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











up vote
1
down vote

favorite












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!










share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    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!










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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!










      share|cite|improve this question













      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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      S. Crim

      7810




      7810






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          0
          down vote













          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$



          In fact,
          $$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
          =mathbb{E}[mu^N].$$

          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
          instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.






          share|cite|improve this answer























          • Is there a proof of this, or a reason that it works?
            – S. Crim
            2 days ago













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997056%2fis-there-an-extension-of-walds-equation-to-the-expectation-of-a-product-of-rand%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote













          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$



          In fact,
          $$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
          =mathbb{E}[mu^N].$$

          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
          instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.






          share|cite|improve this answer























          • Is there a proof of this, or a reason that it works?
            – S. Crim
            2 days ago

















          up vote
          0
          down vote













          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$



          In fact,
          $$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
          =mathbb{E}[mu^N].$$

          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
          instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.






          share|cite|improve this answer























          • Is there a proof of this, or a reason that it works?
            – S. Crim
            2 days ago















          up vote
          0
          down vote










          up vote
          0
          down vote









          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$



          In fact,
          $$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
          =mathbb{E}[mu^N].$$

          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
          instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.






          share|cite|improve this answer














          Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
          $$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$



          In fact,
          $$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
          =mathbb{E}[mu^N].$$

          Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
          instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered 2 days ago









          Daniel Camarena Perez

          54728




          54728












          • Is there a proof of this, or a reason that it works?
            – S. Crim
            2 days ago




















          • Is there a proof of this, or a reason that it works?
            – S. Crim
            2 days ago


















          Is there a proof of this, or a reason that it works?
          – S. Crim
          2 days ago






          Is there a proof of this, or a reason that it works?
          – S. Crim
          2 days ago




















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997056%2fis-there-an-extension-of-walds-equation-to-the-expectation-of-a-product-of-rand%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Plaza Victoria

          Brian Clough

          Cáceres