Use of law of total expectation without checking integrability











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$newcommand{E}{mathbb E}$In basic probability classes, people often use the formula, namely the law of total expectation
$$E[X]=E[E[Xmid Y]]$$
without checking integrability of $X$.



I can't get my head around it, because we sometimes use it to check the integrability of $X$... Wow! We don't even know if $E[Xmid Y]$ exists, yet we use it.



Example.
For a simple example, I wanted to do it properly. For instance, consider nonnegative integrable random variables $Z$ and $X_i$ i.i.d. independent of $Z$. We need to check whether $$sum_{i= 1}^ZX_i $$
is integrable. I can do it like this
$$Eleft[sum_{i= 1}^ZX_i right]=Eleft[Eleft[sum_{i= 1}^ZX_imid Zright] right]=Eleft[sum_{i=1}^Z E[X_i] right]=E[Z]E[X_1]$$
But I thought this is cheating, so I considered $Xmathbb I_{X<K}$ for some constant $K>0$, then we have
begin{align}
sum_{i=1}^Z X_imathbb I_{X_i<K} leq ZK
end{align}

This is clearly integrable. So we can apply the same steps as before "legally" to show integrability and then MCT does the rest of the work
begin{align}
Eleft[sum_{i=1}^Z X_iright]stackrel{text{MCT}}{=}lim_{Ktoinfty}Eleft[sum_{i=1}^Z X_imathbb I_{X_i<K} right]=lim_{Ktoinfty}E[X_1mathbb I_{X_1<K}]E[Z]stackrel{text{MCT}}{=}E[X_1]E[Z]
end{align}

Do something similar always works, giving us the reason why people just use it? Indeed, I guess it is boring to repeat these kind of steps if it always works, therefore my question:




Question. Can we actually apply the law of total expectation to show integrability? If the answer is "yes", is there a general proof? If the answer is "no", are there examples where it fails?











share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    $newcommand{E}{mathbb E}$In basic probability classes, people often use the formula, namely the law of total expectation
    $$E[X]=E[E[Xmid Y]]$$
    without checking integrability of $X$.



    I can't get my head around it, because we sometimes use it to check the integrability of $X$... Wow! We don't even know if $E[Xmid Y]$ exists, yet we use it.



    Example.
    For a simple example, I wanted to do it properly. For instance, consider nonnegative integrable random variables $Z$ and $X_i$ i.i.d. independent of $Z$. We need to check whether $$sum_{i= 1}^ZX_i $$
    is integrable. I can do it like this
    $$Eleft[sum_{i= 1}^ZX_i right]=Eleft[Eleft[sum_{i= 1}^ZX_imid Zright] right]=Eleft[sum_{i=1}^Z E[X_i] right]=E[Z]E[X_1]$$
    But I thought this is cheating, so I considered $Xmathbb I_{X<K}$ for some constant $K>0$, then we have
    begin{align}
    sum_{i=1}^Z X_imathbb I_{X_i<K} leq ZK
    end{align}

    This is clearly integrable. So we can apply the same steps as before "legally" to show integrability and then MCT does the rest of the work
    begin{align}
    Eleft[sum_{i=1}^Z X_iright]stackrel{text{MCT}}{=}lim_{Ktoinfty}Eleft[sum_{i=1}^Z X_imathbb I_{X_i<K} right]=lim_{Ktoinfty}E[X_1mathbb I_{X_1<K}]E[Z]stackrel{text{MCT}}{=}E[X_1]E[Z]
    end{align}

    Do something similar always works, giving us the reason why people just use it? Indeed, I guess it is boring to repeat these kind of steps if it always works, therefore my question:




    Question. Can we actually apply the law of total expectation to show integrability? If the answer is "yes", is there a general proof? If the answer is "no", are there examples where it fails?











    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $newcommand{E}{mathbb E}$In basic probability classes, people often use the formula, namely the law of total expectation
      $$E[X]=E[E[Xmid Y]]$$
      without checking integrability of $X$.



      I can't get my head around it, because we sometimes use it to check the integrability of $X$... Wow! We don't even know if $E[Xmid Y]$ exists, yet we use it.



      Example.
      For a simple example, I wanted to do it properly. For instance, consider nonnegative integrable random variables $Z$ and $X_i$ i.i.d. independent of $Z$. We need to check whether $$sum_{i= 1}^ZX_i $$
      is integrable. I can do it like this
      $$Eleft[sum_{i= 1}^ZX_i right]=Eleft[Eleft[sum_{i= 1}^ZX_imid Zright] right]=Eleft[sum_{i=1}^Z E[X_i] right]=E[Z]E[X_1]$$
      But I thought this is cheating, so I considered $Xmathbb I_{X<K}$ for some constant $K>0$, then we have
      begin{align}
      sum_{i=1}^Z X_imathbb I_{X_i<K} leq ZK
      end{align}

      This is clearly integrable. So we can apply the same steps as before "legally" to show integrability and then MCT does the rest of the work
      begin{align}
      Eleft[sum_{i=1}^Z X_iright]stackrel{text{MCT}}{=}lim_{Ktoinfty}Eleft[sum_{i=1}^Z X_imathbb I_{X_i<K} right]=lim_{Ktoinfty}E[X_1mathbb I_{X_1<K}]E[Z]stackrel{text{MCT}}{=}E[X_1]E[Z]
      end{align}

      Do something similar always works, giving us the reason why people just use it? Indeed, I guess it is boring to repeat these kind of steps if it always works, therefore my question:




      Question. Can we actually apply the law of total expectation to show integrability? If the answer is "yes", is there a general proof? If the answer is "no", are there examples where it fails?











      share|cite|improve this question















      $newcommand{E}{mathbb E}$In basic probability classes, people often use the formula, namely the law of total expectation
      $$E[X]=E[E[Xmid Y]]$$
      without checking integrability of $X$.



      I can't get my head around it, because we sometimes use it to check the integrability of $X$... Wow! We don't even know if $E[Xmid Y]$ exists, yet we use it.



      Example.
      For a simple example, I wanted to do it properly. For instance, consider nonnegative integrable random variables $Z$ and $X_i$ i.i.d. independent of $Z$. We need to check whether $$sum_{i= 1}^ZX_i $$
      is integrable. I can do it like this
      $$Eleft[sum_{i= 1}^ZX_i right]=Eleft[Eleft[sum_{i= 1}^ZX_imid Zright] right]=Eleft[sum_{i=1}^Z E[X_i] right]=E[Z]E[X_1]$$
      But I thought this is cheating, so I considered $Xmathbb I_{X<K}$ for some constant $K>0$, then we have
      begin{align}
      sum_{i=1}^Z X_imathbb I_{X_i<K} leq ZK
      end{align}

      This is clearly integrable. So we can apply the same steps as before "legally" to show integrability and then MCT does the rest of the work
      begin{align}
      Eleft[sum_{i=1}^Z X_iright]stackrel{text{MCT}}{=}lim_{Ktoinfty}Eleft[sum_{i=1}^Z X_imathbb I_{X_i<K} right]=lim_{Ktoinfty}E[X_1mathbb I_{X_1<K}]E[Z]stackrel{text{MCT}}{=}E[X_1]E[Z]
      end{align}

      Do something similar always works, giving us the reason why people just use it? Indeed, I guess it is boring to repeat these kind of steps if it always works, therefore my question:




      Question. Can we actually apply the law of total expectation to show integrability? If the answer is "yes", is there a general proof? If the answer is "no", are there examples where it fails?








      probability-theory conditional-expectation expected-value






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      edited Nov 17 at 15:50

























      asked Nov 17 at 15:43









      Shashi

      6,9191527




      6,9191527






















          1 Answer
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          All the usual laws of conditional expectation work for nonnegative random variables, integrable or not, if you allow the value $+infty$. For instance, if $X$ is nonnegative, $E[X mid Y]$ will always exist, but potentially could equal $+infty$ with positive probability. And if $X$ is nonnegative and not integrable, then you will get $E[E[X mid Y]] = +infty$ as well (i.e. $E[X mid Y]$ is nonnegative and not integrable). These cases are often left as exercises since it is a little tedious to write out the proofs, but you can show it by considering "cut off" random variables like $X_n = X 1_{X le n}$ and using monotone convergence.



          The properties do not necessarily hold for signed random variables that may not be integrable. However, you can determine whether a random variable $X$ is integrable by computing $E[|X| mid Y]$ and seeing if $E[E[|X| mid Y]] < infty$, and if so then you will know that $E[X] = E[E[X mid Y]]$.






          share|cite|improve this answer





















          • So it actually makes sense to use the law of total expectation for checking integrability. Many thanks! I suddenly remember that there is an exercise like that in Rick Durret's book. Silly me that I forgot about it. Thanks again!
            – Shashi
            Nov 17 at 16:05











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          1 Answer
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          active

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          up vote
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          down vote



          accepted










          All the usual laws of conditional expectation work for nonnegative random variables, integrable or not, if you allow the value $+infty$. For instance, if $X$ is nonnegative, $E[X mid Y]$ will always exist, but potentially could equal $+infty$ with positive probability. And if $X$ is nonnegative and not integrable, then you will get $E[E[X mid Y]] = +infty$ as well (i.e. $E[X mid Y]$ is nonnegative and not integrable). These cases are often left as exercises since it is a little tedious to write out the proofs, but you can show it by considering "cut off" random variables like $X_n = X 1_{X le n}$ and using monotone convergence.



          The properties do not necessarily hold for signed random variables that may not be integrable. However, you can determine whether a random variable $X$ is integrable by computing $E[|X| mid Y]$ and seeing if $E[E[|X| mid Y]] < infty$, and if so then you will know that $E[X] = E[E[X mid Y]]$.






          share|cite|improve this answer





















          • So it actually makes sense to use the law of total expectation for checking integrability. Many thanks! I suddenly remember that there is an exercise like that in Rick Durret's book. Silly me that I forgot about it. Thanks again!
            – Shashi
            Nov 17 at 16:05















          up vote
          1
          down vote



          accepted










          All the usual laws of conditional expectation work for nonnegative random variables, integrable or not, if you allow the value $+infty$. For instance, if $X$ is nonnegative, $E[X mid Y]$ will always exist, but potentially could equal $+infty$ with positive probability. And if $X$ is nonnegative and not integrable, then you will get $E[E[X mid Y]] = +infty$ as well (i.e. $E[X mid Y]$ is nonnegative and not integrable). These cases are often left as exercises since it is a little tedious to write out the proofs, but you can show it by considering "cut off" random variables like $X_n = X 1_{X le n}$ and using monotone convergence.



          The properties do not necessarily hold for signed random variables that may not be integrable. However, you can determine whether a random variable $X$ is integrable by computing $E[|X| mid Y]$ and seeing if $E[E[|X| mid Y]] < infty$, and if so then you will know that $E[X] = E[E[X mid Y]]$.






          share|cite|improve this answer





















          • So it actually makes sense to use the law of total expectation for checking integrability. Many thanks! I suddenly remember that there is an exercise like that in Rick Durret's book. Silly me that I forgot about it. Thanks again!
            – Shashi
            Nov 17 at 16:05













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          All the usual laws of conditional expectation work for nonnegative random variables, integrable or not, if you allow the value $+infty$. For instance, if $X$ is nonnegative, $E[X mid Y]$ will always exist, but potentially could equal $+infty$ with positive probability. And if $X$ is nonnegative and not integrable, then you will get $E[E[X mid Y]] = +infty$ as well (i.e. $E[X mid Y]$ is nonnegative and not integrable). These cases are often left as exercises since it is a little tedious to write out the proofs, but you can show it by considering "cut off" random variables like $X_n = X 1_{X le n}$ and using monotone convergence.



          The properties do not necessarily hold for signed random variables that may not be integrable. However, you can determine whether a random variable $X$ is integrable by computing $E[|X| mid Y]$ and seeing if $E[E[|X| mid Y]] < infty$, and if so then you will know that $E[X] = E[E[X mid Y]]$.






          share|cite|improve this answer












          All the usual laws of conditional expectation work for nonnegative random variables, integrable or not, if you allow the value $+infty$. For instance, if $X$ is nonnegative, $E[X mid Y]$ will always exist, but potentially could equal $+infty$ with positive probability. And if $X$ is nonnegative and not integrable, then you will get $E[E[X mid Y]] = +infty$ as well (i.e. $E[X mid Y]$ is nonnegative and not integrable). These cases are often left as exercises since it is a little tedious to write out the proofs, but you can show it by considering "cut off" random variables like $X_n = X 1_{X le n}$ and using monotone convergence.



          The properties do not necessarily hold for signed random variables that may not be integrable. However, you can determine whether a random variable $X$ is integrable by computing $E[|X| mid Y]$ and seeing if $E[E[|X| mid Y]] < infty$, and if so then you will know that $E[X] = E[E[X mid Y]]$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 15:58









          Nate Eldredge

          61.6k680167




          61.6k680167












          • So it actually makes sense to use the law of total expectation for checking integrability. Many thanks! I suddenly remember that there is an exercise like that in Rick Durret's book. Silly me that I forgot about it. Thanks again!
            – Shashi
            Nov 17 at 16:05


















          • So it actually makes sense to use the law of total expectation for checking integrability. Many thanks! I suddenly remember that there is an exercise like that in Rick Durret's book. Silly me that I forgot about it. Thanks again!
            – Shashi
            Nov 17 at 16:05
















          So it actually makes sense to use the law of total expectation for checking integrability. Many thanks! I suddenly remember that there is an exercise like that in Rick Durret's book. Silly me that I forgot about it. Thanks again!
          – Shashi
          Nov 17 at 16:05




          So it actually makes sense to use the law of total expectation for checking integrability. Many thanks! I suddenly remember that there is an exercise like that in Rick Durret's book. Silly me that I forgot about it. Thanks again!
          – Shashi
          Nov 17 at 16:05


















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