Showing that $X + Y$ and $X/Y$ are independent given that $X$ and $Y$ are i.i.d with a PDF












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Let $X$ and $Y$ be i.i.d with pdfs given by



$$f_{X}(x) = begin{cases}
text{exp}(-x), & text{ if } x > 0, \
0 & text{ otherwise},
end{cases} $$



and



$$f_{Y}(y) = begin{cases}
text{exp}(-y), & text{ if } x > 0, \
0 & text{ otherwise}.
end{cases} $$



Show that $X + Y$ and $X/Y$ are independent.




My attempt:



By independence, we have



$$f_{X, Y}(x, y) = begin{cases}
text{exp}(-(x + y)) & text{ if } x+y > 0 \
0 & text{ otherwise.}
end{cases} $$



Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by



$$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$



$$= -frac{X + Y}{Y^{2}} $$



In terms of $Y_{1}$ and $Y_{2}$, we have



$$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$



Therefore, our new probability density function is given by



$$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$



$$= begin{cases}
dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
0 & text{ otherwise.}
end{cases} $$



So, we compute the marginal densities as follows:



$$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$



but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)






probability probability-theory







share|cite|improve this question



























    0














    Let $X$ and $Y$ be i.i.d with pdfs given by



    $$f_{X}(x) = begin{cases}
    text{exp}(-x), & text{ if } x > 0, \
    0 & text{ otherwise},
    end{cases} $$



    and



    $$f_{Y}(y) = begin{cases}
    text{exp}(-y), & text{ if } x > 0, \
    0 & text{ otherwise}.
    end{cases} $$



    Show that $X + Y$ and $X/Y$ are independent.




    My attempt:



    By independence, we have



    $$f_{X, Y}(x, y) = begin{cases}
    text{exp}(-(x + y)) & text{ if } x+y > 0 \
    0 & text{ otherwise.}
    end{cases} $$



    Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by



    $$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$



    $$= -frac{X + Y}{Y^{2}} $$



    In terms of $Y_{1}$ and $Y_{2}$, we have



    $$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$



    Therefore, our new probability density function is given by



    $$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$



    $$= begin{cases}
    dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
    0 & text{ otherwise.}
    end{cases} $$



    So, we compute the marginal densities as follows:



    $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$



    but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)






    probability probability-theory







    share|cite|improve this question

























      0












      0








      0







      Let $X$ and $Y$ be i.i.d with pdfs given by



      $$f_{X}(x) = begin{cases}
      text{exp}(-x), & text{ if } x > 0, \
      0 & text{ otherwise},
      end{cases} $$



      and



      $$f_{Y}(y) = begin{cases}
      text{exp}(-y), & text{ if } x > 0, \
      0 & text{ otherwise}.
      end{cases} $$



      Show that $X + Y$ and $X/Y$ are independent.




      My attempt:



      By independence, we have



      $$f_{X, Y}(x, y) = begin{cases}
      text{exp}(-(x + y)) & text{ if } x+y > 0 \
      0 & text{ otherwise.}
      end{cases} $$



      Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by



      $$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$



      $$= -frac{X + Y}{Y^{2}} $$



      In terms of $Y_{1}$ and $Y_{2}$, we have



      $$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$



      Therefore, our new probability density function is given by



      $$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$



      $$= begin{cases}
      dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
      0 & text{ otherwise.}
      end{cases} $$



      So, we compute the marginal densities as follows:



      $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$



      but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)






      probability probability-theory







      share|cite|improve this question













      Let $X$ and $Y$ be i.i.d with pdfs given by



      $$f_{X}(x) = begin{cases}
      text{exp}(-x), & text{ if } x > 0, \
      0 & text{ otherwise},
      end{cases} $$



      and



      $$f_{Y}(y) = begin{cases}
      text{exp}(-y), & text{ if } x > 0, \
      0 & text{ otherwise}.
      end{cases} $$



      Show that $X + Y$ and $X/Y$ are independent.




      My attempt:



      By independence, we have



      $$f_{X, Y}(x, y) = begin{cases}
      text{exp}(-(x + y)) & text{ if } x+y > 0 \
      0 & text{ otherwise.}
      end{cases} $$



      Let $Y_{1} = g_{1}(X, Y) = X + Y$ and $Y_{2} = g_{2}(X, Y) = X/Y$. Solving for $X$ and $Y$ gives us the solutions $X = Y_{1}Y_{2}/(Y_{2} + 1)$ and $Y = Y_{1}/(Y_{2} + 1)$. Thus, our Jacobian is given by



      $$J = frac{partial g_{1}}{partial X}frac{partial g_{2}}{partial Y} - frac{partial g_{1}}{partial Y}frac{partial g_{2}}{partial X} $$



      $$= -frac{X + Y}{Y^{2}} $$



      In terms of $Y_{1}$ and $Y_{2}$, we have



      $$-frac{X + Y}{Y^{2}} = -frac{(Y_{2} + 1)^{2}}{Y_{1}}.$$



      Therefore, our new probability density function is given by



      $$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = f(x_{1}, x_{2}) cdot |J|^{-1} $$



      $$= begin{cases}
      dfrac{y_{1}}{(y_{2} + 1)^{2}} cdot text{exp}(-y_{1}) & text{ if } y_{1} > 0 \[1em]
      0 & text{ otherwise.}
      end{cases} $$



      So, we compute the marginal densities as follows:



      $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{-infty}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}},$$



      but the integral on the right diverges. Where did I go wrong? Can't figure it out. Also, did I do anything else wrong (in particular, are my domains for $f_{X, Y}$ and $f_{Y_{1}, Y_{2}}$ correct?)






      probability probability-theory




      probability probability-theory






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      asked Nov 24 at 4:14









      stackofhay42

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          We have if $y_1 > 0$, then



          $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$






          share|cite|improve this answer





















          • Why is the lower bound $0$?
            – stackofhay42
            Nov 24 at 4:26










          • $y_2$ only take positive values right since it is a division of two positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:27










          • Ok. Then, $y_{1}$ must only take positive values too, right?
            – stackofhay42
            Nov 24 at 4:28










          • yes, addition of positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:29











          Your Answer





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          We have if $y_1 > 0$, then



          $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$






          share|cite|improve this answer





















          • Why is the lower bound $0$?
            – stackofhay42
            Nov 24 at 4:26










          • $y_2$ only take positive values right since it is a division of two positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:27










          • Ok. Then, $y_{1}$ must only take positive values too, right?
            – stackofhay42
            Nov 24 at 4:28










          • yes, addition of positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:29
















          1














          We have if $y_1 > 0$, then



          $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$






          share|cite|improve this answer





















          • Why is the lower bound $0$?
            – stackofhay42
            Nov 24 at 4:26










          • $y_2$ only take positive values right since it is a division of two positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:27










          • Ok. Then, $y_{1}$ must only take positive values too, right?
            – stackofhay42
            Nov 24 at 4:28










          • yes, addition of positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:29














          1












          1








          1






          We have if $y_1 > 0$, then



          $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$






          share|cite|improve this answer












          We have if $y_1 > 0$, then



          $$f_{Y_{1}}(y_{1}) = int_{-infty}^{infty} f_{Y_{1},Y_{2}}(y_{1}, y_{2}) mathop{dy_{2}} = y_{1}text{exp}(-y_{1}) cdot int_{color{red}0}^{infty} frac{1}{(y_{2} + 1)^{2}} mathop{dy_{2}} = y_1 exp(-y_1)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 4:22









          Siong Thye Goh

          98.9k1464116




          98.9k1464116












          • Why is the lower bound $0$?
            – stackofhay42
            Nov 24 at 4:26










          • $y_2$ only take positive values right since it is a division of two positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:27










          • Ok. Then, $y_{1}$ must only take positive values too, right?
            – stackofhay42
            Nov 24 at 4:28










          • yes, addition of positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:29


















          • Why is the lower bound $0$?
            – stackofhay42
            Nov 24 at 4:26










          • $y_2$ only take positive values right since it is a division of two positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:27










          • Ok. Then, $y_{1}$ must only take positive values too, right?
            – stackofhay42
            Nov 24 at 4:28










          • yes, addition of positive numbers.
            – Siong Thye Goh
            Nov 24 at 4:29
















          Why is the lower bound $0$?
          – stackofhay42
          Nov 24 at 4:26




          Why is the lower bound $0$?
          – stackofhay42
          Nov 24 at 4:26












          $y_2$ only take positive values right since it is a division of two positive numbers.
          – Siong Thye Goh
          Nov 24 at 4:27




          $y_2$ only take positive values right since it is a division of two positive numbers.
          – Siong Thye Goh
          Nov 24 at 4:27












          Ok. Then, $y_{1}$ must only take positive values too, right?
          – stackofhay42
          Nov 24 at 4:28




          Ok. Then, $y_{1}$ must only take positive values too, right?
          – stackofhay42
          Nov 24 at 4:28












          yes, addition of positive numbers.
          – Siong Thye Goh
          Nov 24 at 4:29




          yes, addition of positive numbers.
          – Siong Thye Goh
          Nov 24 at 4:29


















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