Expected value of outer product of multivariate normal random vector with itself












1












$begingroup$


Let's say I have a random vector $boldsymbol{t}$ that is distributed according to a multivariate normal distribution:
$$
boldsymbol{t} sim mathcal{N}(boldsymbol{mu}, boldsymbol{Psi})
$$
I now want to find the expected value of the outer product of this random vector:
$$
mathbb{E}left[boldsymbol{t}boldsymbol{t}^intercalright]
$$
Is there a closed-form solution to this problem? In my studies, I have stumbled across the Wishart distribution. Might this be a way to tackle this problem?






probability random-variables normal-distribution expected-value outer-product







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    1












    $begingroup$


    Let's say I have a random vector $boldsymbol{t}$ that is distributed according to a multivariate normal distribution:
    $$
    boldsymbol{t} sim mathcal{N}(boldsymbol{mu}, boldsymbol{Psi})
    $$
    I now want to find the expected value of the outer product of this random vector:
    $$
    mathbb{E}left[boldsymbol{t}boldsymbol{t}^intercalright]
    $$
    Is there a closed-form solution to this problem? In my studies, I have stumbled across the Wishart distribution. Might this be a way to tackle this problem?






    probability random-variables normal-distribution expected-value outer-product







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let's say I have a random vector $boldsymbol{t}$ that is distributed according to a multivariate normal distribution:
      $$
      boldsymbol{t} sim mathcal{N}(boldsymbol{mu}, boldsymbol{Psi})
      $$
      I now want to find the expected value of the outer product of this random vector:
      $$
      mathbb{E}left[boldsymbol{t}boldsymbol{t}^intercalright]
      $$
      Is there a closed-form solution to this problem? In my studies, I have stumbled across the Wishart distribution. Might this be a way to tackle this problem?






      probability random-variables normal-distribution expected-value outer-product







      share|cite|improve this question









      $endgroup$




      Let's say I have a random vector $boldsymbol{t}$ that is distributed according to a multivariate normal distribution:
      $$
      boldsymbol{t} sim mathcal{N}(boldsymbol{mu}, boldsymbol{Psi})
      $$
      I now want to find the expected value of the outer product of this random vector:
      $$
      mathbb{E}left[boldsymbol{t}boldsymbol{t}^intercalright]
      $$
      Is there a closed-form solution to this problem? In my studies, I have stumbled across the Wishart distribution. Might this be a way to tackle this problem?






      probability random-variables normal-distribution expected-value outer-product




      probability random-variables normal-distribution expected-value outer-product






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      asked Jan 30 at 7:41









      fujifuji

      1083




      1083






















          1 Answer
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          1












          $begingroup$

          First, note that
          $$
          left(t-muright)left(t-muright)^{intercal}=tt^{intercal}-tmu^{intercal}-mu t^{intercal}+mumu^{intercal}.
          $$
          Therefore,
          begin{align*}
          operatorname{Var}(t) & =mathbb{E}left[left(t-muright)left(t-muright)^{intercal}right]\
          & =mathbb{E}left[tt^{intercal}right]-mathbb{E}left[tright]mu^{intercal}-mumathbb{E}left[t^{intercal}right]+mumu^{intercal}\
          & =mathbb{E}left[tt^{intercal}right]-mumu^{intercal}.
          end{align*}
          You already know that $operatorname{Var}(t)=Psi$.
          Moving some terms around, $mathbb{E}left[tt^{intercal}right]=Psi+mumu^{intercal}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the second equation, are $mathbb{E}[t]mu^intercal$ and $mumathbb{E}[t^intercal]$ the same as $mumu^intercal$, or how do you perform the simplification for the last line?
            $endgroup$
            – fuji
            Jan 30 at 8:02








          • 1




            $begingroup$
            @fuji: $mathbb{E}[t] = mu$ by definition of the random variable $t$ (it is normal with mean $mu$). As for the transpose, $mathbb{E}[t^intercal] = mathbb{E}[t]^intercal$ for any random vector $t$.
            $endgroup$
            – parsiad
            Jan 31 at 1:06












          • $begingroup$
            Thanks for the clarification!
            $endgroup$
            – fuji
            Jan 31 at 8:23












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          1 Answer
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          active

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          1












          $begingroup$

          First, note that
          $$
          left(t-muright)left(t-muright)^{intercal}=tt^{intercal}-tmu^{intercal}-mu t^{intercal}+mumu^{intercal}.
          $$
          Therefore,
          begin{align*}
          operatorname{Var}(t) & =mathbb{E}left[left(t-muright)left(t-muright)^{intercal}right]\
          & =mathbb{E}left[tt^{intercal}right]-mathbb{E}left[tright]mu^{intercal}-mumathbb{E}left[t^{intercal}right]+mumu^{intercal}\
          & =mathbb{E}left[tt^{intercal}right]-mumu^{intercal}.
          end{align*}
          You already know that $operatorname{Var}(t)=Psi$.
          Moving some terms around, $mathbb{E}left[tt^{intercal}right]=Psi+mumu^{intercal}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the second equation, are $mathbb{E}[t]mu^intercal$ and $mumathbb{E}[t^intercal]$ the same as $mumu^intercal$, or how do you perform the simplification for the last line?
            $endgroup$
            – fuji
            Jan 30 at 8:02








          • 1




            $begingroup$
            @fuji: $mathbb{E}[t] = mu$ by definition of the random variable $t$ (it is normal with mean $mu$). As for the transpose, $mathbb{E}[t^intercal] = mathbb{E}[t]^intercal$ for any random vector $t$.
            $endgroup$
            – parsiad
            Jan 31 at 1:06












          • $begingroup$
            Thanks for the clarification!
            $endgroup$
            – fuji
            Jan 31 at 8:23
















          1












          $begingroup$

          First, note that
          $$
          left(t-muright)left(t-muright)^{intercal}=tt^{intercal}-tmu^{intercal}-mu t^{intercal}+mumu^{intercal}.
          $$
          Therefore,
          begin{align*}
          operatorname{Var}(t) & =mathbb{E}left[left(t-muright)left(t-muright)^{intercal}right]\
          & =mathbb{E}left[tt^{intercal}right]-mathbb{E}left[tright]mu^{intercal}-mumathbb{E}left[t^{intercal}right]+mumu^{intercal}\
          & =mathbb{E}left[tt^{intercal}right]-mumu^{intercal}.
          end{align*}
          You already know that $operatorname{Var}(t)=Psi$.
          Moving some terms around, $mathbb{E}left[tt^{intercal}right]=Psi+mumu^{intercal}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the second equation, are $mathbb{E}[t]mu^intercal$ and $mumathbb{E}[t^intercal]$ the same as $mumu^intercal$, or how do you perform the simplification for the last line?
            $endgroup$
            – fuji
            Jan 30 at 8:02








          • 1




            $begingroup$
            @fuji: $mathbb{E}[t] = mu$ by definition of the random variable $t$ (it is normal with mean $mu$). As for the transpose, $mathbb{E}[t^intercal] = mathbb{E}[t]^intercal$ for any random vector $t$.
            $endgroup$
            – parsiad
            Jan 31 at 1:06












          • $begingroup$
            Thanks for the clarification!
            $endgroup$
            – fuji
            Jan 31 at 8:23














          1












          1








          1





          $begingroup$

          First, note that
          $$
          left(t-muright)left(t-muright)^{intercal}=tt^{intercal}-tmu^{intercal}-mu t^{intercal}+mumu^{intercal}.
          $$
          Therefore,
          begin{align*}
          operatorname{Var}(t) & =mathbb{E}left[left(t-muright)left(t-muright)^{intercal}right]\
          & =mathbb{E}left[tt^{intercal}right]-mathbb{E}left[tright]mu^{intercal}-mumathbb{E}left[t^{intercal}right]+mumu^{intercal}\
          & =mathbb{E}left[tt^{intercal}right]-mumu^{intercal}.
          end{align*}
          You already know that $operatorname{Var}(t)=Psi$.
          Moving some terms around, $mathbb{E}left[tt^{intercal}right]=Psi+mumu^{intercal}$.






          share|cite|improve this answer









          $endgroup$



          First, note that
          $$
          left(t-muright)left(t-muright)^{intercal}=tt^{intercal}-tmu^{intercal}-mu t^{intercal}+mumu^{intercal}.
          $$
          Therefore,
          begin{align*}
          operatorname{Var}(t) & =mathbb{E}left[left(t-muright)left(t-muright)^{intercal}right]\
          & =mathbb{E}left[tt^{intercal}right]-mathbb{E}left[tright]mu^{intercal}-mumathbb{E}left[t^{intercal}right]+mumu^{intercal}\
          & =mathbb{E}left[tt^{intercal}right]-mumu^{intercal}.
          end{align*}
          You already know that $operatorname{Var}(t)=Psi$.
          Moving some terms around, $mathbb{E}left[tt^{intercal}right]=Psi+mumu^{intercal}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 7:51









          parsiadparsiad

          18.6k32453




          18.6k32453












          • $begingroup$
            In the second equation, are $mathbb{E}[t]mu^intercal$ and $mumathbb{E}[t^intercal]$ the same as $mumu^intercal$, or how do you perform the simplification for the last line?
            $endgroup$
            – fuji
            Jan 30 at 8:02








          • 1




            $begingroup$
            @fuji: $mathbb{E}[t] = mu$ by definition of the random variable $t$ (it is normal with mean $mu$). As for the transpose, $mathbb{E}[t^intercal] = mathbb{E}[t]^intercal$ for any random vector $t$.
            $endgroup$
            – parsiad
            Jan 31 at 1:06












          • $begingroup$
            Thanks for the clarification!
            $endgroup$
            – fuji
            Jan 31 at 8:23


















          • $begingroup$
            In the second equation, are $mathbb{E}[t]mu^intercal$ and $mumathbb{E}[t^intercal]$ the same as $mumu^intercal$, or how do you perform the simplification for the last line?
            $endgroup$
            – fuji
            Jan 30 at 8:02








          • 1




            $begingroup$
            @fuji: $mathbb{E}[t] = mu$ by definition of the random variable $t$ (it is normal with mean $mu$). As for the transpose, $mathbb{E}[t^intercal] = mathbb{E}[t]^intercal$ for any random vector $t$.
            $endgroup$
            – parsiad
            Jan 31 at 1:06












          • $begingroup$
            Thanks for the clarification!
            $endgroup$
            – fuji
            Jan 31 at 8:23
















          $begingroup$
          In the second equation, are $mathbb{E}[t]mu^intercal$ and $mumathbb{E}[t^intercal]$ the same as $mumu^intercal$, or how do you perform the simplification for the last line?
          $endgroup$
          – fuji
          Jan 30 at 8:02






          $begingroup$
          In the second equation, are $mathbb{E}[t]mu^intercal$ and $mumathbb{E}[t^intercal]$ the same as $mumu^intercal$, or how do you perform the simplification for the last line?
          $endgroup$
          – fuji
          Jan 30 at 8:02






          1




          1




          $begingroup$
          @fuji: $mathbb{E}[t] = mu$ by definition of the random variable $t$ (it is normal with mean $mu$). As for the transpose, $mathbb{E}[t^intercal] = mathbb{E}[t]^intercal$ for any random vector $t$.
          $endgroup$
          – parsiad
          Jan 31 at 1:06






          $begingroup$
          @fuji: $mathbb{E}[t] = mu$ by definition of the random variable $t$ (it is normal with mean $mu$). As for the transpose, $mathbb{E}[t^intercal] = mathbb{E}[t]^intercal$ for any random vector $t$.
          $endgroup$
          – parsiad
          Jan 31 at 1:06














          $begingroup$
          Thanks for the clarification!
          $endgroup$
          – fuji
          Jan 31 at 8:23




          $begingroup$
          Thanks for the clarification!
          $endgroup$
          – fuji
          Jan 31 at 8:23


















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