1}|mathcal{F}_n)+bE(Y_{n+1}|mathcal{F}_n)=aX_n+bY_n$






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    1 Answer
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    1 Answer
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    active

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    up vote
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    down vote













    $E(aX_{n+1}+bY_{n+1}|mathcal{F}_n)=aE(X_{n+1}|mathcal{F}_n)+bE(Y_{n+1}|mathcal{F}_n)=aX_n+bY_n$






    share|cite|improve this answer

























      up vote
      1
      down vote













      $E(aX_{n+1}+bY_{n+1}|mathcal{F}_n)=aE(X_{n+1}|mathcal{F}_n)+bE(Y_{n+1}|mathcal{F}_n)=aX_n+bY_n$






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        $E(aX_{n+1}+bY_{n+1}|mathcal{F}_n)=aE(X_{n+1}|mathcal{F}_n)+bE(Y_{n+1}|mathcal{F}_n)=aX_n+bY_n$






        share|cite|improve this answer












        $E(aX_{n+1}+bY_{n+1}|mathcal{F}_n)=aE(X_{n+1}|mathcal{F}_n)+bE(Y_{n+1}|mathcal{F}_n)=aX_n+bY_n$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 18:04









        John_Wick

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        1,199111















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