Expectation of Product of Subset of Jointly Gaussian Random Vector

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I have a scenario, where I know that $boldsymbolz sim mathcalN( boldsymbolmu, boldsymbolSigma ) $ and I need to compute $ mathbbE( z_i^2 z_j ) $, where $z_i$ is the i-th component of vector $boldsymbolz$.



Is there any way of computing this, assuming that $boldsymbolz$ is a random vector of dimension n>2, say e.g. n=5?



If there is not a general way of computing this, is there anything for the special case of $boldsymbolmu=boldsymbol0$?



The closest thing I found so far is Isserlis' theorem, but it still seems to handle a slightly different case.




probability probability-theory expectation




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    I have a scenario, where I know that $boldsymbolz sim mathcalN( boldsymbolmu, boldsymbolSigma ) $ and I need to compute $ mathbbE( z_i^2 z_j ) $, where $z_i$ is the i-th component of vector $boldsymbolz$.



    Is there any way of computing this, assuming that $boldsymbolz$ is a random vector of dimension n>2, say e.g. n=5?



    If there is not a general way of computing this, is there anything for the special case of $boldsymbolmu=boldsymbol0$?



    The closest thing I found so far is Isserlis' theorem, but it still seems to handle a slightly different case.




    probability probability-theory expectation




    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have a scenario, where I know that $boldsymbolz sim mathcalN( boldsymbolmu, boldsymbolSigma ) $ and I need to compute $ mathbbE( z_i^2 z_j ) $, where $z_i$ is the i-th component of vector $boldsymbolz$.



      Is there any way of computing this, assuming that $boldsymbolz$ is a random vector of dimension n>2, say e.g. n=5?



      If there is not a general way of computing this, is there anything for the special case of $boldsymbolmu=boldsymbol0$?



      The closest thing I found so far is Isserlis' theorem, but it still seems to handle a slightly different case.




      probability probability-theory expectation




      share|cite|improve this question












      I have a scenario, where I know that $boldsymbolz sim mathcalN( boldsymbolmu, boldsymbolSigma ) $ and I need to compute $ mathbbE( z_i^2 z_j ) $, where $z_i$ is the i-th component of vector $boldsymbolz$.



      Is there any way of computing this, assuming that $boldsymbolz$ is a random vector of dimension n>2, say e.g. n=5?



      If there is not a general way of computing this, is there anything for the special case of $boldsymbolmu=boldsymbol0$?



      The closest thing I found so far is Isserlis' theorem, but it still seems to handle a slightly different case.





      probability probability-theory expectation





      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 22 at 9:26









      bonanza

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          We can find $a,b$ such that $z_i$ and $az_i+bz_j$ are independent: just make the covariance $0$. [Since they are jointly normal this is enough to say they are independent]. Now $Ez_i^2(az_i+bz_j)=Ez_i^2E(az_i+bz_j)$. The right side can be computed and the left side is $Ez_i^2(az_i+bz_j)=aEz_i^3+bEz_i^2 z_j$. From this you can compute $Ez_i^2z_j$.






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          • Thanks! But how to find corresponding a and b?
            – bonanza
            Aug 22 at 10:25










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          We can find $a,b$ such that $z_i$ and $az_i+bz_j$ are independent: just make the covariance $0$. [Since they are jointly normal this is enough to say they are independent]. Now $Ez_i^2(az_i+bz_j)=Ez_i^2E(az_i+bz_j)$. The right side can be computed and the left side is $Ez_i^2(az_i+bz_j)=aEz_i^3+bEz_i^2 z_j$. From this you can compute $Ez_i^2z_j$.






          share|cite|improve this answer






















          • Thanks! But how to find corresponding a and b?
            – bonanza
            Aug 22 at 10:25














          up vote
          0
          down vote













          We can find $a,b$ such that $z_i$ and $az_i+bz_j$ are independent: just make the covariance $0$. [Since they are jointly normal this is enough to say they are independent]. Now $Ez_i^2(az_i+bz_j)=Ez_i^2E(az_i+bz_j)$. The right side can be computed and the left side is $Ez_i^2(az_i+bz_j)=aEz_i^3+bEz_i^2 z_j$. From this you can compute $Ez_i^2z_j$.






          share|cite|improve this answer






















          • Thanks! But how to find corresponding a and b?
            – bonanza
            Aug 22 at 10:25












          up vote
          0
          down vote










          up vote
          0
          down vote









          We can find $a,b$ such that $z_i$ and $az_i+bz_j$ are independent: just make the covariance $0$. [Since they are jointly normal this is enough to say they are independent]. Now $Ez_i^2(az_i+bz_j)=Ez_i^2E(az_i+bz_j)$. The right side can be computed and the left side is $Ez_i^2(az_i+bz_j)=aEz_i^3+bEz_i^2 z_j$. From this you can compute $Ez_i^2z_j$.






          share|cite|improve this answer














          We can find $a,b$ such that $z_i$ and $az_i+bz_j$ are independent: just make the covariance $0$. [Since they are jointly normal this is enough to say they are independent]. Now $Ez_i^2(az_i+bz_j)=Ez_i^2E(az_i+bz_j)$. The right side can be computed and the left side is $Ez_i^2(az_i+bz_j)=aEz_i^3+bEz_i^2 z_j$. From this you can compute $Ez_i^2z_j$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 22 at 9:53

























          answered Aug 22 at 9:42









          Kavi Rama Murthy

          23.5k2933




          23.5k2933











          • Thanks! But how to find corresponding a and b?
            – bonanza
            Aug 22 at 10:25
















          • Thanks! But how to find corresponding a and b?
            – bonanza
            Aug 22 at 10:25















          Thanks! But how to find corresponding a and b?
          – bonanza
          Aug 22 at 10:25




          Thanks! But how to find corresponding a and b?
          – bonanza
          Aug 22 at 10:25












           

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