Why does expectation of a transformation of one random variable appear different from expectation of one...












3














Let's say we have a continuous random variable $X_{1}$ with a probability density function $p_{X_{1}}$. The expectation is: $$E(X_1) = int_{X_{1}} dx_{1} Big ( x_{1} cdot p_{X_{1}}(x_{1}) Big)$$.



Now say we have a random variable $Y_{1}$, which is obtained as a transformation $g_{1}:X_{1} rightarrow Y_{1}$ where $y_{1} = (x_{1} - 4)^{2} $. The expectation for $Y_{1}$ is: $$E(Y_{1})=int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big) $$.



Why is the expectation not: $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$? Does it mean $p_{Y_{1}}big((x_{1}-4)^2big) = p_{X_{1}}(x_{1}) $ and if so why is that the case?






probability random-variables expectation transformation random-functions







share|cite|improve this question




















  • 1




    That's the Law of the unconscious statistician, incidentally.
    – Clement C.
    Jun 13 '17 at 14:40










  • Wait how can you prove $p_{Y_1}=p_{x_1}$? Are your sure that is correct, you use different notation than I am used to but this seems wrong. if $p_{Y_1}=p_{x_1}$ than the change of being in an interval ${a,b}$ is equal for both distributions which seems weird at least
    – zen
    Jun 13 '17 at 15:18












  • @zen I used the method of transformations $p_{Y_{1}} = p_{X_{1}}(h_{1}) cdot det(boldsymbol{J})$ where $h_{1}$ is the inverse of $g_{1}$ and $J = frac{d}{dy_{1}}h_{1}$, but could have made a mistake.
    – A.L. Verminburger
    Jun 13 '17 at 15:26












  • I am going to remove the bit $p_{Y_{1}} = p_{X_{1}}$ to improve clarity of the question.
    – A.L. Verminburger
    Jun 14 '17 at 6:45
















3














Let's say we have a continuous random variable $X_{1}$ with a probability density function $p_{X_{1}}$. The expectation is: $$E(X_1) = int_{X_{1}} dx_{1} Big ( x_{1} cdot p_{X_{1}}(x_{1}) Big)$$.



Now say we have a random variable $Y_{1}$, which is obtained as a transformation $g_{1}:X_{1} rightarrow Y_{1}$ where $y_{1} = (x_{1} - 4)^{2} $. The expectation for $Y_{1}$ is: $$E(Y_{1})=int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big) $$.



Why is the expectation not: $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$? Does it mean $p_{Y_{1}}big((x_{1}-4)^2big) = p_{X_{1}}(x_{1}) $ and if so why is that the case?






probability random-variables expectation transformation random-functions







share|cite|improve this question




















  • 1




    That's the Law of the unconscious statistician, incidentally.
    – Clement C.
    Jun 13 '17 at 14:40










  • Wait how can you prove $p_{Y_1}=p_{x_1}$? Are your sure that is correct, you use different notation than I am used to but this seems wrong. if $p_{Y_1}=p_{x_1}$ than the change of being in an interval ${a,b}$ is equal for both distributions which seems weird at least
    – zen
    Jun 13 '17 at 15:18












  • @zen I used the method of transformations $p_{Y_{1}} = p_{X_{1}}(h_{1}) cdot det(boldsymbol{J})$ where $h_{1}$ is the inverse of $g_{1}$ and $J = frac{d}{dy_{1}}h_{1}$, but could have made a mistake.
    – A.L. Verminburger
    Jun 13 '17 at 15:26












  • I am going to remove the bit $p_{Y_{1}} = p_{X_{1}}$ to improve clarity of the question.
    – A.L. Verminburger
    Jun 14 '17 at 6:45














3












3








3







Let's say we have a continuous random variable $X_{1}$ with a probability density function $p_{X_{1}}$. The expectation is: $$E(X_1) = int_{X_{1}} dx_{1} Big ( x_{1} cdot p_{X_{1}}(x_{1}) Big)$$.



Now say we have a random variable $Y_{1}$, which is obtained as a transformation $g_{1}:X_{1} rightarrow Y_{1}$ where $y_{1} = (x_{1} - 4)^{2} $. The expectation for $Y_{1}$ is: $$E(Y_{1})=int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big) $$.



Why is the expectation not: $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$? Does it mean $p_{Y_{1}}big((x_{1}-4)^2big) = p_{X_{1}}(x_{1}) $ and if so why is that the case?






probability random-variables expectation transformation random-functions







share|cite|improve this question















Let's say we have a continuous random variable $X_{1}$ with a probability density function $p_{X_{1}}$. The expectation is: $$E(X_1) = int_{X_{1}} dx_{1} Big ( x_{1} cdot p_{X_{1}}(x_{1}) Big)$$.



Now say we have a random variable $Y_{1}$, which is obtained as a transformation $g_{1}:X_{1} rightarrow Y_{1}$ where $y_{1} = (x_{1} - 4)^{2} $. The expectation for $Y_{1}$ is: $$E(Y_{1})=int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big) $$.



Why is the expectation not: $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$? Does it mean $p_{Y_{1}}big((x_{1}-4)^2big) = p_{X_{1}}(x_{1}) $ and if so why is that the case?






probability random-variables expectation transformation random-functions




probability random-variables expectation transformation random-functions






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share|cite|improve this question













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edited Jun 14 '17 at 6:51







A.L. Verminburger

















asked Jun 13 '17 at 14:37









A.L. VerminburgerA.L. Verminburger

313311




313311








  • 1




    That's the Law of the unconscious statistician, incidentally.
    – Clement C.
    Jun 13 '17 at 14:40










  • Wait how can you prove $p_{Y_1}=p_{x_1}$? Are your sure that is correct, you use different notation than I am used to but this seems wrong. if $p_{Y_1}=p_{x_1}$ than the change of being in an interval ${a,b}$ is equal for both distributions which seems weird at least
    – zen
    Jun 13 '17 at 15:18












  • @zen I used the method of transformations $p_{Y_{1}} = p_{X_{1}}(h_{1}) cdot det(boldsymbol{J})$ where $h_{1}$ is the inverse of $g_{1}$ and $J = frac{d}{dy_{1}}h_{1}$, but could have made a mistake.
    – A.L. Verminburger
    Jun 13 '17 at 15:26












  • I am going to remove the bit $p_{Y_{1}} = p_{X_{1}}$ to improve clarity of the question.
    – A.L. Verminburger
    Jun 14 '17 at 6:45














  • 1




    That's the Law of the unconscious statistician, incidentally.
    – Clement C.
    Jun 13 '17 at 14:40










  • Wait how can you prove $p_{Y_1}=p_{x_1}$? Are your sure that is correct, you use different notation than I am used to but this seems wrong. if $p_{Y_1}=p_{x_1}$ than the change of being in an interval ${a,b}$ is equal for both distributions which seems weird at least
    – zen
    Jun 13 '17 at 15:18












  • @zen I used the method of transformations $p_{Y_{1}} = p_{X_{1}}(h_{1}) cdot det(boldsymbol{J})$ where $h_{1}$ is the inverse of $g_{1}$ and $J = frac{d}{dy_{1}}h_{1}$, but could have made a mistake.
    – A.L. Verminburger
    Jun 13 '17 at 15:26












  • I am going to remove the bit $p_{Y_{1}} = p_{X_{1}}$ to improve clarity of the question.
    – A.L. Verminburger
    Jun 14 '17 at 6:45








1




1




That's the Law of the unconscious statistician, incidentally.
– Clement C.
Jun 13 '17 at 14:40




That's the Law of the unconscious statistician, incidentally.
– Clement C.
Jun 13 '17 at 14:40












Wait how can you prove $p_{Y_1}=p_{x_1}$? Are your sure that is correct, you use different notation than I am used to but this seems wrong. if $p_{Y_1}=p_{x_1}$ than the change of being in an interval ${a,b}$ is equal for both distributions which seems weird at least
– zen
Jun 13 '17 at 15:18






Wait how can you prove $p_{Y_1}=p_{x_1}$? Are your sure that is correct, you use different notation than I am used to but this seems wrong. if $p_{Y_1}=p_{x_1}$ than the change of being in an interval ${a,b}$ is equal for both distributions which seems weird at least
– zen
Jun 13 '17 at 15:18














@zen I used the method of transformations $p_{Y_{1}} = p_{X_{1}}(h_{1}) cdot det(boldsymbol{J})$ where $h_{1}$ is the inverse of $g_{1}$ and $J = frac{d}{dy_{1}}h_{1}$, but could have made a mistake.
– A.L. Verminburger
Jun 13 '17 at 15:26






@zen I used the method of transformations $p_{Y_{1}} = p_{X_{1}}(h_{1}) cdot det(boldsymbol{J})$ where $h_{1}$ is the inverse of $g_{1}$ and $J = frac{d}{dy_{1}}h_{1}$, but could have made a mistake.
– A.L. Verminburger
Jun 13 '17 at 15:26














I am going to remove the bit $p_{Y_{1}} = p_{X_{1}}$ to improve clarity of the question.
– A.L. Verminburger
Jun 14 '17 at 6:45




I am going to remove the bit $p_{Y_{1}} = p_{X_{1}}$ to improve clarity of the question.
– A.L. Verminburger
Jun 14 '17 at 6:45










2 Answers
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Because
$$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) neq int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$



you need to substitute the expression for $y_1$ into the differential too.






share|cite|improve this answer





















  • Is $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) = int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big)$ then? And if so, why?
    – A.L. Verminburger
    Jun 14 '17 at 7:34










  • This is true because of the expected value rule.
    – kludg
    Jun 14 '17 at 7:56



















0














$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$



is not true.



This is because by definition of expected value we know that the expected value of a discrete random variable is the probability-weighted average of all possible values. The same principle applies to an absolutely continuous random variable, except that an integral of the variable with respect to its probability density replaces the sum.



And, we also know that a function of a random variable i.e. a transformation of type $g_{1}:X_{1} rightarrow Y_{1}$ is a random variable. In the above case you have a probability measure defined over $mathbb{R}$ (the distribution of $Y$) instead of $mathbb{R}$(the distribution of $X$).



You can compute expectation in below three ways, in general:



Assuming the probability measure $P$ on the sample space $S$ has a pdf $f$. And, random variable $X$ is mapping from probability space $(S,P)$ to $mathbb{R}$ and function $g$ mapping $mathbb{R}$ into $mathbb{R}$, and $g(X) = Y$, then expectation of $Y$ can be calculated in following ways:



$E(Y)= int_{Y} Big( y cdot p_{Y}(y) Big) dy$



$ =int_{X} Big( g(X)cdot p_{X}(X) Big)dx$



$ = int_{sin{S}} g(X(s))cdot f(s)ds$






share|cite|improve this answer























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    2 Answers
    2






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    2 Answers
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    active

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    Because
    $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) neq int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$



    you need to substitute the expression for $y_1$ into the differential too.






    share|cite|improve this answer





















    • Is $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) = int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big)$ then? And if so, why?
      – A.L. Verminburger
      Jun 14 '17 at 7:34










    • This is true because of the expected value rule.
      – kludg
      Jun 14 '17 at 7:56
















    1














    Because
    $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) neq int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$



    you need to substitute the expression for $y_1$ into the differential too.






    share|cite|improve this answer





















    • Is $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) = int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big)$ then? And if so, why?
      – A.L. Verminburger
      Jun 14 '17 at 7:34










    • This is true because of the expected value rule.
      – kludg
      Jun 14 '17 at 7:56














    1












    1








    1






    Because
    $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) neq int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$



    you need to substitute the expression for $y_1$ into the differential too.






    share|cite|improve this answer












    Because
    $$E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) neq int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$$



    you need to substitute the expression for $y_1$ into the differential too.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 14 '17 at 7:00









    kludgkludg

    1,183611




    1,183611












    • Is $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) = int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big)$ then? And if so, why?
      – A.L. Verminburger
      Jun 14 '17 at 7:34










    • This is true because of the expected value rule.
      – kludg
      Jun 14 '17 at 7:56


















    • Is $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) = int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big)$ then? And if so, why?
      – A.L. Verminburger
      Jun 14 '17 at 7:34










    • This is true because of the expected value rule.
      – kludg
      Jun 14 '17 at 7:56
















    Is $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) = int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big)$ then? And if so, why?
    – A.L. Verminburger
    Jun 14 '17 at 7:34




    Is $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) = int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{X_{1}}(x_{1}) Big)$ then? And if so, why?
    – A.L. Verminburger
    Jun 14 '17 at 7:34












    This is true because of the expected value rule.
    – kludg
    Jun 14 '17 at 7:56




    This is true because of the expected value rule.
    – kludg
    Jun 14 '17 at 7:56











    0














    $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$



    is not true.



    This is because by definition of expected value we know that the expected value of a discrete random variable is the probability-weighted average of all possible values. The same principle applies to an absolutely continuous random variable, except that an integral of the variable with respect to its probability density replaces the sum.



    And, we also know that a function of a random variable i.e. a transformation of type $g_{1}:X_{1} rightarrow Y_{1}$ is a random variable. In the above case you have a probability measure defined over $mathbb{R}$ (the distribution of $Y$) instead of $mathbb{R}$(the distribution of $X$).



    You can compute expectation in below three ways, in general:



    Assuming the probability measure $P$ on the sample space $S$ has a pdf $f$. And, random variable $X$ is mapping from probability space $(S,P)$ to $mathbb{R}$ and function $g$ mapping $mathbb{R}$ into $mathbb{R}$, and $g(X) = Y$, then expectation of $Y$ can be calculated in following ways:



    $E(Y)= int_{Y} Big( y cdot p_{Y}(y) Big) dy$



    $ =int_{X} Big( g(X)cdot p_{X}(X) Big)dx$



    $ = int_{sin{S}} g(X(s))cdot f(s)ds$






    share|cite|improve this answer




























      0














      $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$



      is not true.



      This is because by definition of expected value we know that the expected value of a discrete random variable is the probability-weighted average of all possible values. The same principle applies to an absolutely continuous random variable, except that an integral of the variable with respect to its probability density replaces the sum.



      And, we also know that a function of a random variable i.e. a transformation of type $g_{1}:X_{1} rightarrow Y_{1}$ is a random variable. In the above case you have a probability measure defined over $mathbb{R}$ (the distribution of $Y$) instead of $mathbb{R}$(the distribution of $X$).



      You can compute expectation in below three ways, in general:



      Assuming the probability measure $P$ on the sample space $S$ has a pdf $f$. And, random variable $X$ is mapping from probability space $(S,P)$ to $mathbb{R}$ and function $g$ mapping $mathbb{R}$ into $mathbb{R}$, and $g(X) = Y$, then expectation of $Y$ can be calculated in following ways:



      $E(Y)= int_{Y} Big( y cdot p_{Y}(y) Big) dy$



      $ =int_{X} Big( g(X)cdot p_{X}(X) Big)dx$



      $ = int_{sin{S}} g(X(s))cdot f(s)ds$






      share|cite|improve this answer


























        0












        0








        0






        $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$



        is not true.



        This is because by definition of expected value we know that the expected value of a discrete random variable is the probability-weighted average of all possible values. The same principle applies to an absolutely continuous random variable, except that an integral of the variable with respect to its probability density replaces the sum.



        And, we also know that a function of a random variable i.e. a transformation of type $g_{1}:X_{1} rightarrow Y_{1}$ is a random variable. In the above case you have a probability measure defined over $mathbb{R}$ (the distribution of $Y$) instead of $mathbb{R}$(the distribution of $X$).



        You can compute expectation in below three ways, in general:



        Assuming the probability measure $P$ on the sample space $S$ has a pdf $f$. And, random variable $X$ is mapping from probability space $(S,P)$ to $mathbb{R}$ and function $g$ mapping $mathbb{R}$ into $mathbb{R}$, and $g(X) = Y$, then expectation of $Y$ can be calculated in following ways:



        $E(Y)= int_{Y} Big( y cdot p_{Y}(y) Big) dy$



        $ =int_{X} Big( g(X)cdot p_{X}(X) Big)dx$



        $ = int_{sin{S}} g(X(s))cdot f(s)ds$






        share|cite|improve this answer














        $E(Y_{1})= int_{Y_{1}} dy_{1} Big( y_{1} cdot p_{Y_{1}}(y_{1}) Big) =int_{X_{1}} dx_{1} Big( (x_{1}-4)^2 cdot p_{Y_{1}}((x_{1}-4)^2) Big)$



        is not true.



        This is because by definition of expected value we know that the expected value of a discrete random variable is the probability-weighted average of all possible values. The same principle applies to an absolutely continuous random variable, except that an integral of the variable with respect to its probability density replaces the sum.



        And, we also know that a function of a random variable i.e. a transformation of type $g_{1}:X_{1} rightarrow Y_{1}$ is a random variable. In the above case you have a probability measure defined over $mathbb{R}$ (the distribution of $Y$) instead of $mathbb{R}$(the distribution of $X$).



        You can compute expectation in below three ways, in general:



        Assuming the probability measure $P$ on the sample space $S$ has a pdf $f$. And, random variable $X$ is mapping from probability space $(S,P)$ to $mathbb{R}$ and function $g$ mapping $mathbb{R}$ into $mathbb{R}$, and $g(X) = Y$, then expectation of $Y$ can be calculated in following ways:



        $E(Y)= int_{Y} Big( y cdot p_{Y}(y) Big) dy$



        $ =int_{X} Big( g(X)cdot p_{X}(X) Big)dx$



        $ = int_{sin{S}} g(X(s))cdot f(s)ds$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 6:05

























        answered Nov 7 '18 at 9:52









        naivenaive

        3114




        3114






























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