Expectation of Independent Bernoulli Trials












1












$begingroup$


Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
$$
mathbb{E}left[ X W X right],
$$



where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.






probability expected-value







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
    $$
    mathbb{E}left[ X W X right],
    $$



    where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.






    probability expected-value







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
      $$
      mathbb{E}left[ X W X right],
      $$



      where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.






      probability expected-value







      share|cite|improve this question









      $endgroup$




      Say there is a sequence of IID bernoulli trials $(x_1,x_2,...,x_N)$. Where $X in mathbb{R}^{Ntimes N}$ is a diagonal matrix with these RV's on the main diagonal and $mathbb{P}left[x_k =1right]=p, 0<kle N in mathbb{N}$. I want to know what is:
      $$
      mathbb{E}left[ X W X right],
      $$



      where $W inmathbb{R}^{Ntimes N}succ 0 $ is some positive definite weightings matrix. Due to this being bernoulli trials I believe the main diagonal after the expectation is excecuted to be $W_{ii} cdot p$ and $W_{ij}cdot p^2 $ for $ineq j$. However, I am unsure.






      probability expected-value




      probability expected-value






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 5 '18 at 11:15









      p32fr4p32fr4

      3717




      3717






















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          $begingroup$

          You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
          begin{align*}
          (X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
          &=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
          &=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
          end{align*}



          The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.



          Hence
          begin{align*}
          mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
          &=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
          &=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
          end{align*}



          Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 11:55






          • 1




            $begingroup$
            True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
            $endgroup$
            – P. Quinton
            Dec 5 '18 at 12:12










          • $begingroup$
            That's great thanks.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 12:19











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          1 Answer
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          1 Answer
          1






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          active

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          active

          oldest

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          1












          $begingroup$

          You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
          begin{align*}
          (X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
          &=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
          &=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
          end{align*}



          The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.



          Hence
          begin{align*}
          mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
          &=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
          &=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
          end{align*}



          Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 11:55






          • 1




            $begingroup$
            True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
            $endgroup$
            – P. Quinton
            Dec 5 '18 at 12:12










          • $begingroup$
            That's great thanks.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 12:19
















          1












          $begingroup$

          You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
          begin{align*}
          (X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
          &=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
          &=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
          end{align*}



          The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.



          Hence
          begin{align*}
          mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
          &=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
          &=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
          end{align*}



          Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 11:55






          • 1




            $begingroup$
            True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
            $endgroup$
            – P. Quinton
            Dec 5 '18 at 12:12










          • $begingroup$
            That's great thanks.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 12:19














          1












          1








          1





          $begingroup$

          You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
          begin{align*}
          (X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
          &=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
          &=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
          end{align*}



          The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.



          Hence
          begin{align*}
          mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
          &=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
          &=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
          end{align*}



          Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.






          share|cite|improve this answer











          $endgroup$



          You can explicitly give the elements of the product $X W X$ as, for element at coordinate $(i,j)$
          begin{align*}
          (X W X)_{ij} &= sum_{k} X_{ik} (WX)_{kj}\
          &=sum_{k} X_{ik} sum_{ell} W_{kell} X_{ell j}\
          &=sum_k sum_ell W_{kell} X_{ik}X_{ell j}
          end{align*}



          The expectation is linear hence we only look at $mathbb E[X_{ab} X_{cd}]$ which is $p$ if $(a,b)=(c,d)$ and $p^2$ otherwise.



          Hence
          begin{align*}
          mathbb E[(XWX)_{ij}] &= sum_k sum_ell W_{kell} [pmathbb 1(i=ell land k=j) + p^2 (1-mathbb 1(i=ell land k=j))]\
          &=sum_k sum_ell W_{kell} [mathbb 1(i=ell land k=j)(p-p^2) + p^2)]\
          &=p(1-p)W_{ij} + p^2 sum_k sum_ell W_{kell}
          end{align*}



          Or in matrix notation $mathbb E[XWX]=p(1-p)W + p^2 mathbf 1 W mathbf 1$ where in this case $mathbf 1$ is the all one matrix.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 12:11

























          answered Dec 5 '18 at 11:44









          P. QuintonP. Quinton

          1,8011213




          1,8011213












          • $begingroup$
            $X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 11:55






          • 1




            $begingroup$
            True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
            $endgroup$
            – P. Quinton
            Dec 5 '18 at 12:12










          • $begingroup$
            That's great thanks.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 12:19


















          • $begingroup$
            $X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 11:55






          • 1




            $begingroup$
            True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
            $endgroup$
            – P. Quinton
            Dec 5 '18 at 12:12










          • $begingroup$
            That's great thanks.
            $endgroup$
            – p32fr4
            Dec 5 '18 at 12:19
















          $begingroup$
          $X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
          $endgroup$
          – p32fr4
          Dec 5 '18 at 11:55




          $begingroup$
          $X$ was a diagonal matrix so I don't believe there is a dimensional issue. Either way i also have a column vector multiplying each side ( I didn't post this in the problem above). So it should boil down to a variation of you answer thank you.
          $endgroup$
          – p32fr4
          Dec 5 '18 at 11:55




          1




          1




          $begingroup$
          True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
          $endgroup$
          – P. Quinton
          Dec 5 '18 at 12:12




          $begingroup$
          True, I fixed my answer so that it answers the problem, the trace trick cannot be used anymore and so I had to do it explicitly.
          $endgroup$
          – P. Quinton
          Dec 5 '18 at 12:12












          $begingroup$
          That's great thanks.
          $endgroup$
          – p32fr4
          Dec 5 '18 at 12:19




          $begingroup$
          That's great thanks.
          $endgroup$
          – p32fr4
          Dec 5 '18 at 12:19


















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