Is there an extension of Wald's Equation to the expectation of a product of random variables?












1












$begingroup$


So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:



$mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$



Under the assumption that both have finite expectation.



My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:



$mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$



Or is the outcome something comparable to the Wald's Equation?



Any help is appreciated!






probability probability-theory expected-value







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$endgroup$

















    1












    $begingroup$


    So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:



    $mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$



    Under the assumption that both have finite expectation.



    My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:



    $mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$



    Or is the outcome something comparable to the Wald's Equation?



    Any help is appreciated!






    probability probability-theory expected-value







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:



      $mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$



      Under the assumption that both have finite expectation.



      My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:



      $mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$



      Or is the outcome something comparable to the Wald's Equation?



      Any help is appreciated!






      probability probability-theory expected-value







      share|cite|improve this question









      $endgroup$




      So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:



      $mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$



      Under the assumption that both have finite expectation.



      My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:



      $mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$



      Or is the outcome something comparable to the Wald's Equation?



      Any help is appreciated!






      probability probability-theory expected-value




      probability probability-theory expected-value






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 13 '18 at 17:54









      S. CrimS. Crim

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          $begingroup$

          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable, $N$ is integer valued and $ {(mathbb{E}|X_1|)}^N$ is integrable. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright] =mathbb{E}left[{mathbb{E}[X_1]}^Nright].$$
          In fact,
          $$mathbb{E}left[ |Pi_{i=1}^NX_i| right] =mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N |X_i|right]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N |X_i|right]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n |X_i|right]=sum_{n=1}^infty {(mathbb{E}|X_1|)}^n mathbb{P}(N=n)
          =mathbb{E}left[ {(mathbb{E}|X_1|)}^Nright].$$
          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis of integrability over
          $ {(mathbb{E}|X_1|)}^N$. Since
          $$ left|{mathbb{E}[X_1]}^N right| le {(mathbb{E}|X_1|)}^N $$
          we have ${mathbb{E}[X_1]}^N$ is integrable and the claim is verified.






          share|cite|improve this answer











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          • $begingroup$
            Is there a proof of this, or a reason that it works?
            $endgroup$
            – S. Crim
            Nov 14 '18 at 6:38













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          $begingroup$

          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable, $N$ is integer valued and $ {(mathbb{E}|X_1|)}^N$ is integrable. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright] =mathbb{E}left[{mathbb{E}[X_1]}^Nright].$$
          In fact,
          $$mathbb{E}left[ |Pi_{i=1}^NX_i| right] =mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N |X_i|right]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N |X_i|right]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n |X_i|right]=sum_{n=1}^infty {(mathbb{E}|X_1|)}^n mathbb{P}(N=n)
          =mathbb{E}left[ {(mathbb{E}|X_1|)}^Nright].$$
          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis of integrability over
          $ {(mathbb{E}|X_1|)}^N$. Since
          $$ left|{mathbb{E}[X_1]}^N right| le {(mathbb{E}|X_1|)}^N $$
          we have ${mathbb{E}[X_1]}^N$ is integrable and the claim is verified.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is there a proof of this, or a reason that it works?
            $endgroup$
            – S. Crim
            Nov 14 '18 at 6:38


















          0












          $begingroup$

          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable, $N$ is integer valued and $ {(mathbb{E}|X_1|)}^N$ is integrable. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright] =mathbb{E}left[{mathbb{E}[X_1]}^Nright].$$
          In fact,
          $$mathbb{E}left[ |Pi_{i=1}^NX_i| right] =mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N |X_i|right]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N |X_i|right]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n |X_i|right]=sum_{n=1}^infty {(mathbb{E}|X_1|)}^n mathbb{P}(N=n)
          =mathbb{E}left[ {(mathbb{E}|X_1|)}^Nright].$$
          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis of integrability over
          $ {(mathbb{E}|X_1|)}^N$. Since
          $$ left|{mathbb{E}[X_1]}^N right| le {(mathbb{E}|X_1|)}^N $$
          we have ${mathbb{E}[X_1]}^N$ is integrable and the claim is verified.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is there a proof of this, or a reason that it works?
            $endgroup$
            – S. Crim
            Nov 14 '18 at 6:38
















          0












          0








          0





          $begingroup$

          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable, $N$ is integer valued and $ {(mathbb{E}|X_1|)}^N$ is integrable. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright] =mathbb{E}left[{mathbb{E}[X_1]}^Nright].$$
          In fact,
          $$mathbb{E}left[ |Pi_{i=1}^NX_i| right] =mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N |X_i|right]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N |X_i|right]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n |X_i|right]=sum_{n=1}^infty {(mathbb{E}|X_1|)}^n mathbb{P}(N=n)
          =mathbb{E}left[ {(mathbb{E}|X_1|)}^Nright].$$
          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis of integrability over
          $ {(mathbb{E}|X_1|)}^N$. Since
          $$ left|{mathbb{E}[X_1]}^N right| le {(mathbb{E}|X_1|)}^N $$
          we have ${mathbb{E}[X_1]}^N$ is integrable and the claim is verified.






          share|cite|improve this answer











          $endgroup$



          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable, $N$ is integer valued and $ {(mathbb{E}|X_1|)}^N$ is integrable. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright] =mathbb{E}left[{mathbb{E}[X_1]}^Nright].$$
          In fact,
          $$mathbb{E}left[ |Pi_{i=1}^NX_i| right] =mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N |X_i|right]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N |X_i|right]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n |X_i|right]=sum_{n=1}^infty {(mathbb{E}|X_1|)}^n mathbb{P}(N=n)
          =mathbb{E}left[ {(mathbb{E}|X_1|)}^Nright].$$
          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis of integrability over
          $ {(mathbb{E}|X_1|)}^N$. Since
          $$ left|{mathbb{E}[X_1]}^N right| le {(mathbb{E}|X_1|)}^N $$
          we have ${mathbb{E}[X_1]}^N$ is integrable and the claim is verified.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 17:10

























          answered Nov 14 '18 at 5:33









          Daniel Camarena PerezDaniel Camarena Perez

          57738




          57738












          • $begingroup$
            Is there a proof of this, or a reason that it works?
            $endgroup$
            – S. Crim
            Nov 14 '18 at 6:38




















          • $begingroup$
            Is there a proof of this, or a reason that it works?
            $endgroup$
            – S. Crim
            Nov 14 '18 at 6:38


















          $begingroup$
          Is there a proof of this, or a reason that it works?
          $endgroup$
          – S. Crim
          Nov 14 '18 at 6:38






          $begingroup$
          Is there a proof of this, or a reason that it works?
          $endgroup$
          – S. Crim
          Nov 14 '18 at 6:38




















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