Expectation of sum of independent Poisson distributions












0














I have three independent Poisson distributions:




  • $X_1 sim mathcal{P}(15)$,


  • $X_2 sim mathcal{P}(21)$ &


  • $X_3 sim mathcal{P}(10)$.



I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.





For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.



I would appreciate any help.










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    0














    I have three independent Poisson distributions:




    • $X_1 sim mathcal{P}(15)$,


    • $X_2 sim mathcal{P}(21)$ &


    • $X_3 sim mathcal{P}(10)$.



    I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.





    For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.



    I would appreciate any help.










    share|cite|improve this question



























      0












      0








      0







      I have three independent Poisson distributions:




      • $X_1 sim mathcal{P}(15)$,


      • $X_2 sim mathcal{P}(21)$ &


      • $X_3 sim mathcal{P}(10)$.



      I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.





      For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.



      I would appreciate any help.










      share|cite|improve this question















      I have three independent Poisson distributions:




      • $X_1 sim mathcal{P}(15)$,


      • $X_2 sim mathcal{P}(21)$ &


      • $X_3 sim mathcal{P}(10)$.



      I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.





      For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.



      I would appreciate any help.







      self-learning poisson-distribution expected-value






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      edited Nov 19 '18 at 3:06









      Tianlalu

      3,09621038




      3,09621038










      asked Nov 19 '18 at 2:26









      vic12

      254




      254






















          2 Answers
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          Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.



          $X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$



          https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)



          Now that you know the distribution of your sum you can take it from here!






          share|cite|improve this answer





























            0














            $E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$



            $V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.



            So you are correct






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              0














              Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.



              $X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$



              https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)



              Now that you know the distribution of your sum you can take it from here!






              share|cite|improve this answer


























                0














                Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.



                $X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$



                https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)



                Now that you know the distribution of your sum you can take it from here!






                share|cite|improve this answer
























                  0












                  0








                  0






                  Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.



                  $X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$



                  https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)



                  Now that you know the distribution of your sum you can take it from here!






                  share|cite|improve this answer












                  Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.



                  $X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$



                  https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)



                  Now that you know the distribution of your sum you can take it from here!







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 '18 at 2:36









                  KnowsNothing

                  355




                  355























                      0














                      $E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$



                      $V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.



                      So you are correct






                      share|cite|improve this answer


























                        0














                        $E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$



                        $V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.



                        So you are correct






                        share|cite|improve this answer
























                          0












                          0








                          0






                          $E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$



                          $V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.



                          So you are correct






                          share|cite|improve this answer












                          $E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$



                          $V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.



                          So you are correct







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 19 '18 at 2:37









                          pfmr1995

                          93




                          93






























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