Expectation of the function of multiple random variables












0












$begingroup$


I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
$$
E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
$$



we can also obtain the expected value of a function of a single variable by following the workings here
$$
E[g(x)] = int g(x) P_{X}(x) dx
$$



If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
$$
int f(x,y) P_X(x)dx
$$



and wether this expression has any meaning?






probability probability-theory random-variables







share|cite|improve this question











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    0












    $begingroup$


    I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
    $$
    E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
    $$



    we can also obtain the expected value of a function of a single variable by following the workings here
    $$
    E[g(x)] = int g(x) P_{X}(x) dx
    $$



    If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
    $$
    int f(x,y) P_X(x)dx
    $$



    and wether this expression has any meaning?






    probability probability-theory random-variables







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
      $$
      E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
      $$



      we can also obtain the expected value of a function of a single variable by following the workings here
      $$
      E[g(x)] = int g(x) P_{X}(x) dx
      $$



      If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
      $$
      int f(x,y) P_X(x)dx
      $$



      and wether this expression has any meaning?






      probability probability-theory random-variables







      share|cite|improve this question











      $endgroup$




      I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
      $$
      E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
      $$



      we can also obtain the expected value of a function of a single variable by following the workings here
      $$
      E[g(x)] = int g(x) P_{X}(x) dx
      $$



      If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
      $$
      int f(x,y) P_X(x)dx
      $$



      and wether this expression has any meaning?






      probability probability-theory random-variables




      probability probability-theory random-variables






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 11:55







      Tom

















      asked Jan 8 at 11:48









      TomTom

      266




      266






















          1 Answer
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          $begingroup$

          The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.



          So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
            $endgroup$
            – Tom
            Jan 8 at 12:10












          • $begingroup$
            There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
            $endgroup$
            – drhab
            Jan 8 at 12:18










          • $begingroup$
            $mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
            $endgroup$
            – drhab
            Jan 8 at 12:25











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.



          So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
            $endgroup$
            – Tom
            Jan 8 at 12:10












          • $begingroup$
            There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
            $endgroup$
            – drhab
            Jan 8 at 12:18










          • $begingroup$
            $mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
            $endgroup$
            – drhab
            Jan 8 at 12:25
















          2












          $begingroup$

          The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.



          So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
            $endgroup$
            – Tom
            Jan 8 at 12:10












          • $begingroup$
            There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
            $endgroup$
            – drhab
            Jan 8 at 12:18










          • $begingroup$
            $mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
            $endgroup$
            – drhab
            Jan 8 at 12:25














          2












          2








          2





          $begingroup$

          The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.



          So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.






          share|cite|improve this answer









          $endgroup$



          The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.



          So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 12:01









          drhabdrhab

          99.7k544130




          99.7k544130












          • $begingroup$
            Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
            $endgroup$
            – Tom
            Jan 8 at 12:10












          • $begingroup$
            There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
            $endgroup$
            – drhab
            Jan 8 at 12:18










          • $begingroup$
            $mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
            $endgroup$
            – drhab
            Jan 8 at 12:25


















          • $begingroup$
            Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
            $endgroup$
            – Tom
            Jan 8 at 12:10












          • $begingroup$
            There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
            $endgroup$
            – drhab
            Jan 8 at 12:18










          • $begingroup$
            $mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
            $endgroup$
            – drhab
            Jan 8 at 12:25
















          $begingroup$
          Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
          $endgroup$
          – Tom
          Jan 8 at 12:10






          $begingroup$
          Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
          $endgroup$
          – Tom
          Jan 8 at 12:10














          $begingroup$
          There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
          $endgroup$
          – drhab
          Jan 8 at 12:18




          $begingroup$
          There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
          $endgroup$
          – drhab
          Jan 8 at 12:18












          $begingroup$
          $mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
          $endgroup$
          – drhab
          Jan 8 at 12:25




          $begingroup$
          $mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
          $endgroup$
          – drhab
          Jan 8 at 12:25


















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