Finding the PDF of the cubed difference between two independent standard normal random variables

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Let $X_1$ and $X_2$ be independent $operatornameN(0,1)$ random variables. Find the PDF of $(X_1-X_2)^3 / 2$ .




I see that if it were the squared difference then I could use chi-squared with $1$ degree of freedom—but this one is a cubed difference. Can I follow the same approach? Any suggestions appreciated!






probability density-function







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    "Can I follow the same approach?" Which "same approach"?
    – Did
    Aug 31 at 5:39














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Let $X_1$ and $X_2$ be independent $operatornameN(0,1)$ random variables. Find the PDF of $(X_1-X_2)^3 / 2$ .




I see that if it were the squared difference then I could use chi-squared with $1$ degree of freedom—but this one is a cubed difference. Can I follow the same approach? Any suggestions appreciated!






probability density-function







share|cite|improve this question



















  • 1




    "Can I follow the same approach?" Which "same approach"?
    – Did
    Aug 31 at 5:39












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite












Let $X_1$ and $X_2$ be independent $operatornameN(0,1)$ random variables. Find the PDF of $(X_1-X_2)^3 / 2$ .




I see that if it were the squared difference then I could use chi-squared with $1$ degree of freedom—but this one is a cubed difference. Can I follow the same approach? Any suggestions appreciated!






probability density-function







share|cite|improve this question
















Let $X_1$ and $X_2$ be independent $operatornameN(0,1)$ random variables. Find the PDF of $(X_1-X_2)^3 / 2$ .




I see that if it were the squared difference then I could use chi-squared with $1$ degree of freedom—but this one is a cubed difference. Can I follow the same approach? Any suggestions appreciated!






probability density-function




probability density-function






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edited Aug 31 at 15:05









Jendrik Stelzner

7,69121137




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asked Aug 31 at 4:53









Josh

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  • 1




    "Can I follow the same approach?" Which "same approach"?
    – Did
    Aug 31 at 5:39












  • 1




    "Can I follow the same approach?" Which "same approach"?
    – Did
    Aug 31 at 5:39







1




1




"Can I follow the same approach?" Which "same approach"?
– Did
Aug 31 at 5:39




"Can I follow the same approach?" Which "same approach"?
– Did
Aug 31 at 5:39










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If $Y=X_1-X_2$ then $Y$ is normal with mean $0$ and variance 2, so you can write down the density of $Y$. Can you now compute that density of $frac Y^3 2$?.






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    If $Y=X_1-X_2$ then $Y$ is normal with mean $0$ and variance 2, so you can write down the density of $Y$. Can you now compute that density of $frac Y^3 2$?.






    share|cite|improve this answer
























      up vote
      2
      down vote













      If $Y=X_1-X_2$ then $Y$ is normal with mean $0$ and variance 2, so you can write down the density of $Y$. Can you now compute that density of $frac Y^3 2$?.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        If $Y=X_1-X_2$ then $Y$ is normal with mean $0$ and variance 2, so you can write down the density of $Y$. Can you now compute that density of $frac Y^3 2$?.






        share|cite|improve this answer












        If $Y=X_1-X_2$ then $Y$ is normal with mean $0$ and variance 2, so you can write down the density of $Y$. Can you now compute that density of $frac Y^3 2$?.







        share|cite|improve this answer












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        answered Aug 31 at 5:30









        Kavi Rama Murthy

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