Given $ X sim exp(lambda) $ find the density function of $Y=frac{1}{1-X}$












0












$begingroup$



Given $ X sim exp(lambda) $ find the density function of
$Y=frac{1}{1-X}$




So I tried doing the following and gut stuck in the way:



$ F_Y(y)=P(Yleq y)=P(frac{1}{1-x}leq y)$



Now I though about isolating X, which gives me $ F_x(text{something}) $ and by deriving it, get the desired density of Y, but I couldn't figure out how to handle the inequality signs (because I can't simply multiple both sides by $1-x$).



What is so wrong with my way of thinking?






probability density-function







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Given $ X sim exp(lambda) $ find the density function of
    $Y=frac{1}{1-X}$




    So I tried doing the following and gut stuck in the way:



    $ F_Y(y)=P(Yleq y)=P(frac{1}{1-x}leq y)$



    Now I though about isolating X, which gives me $ F_x(text{something}) $ and by deriving it, get the desired density of Y, but I couldn't figure out how to handle the inequality signs (because I can't simply multiple both sides by $1-x$).



    What is so wrong with my way of thinking?






    probability density-function







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      3



      $begingroup$



      Given $ X sim exp(lambda) $ find the density function of
      $Y=frac{1}{1-X}$




      So I tried doing the following and gut stuck in the way:



      $ F_Y(y)=P(Yleq y)=P(frac{1}{1-x}leq y)$



      Now I though about isolating X, which gives me $ F_x(text{something}) $ and by deriving it, get the desired density of Y, but I couldn't figure out how to handle the inequality signs (because I can't simply multiple both sides by $1-x$).



      What is so wrong with my way of thinking?






      probability density-function







      share|cite|improve this question











      $endgroup$





      Given $ X sim exp(lambda) $ find the density function of
      $Y=frac{1}{1-X}$




      So I tried doing the following and gut stuck in the way:



      $ F_Y(y)=P(Yleq y)=P(frac{1}{1-x}leq y)$



      Now I though about isolating X, which gives me $ F_x(text{something}) $ and by deriving it, get the desired density of Y, but I couldn't figure out how to handle the inequality signs (because I can't simply multiple both sides by $1-x$).



      What is so wrong with my way of thinking?






      probability density-function




      probability density-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 21:35









      Asaf Karagila♦

      309k33441775




      309k33441775










      asked Jan 15 at 21:06









      superuser123superuser123

      48628




      48628






















          1 Answer
          1






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          2












          $begingroup$

          Note that $Y$ takes values in $(-infty, 0) cup [1, infty)$. It might be helpful to take each case separately.



          If $y < 0$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X > 1) = P(1 < X le 1-frac{1}{y}).$$



          If $y in [0, 1)$ then $P(frac{1}{1-X} le y) = P(frac{1}{1-X} le 0) = P(X > 1)$.



          If $y ge 1$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X < 1) + P(X > 1) = P(X le 1 - frac{1}{y}) + P(X > 1).$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks, but could you please clarify your first statement: "please not that Y takes these values" because I am not sure how it works
            $endgroup$
            – superuser123
            Jan 15 at 21:23






          • 1




            $begingroup$
            @superuser123 Note that $X$ is always nonnegative, and think about what values $frac{1}{1-X}$ could be. It may help to plot the function $f(x) = frac{1}{1-x}$.
            $endgroup$
            – angryavian
            Jan 15 at 21:30










          • $begingroup$
            could you please clarify the second case? $y in [0, 1)$, and in general, could you please explain how you knew how to divide to these cases? thanks!
            $endgroup$
            – superuser123
            Jan 16 at 13:22










          • $begingroup$
            @superuser123 Look at a plot of $f(x) = frac{1}{1-x}$ and figure out where the graph lies below the horizontal line $y = 0.2$ or $y=0.8$ or any $y in [0, 1)$.
            $endgroup$
            – angryavian
            Jan 16 at 18:18












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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Note that $Y$ takes values in $(-infty, 0) cup [1, infty)$. It might be helpful to take each case separately.



          If $y < 0$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X > 1) = P(1 < X le 1-frac{1}{y}).$$



          If $y in [0, 1)$ then $P(frac{1}{1-X} le y) = P(frac{1}{1-X} le 0) = P(X > 1)$.



          If $y ge 1$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X < 1) + P(X > 1) = P(X le 1 - frac{1}{y}) + P(X > 1).$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks, but could you please clarify your first statement: "please not that Y takes these values" because I am not sure how it works
            $endgroup$
            – superuser123
            Jan 15 at 21:23






          • 1




            $begingroup$
            @superuser123 Note that $X$ is always nonnegative, and think about what values $frac{1}{1-X}$ could be. It may help to plot the function $f(x) = frac{1}{1-x}$.
            $endgroup$
            – angryavian
            Jan 15 at 21:30










          • $begingroup$
            could you please clarify the second case? $y in [0, 1)$, and in general, could you please explain how you knew how to divide to these cases? thanks!
            $endgroup$
            – superuser123
            Jan 16 at 13:22










          • $begingroup$
            @superuser123 Look at a plot of $f(x) = frac{1}{1-x}$ and figure out where the graph lies below the horizontal line $y = 0.2$ or $y=0.8$ or any $y in [0, 1)$.
            $endgroup$
            – angryavian
            Jan 16 at 18:18
















          2












          $begingroup$

          Note that $Y$ takes values in $(-infty, 0) cup [1, infty)$. It might be helpful to take each case separately.



          If $y < 0$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X > 1) = P(1 < X le 1-frac{1}{y}).$$



          If $y in [0, 1)$ then $P(frac{1}{1-X} le y) = P(frac{1}{1-X} le 0) = P(X > 1)$.



          If $y ge 1$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X < 1) + P(X > 1) = P(X le 1 - frac{1}{y}) + P(X > 1).$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks, but could you please clarify your first statement: "please not that Y takes these values" because I am not sure how it works
            $endgroup$
            – superuser123
            Jan 15 at 21:23






          • 1




            $begingroup$
            @superuser123 Note that $X$ is always nonnegative, and think about what values $frac{1}{1-X}$ could be. It may help to plot the function $f(x) = frac{1}{1-x}$.
            $endgroup$
            – angryavian
            Jan 15 at 21:30










          • $begingroup$
            could you please clarify the second case? $y in [0, 1)$, and in general, could you please explain how you knew how to divide to these cases? thanks!
            $endgroup$
            – superuser123
            Jan 16 at 13:22










          • $begingroup$
            @superuser123 Look at a plot of $f(x) = frac{1}{1-x}$ and figure out where the graph lies below the horizontal line $y = 0.2$ or $y=0.8$ or any $y in [0, 1)$.
            $endgroup$
            – angryavian
            Jan 16 at 18:18














          2












          2








          2





          $begingroup$

          Note that $Y$ takes values in $(-infty, 0) cup [1, infty)$. It might be helpful to take each case separately.



          If $y < 0$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X > 1) = P(1 < X le 1-frac{1}{y}).$$



          If $y in [0, 1)$ then $P(frac{1}{1-X} le y) = P(frac{1}{1-X} le 0) = P(X > 1)$.



          If $y ge 1$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X < 1) + P(X > 1) = P(X le 1 - frac{1}{y}) + P(X > 1).$$






          share|cite|improve this answer











          $endgroup$



          Note that $Y$ takes values in $(-infty, 0) cup [1, infty)$. It might be helpful to take each case separately.



          If $y < 0$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X > 1) = P(1 < X le 1-frac{1}{y}).$$



          If $y in [0, 1)$ then $P(frac{1}{1-X} le y) = P(frac{1}{1-X} le 0) = P(X > 1)$.



          If $y ge 1$, then
          $$P(frac{1}{1-X} le y) = P(frac{1}{1-X} le y, X < 1) + P(X > 1) = P(X le 1 - frac{1}{y}) + P(X > 1).$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 16 at 18:16

























          answered Jan 15 at 21:19









          angryavianangryavian

          43k23482




          43k23482












          • $begingroup$
            thanks, but could you please clarify your first statement: "please not that Y takes these values" because I am not sure how it works
            $endgroup$
            – superuser123
            Jan 15 at 21:23






          • 1




            $begingroup$
            @superuser123 Note that $X$ is always nonnegative, and think about what values $frac{1}{1-X}$ could be. It may help to plot the function $f(x) = frac{1}{1-x}$.
            $endgroup$
            – angryavian
            Jan 15 at 21:30










          • $begingroup$
            could you please clarify the second case? $y in [0, 1)$, and in general, could you please explain how you knew how to divide to these cases? thanks!
            $endgroup$
            – superuser123
            Jan 16 at 13:22










          • $begingroup$
            @superuser123 Look at a plot of $f(x) = frac{1}{1-x}$ and figure out where the graph lies below the horizontal line $y = 0.2$ or $y=0.8$ or any $y in [0, 1)$.
            $endgroup$
            – angryavian
            Jan 16 at 18:18


















          • $begingroup$
            thanks, but could you please clarify your first statement: "please not that Y takes these values" because I am not sure how it works
            $endgroup$
            – superuser123
            Jan 15 at 21:23






          • 1




            $begingroup$
            @superuser123 Note that $X$ is always nonnegative, and think about what values $frac{1}{1-X}$ could be. It may help to plot the function $f(x) = frac{1}{1-x}$.
            $endgroup$
            – angryavian
            Jan 15 at 21:30










          • $begingroup$
            could you please clarify the second case? $y in [0, 1)$, and in general, could you please explain how you knew how to divide to these cases? thanks!
            $endgroup$
            – superuser123
            Jan 16 at 13:22










          • $begingroup$
            @superuser123 Look at a plot of $f(x) = frac{1}{1-x}$ and figure out where the graph lies below the horizontal line $y = 0.2$ or $y=0.8$ or any $y in [0, 1)$.
            $endgroup$
            – angryavian
            Jan 16 at 18:18
















          $begingroup$
          thanks, but could you please clarify your first statement: "please not that Y takes these values" because I am not sure how it works
          $endgroup$
          – superuser123
          Jan 15 at 21:23




          $begingroup$
          thanks, but could you please clarify your first statement: "please not that Y takes these values" because I am not sure how it works
          $endgroup$
          – superuser123
          Jan 15 at 21:23




          1




          1




          $begingroup$
          @superuser123 Note that $X$ is always nonnegative, and think about what values $frac{1}{1-X}$ could be. It may help to plot the function $f(x) = frac{1}{1-x}$.
          $endgroup$
          – angryavian
          Jan 15 at 21:30




          $begingroup$
          @superuser123 Note that $X$ is always nonnegative, and think about what values $frac{1}{1-X}$ could be. It may help to plot the function $f(x) = frac{1}{1-x}$.
          $endgroup$
          – angryavian
          Jan 15 at 21:30












          $begingroup$
          could you please clarify the second case? $y in [0, 1)$, and in general, could you please explain how you knew how to divide to these cases? thanks!
          $endgroup$
          – superuser123
          Jan 16 at 13:22




          $begingroup$
          could you please clarify the second case? $y in [0, 1)$, and in general, could you please explain how you knew how to divide to these cases? thanks!
          $endgroup$
          – superuser123
          Jan 16 at 13:22












          $begingroup$
          @superuser123 Look at a plot of $f(x) = frac{1}{1-x}$ and figure out where the graph lies below the horizontal line $y = 0.2$ or $y=0.8$ or any $y in [0, 1)$.
          $endgroup$
          – angryavian
          Jan 16 at 18:18




          $begingroup$
          @superuser123 Look at a plot of $f(x) = frac{1}{1-x}$ and figure out where the graph lies below the horizontal line $y = 0.2$ or $y=0.8$ or any $y in [0, 1)$.
          $endgroup$
          – angryavian
          Jan 16 at 18:18


















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