Intuition behind Expected Value of the Square of a Random Variable $X$











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Let's look at the case of $X sim Pois(lambda)$.



Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:



$mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.






probability-theory poisson-distribution expected-value







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    up vote
    0
    down vote

    favorite












    Let's look at the case of $X sim Pois(lambda)$.



    Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
    from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:



    $mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.






    probability-theory poisson-distribution expected-value







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let's look at the case of $X sim Pois(lambda)$.



      Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
      from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:



      $mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.






      probability-theory poisson-distribution expected-value







      share|cite|improve this question













      Let's look at the case of $X sim Pois(lambda)$.



      Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
      from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:



      $mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.






      probability-theory poisson-distribution expected-value




      probability-theory poisson-distribution expected-value






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      asked 22 hours ago









      SABOY

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          If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.



          If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.



          Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.



          It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.



          (You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)



          Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.



          The same objection counts for the use of $X(Omega)$.



          If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$



          Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.






          share|cite|improve this answer























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            1 Answer
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            1 Answer
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            active

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            active

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            up vote
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            down vote



            accepted










            If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.



            If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.



            Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.



            It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.



            (You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)



            Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.



            The same objection counts for the use of $X(Omega)$.



            If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$



            Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.






            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.



              If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.



              Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.



              It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.



              (You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)



              Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.



              The same objection counts for the use of $X(Omega)$.



              If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$



              Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.



                If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.



                Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.



                It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.



                (You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)



                Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.



                The same objection counts for the use of $X(Omega)$.



                If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$



                Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.






                share|cite|improve this answer














                If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.



                If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.



                Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.



                It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.



                (You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)



                Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.



                The same objection counts for the use of $X(Omega)$.



                If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$



                Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 6 hours ago

























                answered 22 hours ago









                drhab

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