Discretization of an exponential variable












1












$begingroup$


Given $X=Exp(lambda)$, i have to define $Y=ceil(X)$ in order to prove the link between exponential and geometric variables.



By definition of ceiling $forall xin mathbb{R},exists nin mathbb{N}:xleq n< x+1$, so:



$mathbb{P}(Y=n)=mathbb{P}(n-1< Xleq n)=mathbb{P}(Xleq n)-mathbb{P}(X<n-1)=F_X(n)-F_X(n-1)=Exp(n)-Exp(n-1)=1-e^{-lambda n}-1+e^{-lambda (n-1)}=e^{-lambda n}(e^{lambda}-1)$.



Nevertheless, the result is $e^{-lambda n}(1-e^{-lambda})Rightarrow Ysim Geo(1-e^{-lambda})$.



1) Where i wrong in the passages?



2) Geometric variable has $n-1$ at the exponent, not $n$. So, why that result?



Thanks for any help!






probability probability-distributions exponential-function density-function







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$endgroup$

















    1












    $begingroup$


    Given $X=Exp(lambda)$, i have to define $Y=ceil(X)$ in order to prove the link between exponential and geometric variables.



    By definition of ceiling $forall xin mathbb{R},exists nin mathbb{N}:xleq n< x+1$, so:



    $mathbb{P}(Y=n)=mathbb{P}(n-1< Xleq n)=mathbb{P}(Xleq n)-mathbb{P}(X<n-1)=F_X(n)-F_X(n-1)=Exp(n)-Exp(n-1)=1-e^{-lambda n}-1+e^{-lambda (n-1)}=e^{-lambda n}(e^{lambda}-1)$.



    Nevertheless, the result is $e^{-lambda n}(1-e^{-lambda})Rightarrow Ysim Geo(1-e^{-lambda})$.



    1) Where i wrong in the passages?



    2) Geometric variable has $n-1$ at the exponent, not $n$. So, why that result?



    Thanks for any help!






    probability probability-distributions exponential-function density-function







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Given $X=Exp(lambda)$, i have to define $Y=ceil(X)$ in order to prove the link between exponential and geometric variables.



      By definition of ceiling $forall xin mathbb{R},exists nin mathbb{N}:xleq n< x+1$, so:



      $mathbb{P}(Y=n)=mathbb{P}(n-1< Xleq n)=mathbb{P}(Xleq n)-mathbb{P}(X<n-1)=F_X(n)-F_X(n-1)=Exp(n)-Exp(n-1)=1-e^{-lambda n}-1+e^{-lambda (n-1)}=e^{-lambda n}(e^{lambda}-1)$.



      Nevertheless, the result is $e^{-lambda n}(1-e^{-lambda})Rightarrow Ysim Geo(1-e^{-lambda})$.



      1) Where i wrong in the passages?



      2) Geometric variable has $n-1$ at the exponent, not $n$. So, why that result?



      Thanks for any help!






      probability probability-distributions exponential-function density-function







      share|cite|improve this question









      $endgroup$




      Given $X=Exp(lambda)$, i have to define $Y=ceil(X)$ in order to prove the link between exponential and geometric variables.



      By definition of ceiling $forall xin mathbb{R},exists nin mathbb{N}:xleq n< x+1$, so:



      $mathbb{P}(Y=n)=mathbb{P}(n-1< Xleq n)=mathbb{P}(Xleq n)-mathbb{P}(X<n-1)=F_X(n)-F_X(n-1)=Exp(n)-Exp(n-1)=1-e^{-lambda n}-1+e^{-lambda (n-1)}=e^{-lambda n}(e^{lambda}-1)$.



      Nevertheless, the result is $e^{-lambda n}(1-e^{-lambda})Rightarrow Ysim Geo(1-e^{-lambda})$.



      1) Where i wrong in the passages?



      2) Geometric variable has $n-1$ at the exponent, not $n$. So, why that result?



      Thanks for any help!






      probability probability-distributions exponential-function density-function




      probability probability-distributions exponential-function density-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 6 '18 at 17:19









      Marco PittellaMarco Pittella

      1288




      1288






















          2 Answers
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          0












          $begingroup$

          You did nothing wrong but can represent it differently:



          $e^{-lambda (n-1)}-e^{-lambda n} = e^{-lambda (n-1)}-e^{-lambda (n-1) -lambda} =e^{-lambda (n-1)}(1-e^{-lambda})sim Geo(1-e^{-lambda}) $



          but



          $e^{−λn}(1−e^{−λ})neq e^{−λn}(e^{λ}-1)$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You have $e^{-lambda n}(e^{lambda}-1)$. This is equal to $e^{-lambda(n-1)}-e^{-lambda n}$. Now we can factor out $(e^{-lambda})^{n-1}$



            We have to add $lambda (n-1)$ to both exponents.




            • $-lambda(n-1)+lambda(n-1)=0$

            • $-lambda n+lambda (n-1)=lambda (-n+n-1)=-lambda$


            Therefore we get $(e^{-lambda})^{n-1}cdot (e^0-e^{-lambda})=(e^{-lambda})^{n-1}cdot (1-e^{-lambda})$



            This is a geometric distribution with $p=1-e^{-lambda}$






            share|cite|improve this answer











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              2 Answers
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              active

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              2 Answers
              2






              active

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              active

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              active

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              0












              $begingroup$

              You did nothing wrong but can represent it differently:



              $e^{-lambda (n-1)}-e^{-lambda n} = e^{-lambda (n-1)}-e^{-lambda (n-1) -lambda} =e^{-lambda (n-1)}(1-e^{-lambda})sim Geo(1-e^{-lambda}) $



              but



              $e^{−λn}(1−e^{−λ})neq e^{−λn}(e^{λ}-1)$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You did nothing wrong but can represent it differently:



                $e^{-lambda (n-1)}-e^{-lambda n} = e^{-lambda (n-1)}-e^{-lambda (n-1) -lambda} =e^{-lambda (n-1)}(1-e^{-lambda})sim Geo(1-e^{-lambda}) $



                but



                $e^{−λn}(1−e^{−λ})neq e^{−λn}(e^{λ}-1)$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You did nothing wrong but can represent it differently:



                  $e^{-lambda (n-1)}-e^{-lambda n} = e^{-lambda (n-1)}-e^{-lambda (n-1) -lambda} =e^{-lambda (n-1)}(1-e^{-lambda})sim Geo(1-e^{-lambda}) $



                  but



                  $e^{−λn}(1−e^{−λ})neq e^{−λn}(e^{λ}-1)$.






                  share|cite|improve this answer









                  $endgroup$



                  You did nothing wrong but can represent it differently:



                  $e^{-lambda (n-1)}-e^{-lambda n} = e^{-lambda (n-1)}-e^{-lambda (n-1) -lambda} =e^{-lambda (n-1)}(1-e^{-lambda})sim Geo(1-e^{-lambda}) $



                  but



                  $e^{−λn}(1−e^{−λ})neq e^{−λn}(e^{λ}-1)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 17:35









                  sehiglesehigle

                  1565




                  1565























                      0












                      $begingroup$

                      You have $e^{-lambda n}(e^{lambda}-1)$. This is equal to $e^{-lambda(n-1)}-e^{-lambda n}$. Now we can factor out $(e^{-lambda})^{n-1}$



                      We have to add $lambda (n-1)$ to both exponents.




                      • $-lambda(n-1)+lambda(n-1)=0$

                      • $-lambda n+lambda (n-1)=lambda (-n+n-1)=-lambda$


                      Therefore we get $(e^{-lambda})^{n-1}cdot (e^0-e^{-lambda})=(e^{-lambda})^{n-1}cdot (1-e^{-lambda})$



                      This is a geometric distribution with $p=1-e^{-lambda}$






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        You have $e^{-lambda n}(e^{lambda}-1)$. This is equal to $e^{-lambda(n-1)}-e^{-lambda n}$. Now we can factor out $(e^{-lambda})^{n-1}$



                        We have to add $lambda (n-1)$ to both exponents.




                        • $-lambda(n-1)+lambda(n-1)=0$

                        • $-lambda n+lambda (n-1)=lambda (-n+n-1)=-lambda$


                        Therefore we get $(e^{-lambda})^{n-1}cdot (e^0-e^{-lambda})=(e^{-lambda})^{n-1}cdot (1-e^{-lambda})$



                        This is a geometric distribution with $p=1-e^{-lambda}$






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          You have $e^{-lambda n}(e^{lambda}-1)$. This is equal to $e^{-lambda(n-1)}-e^{-lambda n}$. Now we can factor out $(e^{-lambda})^{n-1}$



                          We have to add $lambda (n-1)$ to both exponents.




                          • $-lambda(n-1)+lambda(n-1)=0$

                          • $-lambda n+lambda (n-1)=lambda (-n+n-1)=-lambda$


                          Therefore we get $(e^{-lambda})^{n-1}cdot (e^0-e^{-lambda})=(e^{-lambda})^{n-1}cdot (1-e^{-lambda})$



                          This is a geometric distribution with $p=1-e^{-lambda}$






                          share|cite|improve this answer











                          $endgroup$



                          You have $e^{-lambda n}(e^{lambda}-1)$. This is equal to $e^{-lambda(n-1)}-e^{-lambda n}$. Now we can factor out $(e^{-lambda})^{n-1}$



                          We have to add $lambda (n-1)$ to both exponents.




                          • $-lambda(n-1)+lambda(n-1)=0$

                          • $-lambda n+lambda (n-1)=lambda (-n+n-1)=-lambda$


                          Therefore we get $(e^{-lambda})^{n-1}cdot (e^0-e^{-lambda})=(e^{-lambda})^{n-1}cdot (1-e^{-lambda})$



                          This is a geometric distribution with $p=1-e^{-lambda}$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 6 '18 at 17:58

























                          answered Dec 6 '18 at 17:51









                          callculuscallculus

                          17.9k31427




                          17.9k31427






























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