CDF of a ratio of exponential variables












6












$begingroup$


Let $X$ and $Y$ be independent exponential variables with rates $alpha$ and $beta$, respectively. Find the CDF of $X/Y$.



I tried out the problem, and wanted to check to see if my answer of: $frac{alpha}{ beta/t + alpha}$ is correct, where $t$ is the time, which we need in our final answer since we need a cdf.



Can someone verify if this is correct?






statistics probability-distributions







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    Let $X$ and $Y$ be independent exponential variables with rates $alpha$ and $beta$, respectively. Find the CDF of $X/Y$.



    I tried out the problem, and wanted to check to see if my answer of: $frac{alpha}{ beta/t + alpha}$ is correct, where $t$ is the time, which we need in our final answer since we need a cdf.



    Can someone verify if this is correct?






    statistics probability-distributions







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      0



      $begingroup$


      Let $X$ and $Y$ be independent exponential variables with rates $alpha$ and $beta$, respectively. Find the CDF of $X/Y$.



      I tried out the problem, and wanted to check to see if my answer of: $frac{alpha}{ beta/t + alpha}$ is correct, where $t$ is the time, which we need in our final answer since we need a cdf.



      Can someone verify if this is correct?






      statistics probability-distributions







      share|cite|improve this question











      $endgroup$




      Let $X$ and $Y$ be independent exponential variables with rates $alpha$ and $beta$, respectively. Find the CDF of $X/Y$.



      I tried out the problem, and wanted to check to see if my answer of: $frac{alpha}{ beta/t + alpha}$ is correct, where $t$ is the time, which we need in our final answer since we need a cdf.



      Can someone verify if this is correct?






      statistics probability-distributions




      statistics probability-distributions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 29 '12 at 0:04









      Srivatsan

      20.9k371125




      20.9k371125










      asked Apr 19 '11 at 4:50









      marymary

      48921448




      48921448






















          3 Answers
          3






          active

          oldest

          votes


















          13












          $begingroup$

          Recall one of the most important characterizations of the exponential distribution:




          The random variable $Y$ is exponentially distributed with rate $beta$ if and only if $P(Ygeqslant y)=mathrm{e}^{-beta y}$ for every $ygeqslant0$.




          Let $Z=X/Y$ and $tgt0$. Conditioning on $X$ and applying our characterization to $y=X/t$, one gets
          $$
          P(Zleqslant t)=P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t}).
          $$
          Now, the density of the distribution of $X$ is $alphamathrm{e}^{-alpha x}$ on $xgeqslant0$, hence for every $gammageqslant0$,
          $$
          E(mathrm{e}^{-gamma X})=int_0^{+infty}alphamathrm{e}^{-(alpha+gamma) x}mathrm{d}x=frac{alpha}{alpha+gamma}left[-mathrm{e}^{-(alpha+gamma) x}right]_{0}^{+infty}=frac{alpha}{alpha+gamma}.
          $$
          Substituting $gamma=beta/t$ yields the formula.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What is the probability distribution function?
            $endgroup$
            – user
            Mar 11 '14 at 4:37






          • 4




            $begingroup$
            @user How to deduce the PDF from the CDF? Tell me...
            $endgroup$
            – Did
            Mar 11 '14 at 6:34










          • $begingroup$
            @Did how did you arrive to $P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t})$. How is the probability and expectation related?
            $endgroup$
            – user35443
            Nov 1 '18 at 13:57










          • $begingroup$
            @user35443 For every $xgeqslant0$, $P(Ygeqslant x/t)=e^{-beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Ygeqslant X/tmid X)=e^{-beta X/t}$ almost surely. Finally, take expectations on both sides.
            $endgroup$
            – Did
            Nov 3 '18 at 14:56












          • $begingroup$
            Makes sense, thanks!
            $endgroup$
            – user35443
            Nov 4 '18 at 18:18



















          2












          $begingroup$

          Here's a slightly different point of view.
          begin{align}
          Prleft( frac X Y ge t right) & = iintlimits_{{,(x,y),:, x,ge,ty,ge,0 ,}} e^{-alpha x} e^{-beta y} (alphabeta,d(x,y)) \[10pt]
          & = int_0^infty left( int_{ty}^infty e^{-alpha x} (alpha,dx) right) e^{-beta y} (beta,dy) \[10pt]
          & = int_0^infty (e^{-alpha ty}) e^{-beta y} (beta,dy) \[10pt]
          & = beta int_0^infty e^{-(alpha t+beta)y} , dy = frac beta {alpha t + beta}.
          end{align}
          This is $1$ minus the c.d.f. Find the c.d.f. and differentiate to get the p.d.f. on the interval $tge0.$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            This is correct. I did the calculation and got the same answer.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Did you use u-substitutions, a lot of them?
              $endgroup$
              – mary
              Apr 19 '11 at 5:02










            • $begingroup$
              No, I used Mathematica :) If you do it by hand and you are not familiar with $int alpha e^{-alpha x} dx$ you might need a lot of substitutions.
              $endgroup$
              – GWu
              Apr 19 '11 at 5:03













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            13












            $begingroup$

            Recall one of the most important characterizations of the exponential distribution:




            The random variable $Y$ is exponentially distributed with rate $beta$ if and only if $P(Ygeqslant y)=mathrm{e}^{-beta y}$ for every $ygeqslant0$.




            Let $Z=X/Y$ and $tgt0$. Conditioning on $X$ and applying our characterization to $y=X/t$, one gets
            $$
            P(Zleqslant t)=P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t}).
            $$
            Now, the density of the distribution of $X$ is $alphamathrm{e}^{-alpha x}$ on $xgeqslant0$, hence for every $gammageqslant0$,
            $$
            E(mathrm{e}^{-gamma X})=int_0^{+infty}alphamathrm{e}^{-(alpha+gamma) x}mathrm{d}x=frac{alpha}{alpha+gamma}left[-mathrm{e}^{-(alpha+gamma) x}right]_{0}^{+infty}=frac{alpha}{alpha+gamma}.
            $$
            Substituting $gamma=beta/t$ yields the formula.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              What is the probability distribution function?
              $endgroup$
              – user
              Mar 11 '14 at 4:37






            • 4




              $begingroup$
              @user How to deduce the PDF from the CDF? Tell me...
              $endgroup$
              – Did
              Mar 11 '14 at 6:34










            • $begingroup$
              @Did how did you arrive to $P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t})$. How is the probability and expectation related?
              $endgroup$
              – user35443
              Nov 1 '18 at 13:57










            • $begingroup$
              @user35443 For every $xgeqslant0$, $P(Ygeqslant x/t)=e^{-beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Ygeqslant X/tmid X)=e^{-beta X/t}$ almost surely. Finally, take expectations on both sides.
              $endgroup$
              – Did
              Nov 3 '18 at 14:56












            • $begingroup$
              Makes sense, thanks!
              $endgroup$
              – user35443
              Nov 4 '18 at 18:18
















            13












            $begingroup$

            Recall one of the most important characterizations of the exponential distribution:




            The random variable $Y$ is exponentially distributed with rate $beta$ if and only if $P(Ygeqslant y)=mathrm{e}^{-beta y}$ for every $ygeqslant0$.




            Let $Z=X/Y$ and $tgt0$. Conditioning on $X$ and applying our characterization to $y=X/t$, one gets
            $$
            P(Zleqslant t)=P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t}).
            $$
            Now, the density of the distribution of $X$ is $alphamathrm{e}^{-alpha x}$ on $xgeqslant0$, hence for every $gammageqslant0$,
            $$
            E(mathrm{e}^{-gamma X})=int_0^{+infty}alphamathrm{e}^{-(alpha+gamma) x}mathrm{d}x=frac{alpha}{alpha+gamma}left[-mathrm{e}^{-(alpha+gamma) x}right]_{0}^{+infty}=frac{alpha}{alpha+gamma}.
            $$
            Substituting $gamma=beta/t$ yields the formula.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              What is the probability distribution function?
              $endgroup$
              – user
              Mar 11 '14 at 4:37






            • 4




              $begingroup$
              @user How to deduce the PDF from the CDF? Tell me...
              $endgroup$
              – Did
              Mar 11 '14 at 6:34










            • $begingroup$
              @Did how did you arrive to $P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t})$. How is the probability and expectation related?
              $endgroup$
              – user35443
              Nov 1 '18 at 13:57










            • $begingroup$
              @user35443 For every $xgeqslant0$, $P(Ygeqslant x/t)=e^{-beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Ygeqslant X/tmid X)=e^{-beta X/t}$ almost surely. Finally, take expectations on both sides.
              $endgroup$
              – Did
              Nov 3 '18 at 14:56












            • $begingroup$
              Makes sense, thanks!
              $endgroup$
              – user35443
              Nov 4 '18 at 18:18














            13












            13








            13





            $begingroup$

            Recall one of the most important characterizations of the exponential distribution:




            The random variable $Y$ is exponentially distributed with rate $beta$ if and only if $P(Ygeqslant y)=mathrm{e}^{-beta y}$ for every $ygeqslant0$.




            Let $Z=X/Y$ and $tgt0$. Conditioning on $X$ and applying our characterization to $y=X/t$, one gets
            $$
            P(Zleqslant t)=P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t}).
            $$
            Now, the density of the distribution of $X$ is $alphamathrm{e}^{-alpha x}$ on $xgeqslant0$, hence for every $gammageqslant0$,
            $$
            E(mathrm{e}^{-gamma X})=int_0^{+infty}alphamathrm{e}^{-(alpha+gamma) x}mathrm{d}x=frac{alpha}{alpha+gamma}left[-mathrm{e}^{-(alpha+gamma) x}right]_{0}^{+infty}=frac{alpha}{alpha+gamma}.
            $$
            Substituting $gamma=beta/t$ yields the formula.






            share|cite|improve this answer











            $endgroup$



            Recall one of the most important characterizations of the exponential distribution:




            The random variable $Y$ is exponentially distributed with rate $beta$ if and only if $P(Ygeqslant y)=mathrm{e}^{-beta y}$ for every $ygeqslant0$.




            Let $Z=X/Y$ and $tgt0$. Conditioning on $X$ and applying our characterization to $y=X/t$, one gets
            $$
            P(Zleqslant t)=P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t}).
            $$
            Now, the density of the distribution of $X$ is $alphamathrm{e}^{-alpha x}$ on $xgeqslant0$, hence for every $gammageqslant0$,
            $$
            E(mathrm{e}^{-gamma X})=int_0^{+infty}alphamathrm{e}^{-(alpha+gamma) x}mathrm{d}x=frac{alpha}{alpha+gamma}left[-mathrm{e}^{-(alpha+gamma) x}right]_{0}^{+infty}=frac{alpha}{alpha+gamma}.
            $$
            Substituting $gamma=beta/t$ yields the formula.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 6 '13 at 6:54

























            answered Apr 19 '11 at 5:46









            DidDid

            247k23222458




            247k23222458












            • $begingroup$
              What is the probability distribution function?
              $endgroup$
              – user
              Mar 11 '14 at 4:37






            • 4




              $begingroup$
              @user How to deduce the PDF from the CDF? Tell me...
              $endgroup$
              – Did
              Mar 11 '14 at 6:34










            • $begingroup$
              @Did how did you arrive to $P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t})$. How is the probability and expectation related?
              $endgroup$
              – user35443
              Nov 1 '18 at 13:57










            • $begingroup$
              @user35443 For every $xgeqslant0$, $P(Ygeqslant x/t)=e^{-beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Ygeqslant X/tmid X)=e^{-beta X/t}$ almost surely. Finally, take expectations on both sides.
              $endgroup$
              – Did
              Nov 3 '18 at 14:56












            • $begingroup$
              Makes sense, thanks!
              $endgroup$
              – user35443
              Nov 4 '18 at 18:18


















            • $begingroup$
              What is the probability distribution function?
              $endgroup$
              – user
              Mar 11 '14 at 4:37






            • 4




              $begingroup$
              @user How to deduce the PDF from the CDF? Tell me...
              $endgroup$
              – Did
              Mar 11 '14 at 6:34










            • $begingroup$
              @Did how did you arrive to $P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t})$. How is the probability and expectation related?
              $endgroup$
              – user35443
              Nov 1 '18 at 13:57










            • $begingroup$
              @user35443 For every $xgeqslant0$, $P(Ygeqslant x/t)=e^{-beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Ygeqslant X/tmid X)=e^{-beta X/t}$ almost surely. Finally, take expectations on both sides.
              $endgroup$
              – Did
              Nov 3 '18 at 14:56












            • $begingroup$
              Makes sense, thanks!
              $endgroup$
              – user35443
              Nov 4 '18 at 18:18
















            $begingroup$
            What is the probability distribution function?
            $endgroup$
            – user
            Mar 11 '14 at 4:37




            $begingroup$
            What is the probability distribution function?
            $endgroup$
            – user
            Mar 11 '14 at 4:37




            4




            4




            $begingroup$
            @user How to deduce the PDF from the CDF? Tell me...
            $endgroup$
            – Did
            Mar 11 '14 at 6:34




            $begingroup$
            @user How to deduce the PDF from the CDF? Tell me...
            $endgroup$
            – Did
            Mar 11 '14 at 6:34












            $begingroup$
            @Did how did you arrive to $P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t})$. How is the probability and expectation related?
            $endgroup$
            – user35443
            Nov 1 '18 at 13:57




            $begingroup$
            @Did how did you arrive to $P(Ygeqslant X/t)=E(mathrm{e}^{-beta X/t})$. How is the probability and expectation related?
            $endgroup$
            – user35443
            Nov 1 '18 at 13:57












            $begingroup$
            @user35443 For every $xgeqslant0$, $P(Ygeqslant x/t)=e^{-beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Ygeqslant X/tmid X)=e^{-beta X/t}$ almost surely. Finally, take expectations on both sides.
            $endgroup$
            – Did
            Nov 3 '18 at 14:56






            $begingroup$
            @user35443 For every $xgeqslant0$, $P(Ygeqslant x/t)=e^{-beta x/t}$. Now, $X$ is independent of $Y$ hence $P(Ygeqslant X/tmid X)=e^{-beta X/t}$ almost surely. Finally, take expectations on both sides.
            $endgroup$
            – Did
            Nov 3 '18 at 14:56














            $begingroup$
            Makes sense, thanks!
            $endgroup$
            – user35443
            Nov 4 '18 at 18:18




            $begingroup$
            Makes sense, thanks!
            $endgroup$
            – user35443
            Nov 4 '18 at 18:18











            2












            $begingroup$

            Here's a slightly different point of view.
            begin{align}
            Prleft( frac X Y ge t right) & = iintlimits_{{,(x,y),:, x,ge,ty,ge,0 ,}} e^{-alpha x} e^{-beta y} (alphabeta,d(x,y)) \[10pt]
            & = int_0^infty left( int_{ty}^infty e^{-alpha x} (alpha,dx) right) e^{-beta y} (beta,dy) \[10pt]
            & = int_0^infty (e^{-alpha ty}) e^{-beta y} (beta,dy) \[10pt]
            & = beta int_0^infty e^{-(alpha t+beta)y} , dy = frac beta {alpha t + beta}.
            end{align}
            This is $1$ minus the c.d.f. Find the c.d.f. and differentiate to get the p.d.f. on the interval $tge0.$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Here's a slightly different point of view.
              begin{align}
              Prleft( frac X Y ge t right) & = iintlimits_{{,(x,y),:, x,ge,ty,ge,0 ,}} e^{-alpha x} e^{-beta y} (alphabeta,d(x,y)) \[10pt]
              & = int_0^infty left( int_{ty}^infty e^{-alpha x} (alpha,dx) right) e^{-beta y} (beta,dy) \[10pt]
              & = int_0^infty (e^{-alpha ty}) e^{-beta y} (beta,dy) \[10pt]
              & = beta int_0^infty e^{-(alpha t+beta)y} , dy = frac beta {alpha t + beta}.
              end{align}
              This is $1$ minus the c.d.f. Find the c.d.f. and differentiate to get the p.d.f. on the interval $tge0.$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Here's a slightly different point of view.
                begin{align}
                Prleft( frac X Y ge t right) & = iintlimits_{{,(x,y),:, x,ge,ty,ge,0 ,}} e^{-alpha x} e^{-beta y} (alphabeta,d(x,y)) \[10pt]
                & = int_0^infty left( int_{ty}^infty e^{-alpha x} (alpha,dx) right) e^{-beta y} (beta,dy) \[10pt]
                & = int_0^infty (e^{-alpha ty}) e^{-beta y} (beta,dy) \[10pt]
                & = beta int_0^infty e^{-(alpha t+beta)y} , dy = frac beta {alpha t + beta}.
                end{align}
                This is $1$ minus the c.d.f. Find the c.d.f. and differentiate to get the p.d.f. on the interval $tge0.$






                share|cite|improve this answer











                $endgroup$



                Here's a slightly different point of view.
                begin{align}
                Prleft( frac X Y ge t right) & = iintlimits_{{,(x,y),:, x,ge,ty,ge,0 ,}} e^{-alpha x} e^{-beta y} (alphabeta,d(x,y)) \[10pt]
                & = int_0^infty left( int_{ty}^infty e^{-alpha x} (alpha,dx) right) e^{-beta y} (beta,dy) \[10pt]
                & = int_0^infty (e^{-alpha ty}) e^{-beta y} (beta,dy) \[10pt]
                & = beta int_0^infty e^{-(alpha t+beta)y} , dy = frac beta {alpha t + beta}.
                end{align}
                This is $1$ minus the c.d.f. Find the c.d.f. and differentiate to get the p.d.f. on the interval $tge0.$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Sep 2 '18 at 18:09

























                answered Dec 21 '17 at 19:32









                Michael HardyMichael Hardy

                1




                1























                    1












                    $begingroup$

                    This is correct. I did the calculation and got the same answer.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Did you use u-substitutions, a lot of them?
                      $endgroup$
                      – mary
                      Apr 19 '11 at 5:02










                    • $begingroup$
                      No, I used Mathematica :) If you do it by hand and you are not familiar with $int alpha e^{-alpha x} dx$ you might need a lot of substitutions.
                      $endgroup$
                      – GWu
                      Apr 19 '11 at 5:03


















                    1












                    $begingroup$

                    This is correct. I did the calculation and got the same answer.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Did you use u-substitutions, a lot of them?
                      $endgroup$
                      – mary
                      Apr 19 '11 at 5:02










                    • $begingroup$
                      No, I used Mathematica :) If you do it by hand and you are not familiar with $int alpha e^{-alpha x} dx$ you might need a lot of substitutions.
                      $endgroup$
                      – GWu
                      Apr 19 '11 at 5:03
















                    1












                    1








                    1





                    $begingroup$

                    This is correct. I did the calculation and got the same answer.






                    share|cite|improve this answer









                    $endgroup$



                    This is correct. I did the calculation and got the same answer.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 19 '11 at 5:01









                    GWuGWu

                    1,204710




                    1,204710












                    • $begingroup$
                      Did you use u-substitutions, a lot of them?
                      $endgroup$
                      – mary
                      Apr 19 '11 at 5:02










                    • $begingroup$
                      No, I used Mathematica :) If you do it by hand and you are not familiar with $int alpha e^{-alpha x} dx$ you might need a lot of substitutions.
                      $endgroup$
                      – GWu
                      Apr 19 '11 at 5:03




















                    • $begingroup$
                      Did you use u-substitutions, a lot of them?
                      $endgroup$
                      – mary
                      Apr 19 '11 at 5:02










                    • $begingroup$
                      No, I used Mathematica :) If you do it by hand and you are not familiar with $int alpha e^{-alpha x} dx$ you might need a lot of substitutions.
                      $endgroup$
                      – GWu
                      Apr 19 '11 at 5:03


















                    $begingroup$
                    Did you use u-substitutions, a lot of them?
                    $endgroup$
                    – mary
                    Apr 19 '11 at 5:02




                    $begingroup$
                    Did you use u-substitutions, a lot of them?
                    $endgroup$
                    – mary
                    Apr 19 '11 at 5:02












                    $begingroup$
                    No, I used Mathematica :) If you do it by hand and you are not familiar with $int alpha e^{-alpha x} dx$ you might need a lot of substitutions.
                    $endgroup$
                    – GWu
                    Apr 19 '11 at 5:03






                    $begingroup$
                    No, I used Mathematica :) If you do it by hand and you are not familiar with $int alpha e^{-alpha x} dx$ you might need a lot of substitutions.
                    $endgroup$
                    – GWu
                    Apr 19 '11 at 5:03




















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