Find probability of $P(X Y < 1)$












1












$begingroup$


Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$



I started off by finding the pdf



$f_X(x)=frac{1}{2} $
when $0<x<2$ Same for $Y$.



I then found their joint PDF via independence:



$f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
Otherwise $0$



$int_0^2 int_0^{1/y} frac{1}{4} dx dy$
But this cannot be solved, so where did i go wrong?






probability probability-theory probability-distributions uniform-distribution







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$



    I started off by finding the pdf



    $f_X(x)=frac{1}{2} $
    when $0<x<2$ Same for $Y$.



    I then found their joint PDF via independence:



    $f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
    Otherwise $0$



    $int_0^2 int_0^{1/y} frac{1}{4} dx dy$
    But this cannot be solved, so where did i go wrong?






    probability probability-theory probability-distributions uniform-distribution







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$



      I started off by finding the pdf



      $f_X(x)=frac{1}{2} $
      when $0<x<2$ Same for $Y$.



      I then found their joint PDF via independence:



      $f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
      Otherwise $0$



      $int_0^2 int_0^{1/y} frac{1}{4} dx dy$
      But this cannot be solved, so where did i go wrong?






      probability probability-theory probability-distributions uniform-distribution







      share|cite|improve this question











      $endgroup$




      Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$



      I started off by finding the pdf



      $f_X(x)=frac{1}{2} $
      when $0<x<2$ Same for $Y$.



      I then found their joint PDF via independence:



      $f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
      Otherwise $0$



      $int_0^2 int_0^{1/y} frac{1}{4} dx dy$
      But this cannot be solved, so where did i go wrong?






      probability probability-theory probability-distributions uniform-distribution




      probability probability-theory probability-distributions uniform-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 9:58









      Davide Giraudo

      126k16150261




      126k16150261










      asked Nov 28 '18 at 21:24









      WintherWinther

      247




      247






















          2 Answers
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          3












          $begingroup$

          Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).



          enter image description here



          $$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$



          Your error was forgetting that the upper limit on $y$ was bounded by $2$.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Alternatively, you can consider the complementing blue area:



            $hspace{2cm}$enter image description here



            $$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
            &=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
            &=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
            &=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
            &=frac14+frac12ln 2.end{align}$$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

              votes









              3












              $begingroup$

              Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).



              enter image description here



              $$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$



              Your error was forgetting that the upper limit on $y$ was bounded by $2$.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).



                enter image description here



                $$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$



                Your error was forgetting that the upper limit on $y$ was bounded by $2$.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).



                  enter image description here



                  $$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$



                  Your error was forgetting that the upper limit on $y$ was bounded by $2$.






                  share|cite|improve this answer











                  $endgroup$



                  Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).



                  enter image description here



                  $$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$



                  Your error was forgetting that the upper limit on $y$ was bounded by $2$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 28 '18 at 21:41

























                  answered Nov 28 '18 at 21:35









                  David G. StorkDavid G. Stork

                  10.9k31432




                  10.9k31432























                      0












                      $begingroup$

                      Alternatively, you can consider the complementing blue area:



                      $hspace{2cm}$enter image description here



                      $$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
                      &=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
                      &=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
                      &=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
                      &=frac14+frac12ln 2.end{align}$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Alternatively, you can consider the complementing blue area:



                        $hspace{2cm}$enter image description here



                        $$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
                        &=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
                        &=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
                        &=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
                        &=frac14+frac12ln 2.end{align}$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Alternatively, you can consider the complementing blue area:



                          $hspace{2cm}$enter image description here



                          $$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
                          &=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
                          &=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
                          &=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
                          &=frac14+frac12ln 2.end{align}$$






                          share|cite|improve this answer









                          $endgroup$



                          Alternatively, you can consider the complementing blue area:



                          $hspace{2cm}$enter image description here



                          $$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
                          &=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
                          &=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
                          &=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
                          &=frac14+frac12ln 2.end{align}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 29 '18 at 12:36









                          farruhotafarruhota

                          20.1k2738




                          20.1k2738






























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