Equality in distribution of Cauchy random variables

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Let $X$ be a Cauchy $C(1)$ random variable and let $Y_a = frac1-aXa-X$, where $a$ is a real number. I need to prove that $(s+t)X$ is equal in distribution to $sX - fractX$.



I calculated the distribution of $Y_a$ and it's equal to $X$ in distribution for every $a$. I also know that $frac-1X = Y_0$. But I can't conclude from above the equality in distribution.




probability probability-theory




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    up vote
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    down vote

    favorite












    Let $X$ be a Cauchy $C(1)$ random variable and let $Y_a = frac1-aXa-X$, where $a$ is a real number. I need to prove that $(s+t)X$ is equal in distribution to $sX - fractX$.



    I calculated the distribution of $Y_a$ and it's equal to $X$ in distribution for every $a$. I also know that $frac-1X = Y_0$. But I can't conclude from above the equality in distribution.




    probability probability-theory




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $X$ be a Cauchy $C(1)$ random variable and let $Y_a = frac1-aXa-X$, where $a$ is a real number. I need to prove that $(s+t)X$ is equal in distribution to $sX - fractX$.



      I calculated the distribution of $Y_a$ and it's equal to $X$ in distribution for every $a$. I also know that $frac-1X = Y_0$. But I can't conclude from above the equality in distribution.




      probability probability-theory




      share|cite|improve this question











      Let $X$ be a Cauchy $C(1)$ random variable and let $Y_a = frac1-aXa-X$, where $a$ is a real number. I need to prove that $(s+t)X$ is equal in distribution to $sX - fractX$.



      I calculated the distribution of $Y_a$ and it's equal to $X$ in distribution for every $a$. I also know that $frac-1X = Y_0$. But I can't conclude from above the equality in distribution.





      probability probability-theory





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









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      MadChemist

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