Let X be a Poisson random variable, show that $E[X^n] = lambda E[(X+1)^{n-1}]$












1












$begingroup$


I am struggling to understand some of the steps of the proof of this problem:




Let X be a Poisson random variable, show that $E[X^n] = lambda
E[(X+1)^{n-1}]$.




The proof:
$$
begin{align*}
E(X^n) &stackrel{(1)}=
e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(2)}=
e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(3)}=
e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^j}{(j-1)!} \ &stackrel{(4)}=
lambda e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^{j-1}}{(j-1)!} \ &stackrel{(5)}=
lambda e^{-lambda} left( sum_{i=1}^infty (i+1)^{n-1} frac{lambda^i}{i!} right) \ &stackrel{(6)}=
lambda E((X+1)^{n-1}).
end{align*}
$$



Assuming the lines are numbered 1-6, I will try to explain and show where I'm stuck (the bolded parts):




  1. This is just by definition, but I don't understand why it is necessarily going from 0 to infinity, and why does it look for $P(X=i), 0leq ileq infty $ (hope my questions was clear)


  2. Since for $j=0$ it doesn't contribute to the sum, I can omit it and count from 1


  3. This line is not clear to me, why can I reduce the power and the denominator? doesn't make much sense to me

  4. Nothing special here

  5. We change the lower bound again to count from 0, why again I can simply add 1 to the power and denominator?

  6. Why is this conclusion correct?


Thanks in advance!






probability summation expected-value







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am struggling to understand some of the steps of the proof of this problem:




    Let X be a Poisson random variable, show that $E[X^n] = lambda
    E[(X+1)^{n-1}]$    .




    The proof:
    $$
    begin{align*}
    E(X^n) &stackrel{(1)}=
    e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(2)}=
    e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(3)}=
    e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^j}{(j-1)!} \ &stackrel{(4)}=
    lambda e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^{j-1}}{(j-1)!} \ &stackrel{(5)}=
    lambda e^{-lambda} left( sum_{i=1}^infty (i+1)^{n-1} frac{lambda^i}{i!} right) \ &stackrel{(6)}=
    lambda E((X+1)^{n-1}).
    end{align*}
    $$



    Assuming the lines are numbered 1-6, I will try to explain and show where I'm stuck (the bolded parts):




    1. This is just by definition, but I don't understand why it is necessarily going from 0 to infinity, and why does it look for $P(X=i), 0leq ileq infty $ (hope my questions was clear)


    2. Since for $j=0$ it doesn't contribute to the sum, I can omit it and count from 1


    3. This line is not clear to me, why can I reduce the power and the denominator? doesn't make much sense to me

    4. Nothing special here

    5. We change the lower bound again to count from 0, why again I can simply add 1 to the power and denominator?

    6. Why is this conclusion correct?


    Thanks in advance!






    probability summation expected-value







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am struggling to understand some of the steps of the proof of this problem:




      Let X be a Poisson random variable, show that $E[X^n] = lambda
      E[(X+1)^{n-1}]$      .




      The proof:
      $$
      begin{align*}
      E(X^n) &stackrel{(1)}=
      e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(2)}=
      e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(3)}=
      e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^j}{(j-1)!} \ &stackrel{(4)}=
      lambda e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^{j-1}}{(j-1)!} \ &stackrel{(5)}=
      lambda e^{-lambda} left( sum_{i=1}^infty (i+1)^{n-1} frac{lambda^i}{i!} right) \ &stackrel{(6)}=
      lambda E((X+1)^{n-1}).
      end{align*}
      $$



      Assuming the lines are numbered 1-6, I will try to explain and show where I'm stuck (the bolded parts):




      1. This is just by definition, but I don't understand why it is necessarily going from 0 to infinity, and why does it look for $P(X=i), 0leq ileq infty $ (hope my questions was clear)


      2. Since for $j=0$ it doesn't contribute to the sum, I can omit it and count from 1


      3. This line is not clear to me, why can I reduce the power and the denominator? doesn't make much sense to me

      4. Nothing special here

      5. We change the lower bound again to count from 0, why again I can simply add 1 to the power and denominator?

      6. Why is this conclusion correct?


      Thanks in advance!






      probability summation expected-value







      share|cite|improve this question











      $endgroup$




      I am struggling to understand some of the steps of the proof of this problem:




      Let X be a Poisson random variable, show that $E[X^n] = lambda
      E[(X+1)^{n-1}]$      .




      The proof:
      $$
      begin{align*}
      E(X^n) &stackrel{(1)}=
      e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(2)}=
      e^{-lambda} sum_{j=0}^infty j^n frac{lambda^j}{j!} \ &stackrel{(3)}=
      e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^j}{(j-1)!} \ &stackrel{(4)}=
      lambda e^{-lambda} sum_{j=0}^infty j^{n-1} frac{lambda^{j-1}}{(j-1)!} \ &stackrel{(5)}=
      lambda e^{-lambda} left( sum_{i=1}^infty (i+1)^{n-1} frac{lambda^i}{i!} right) \ &stackrel{(6)}=
      lambda E((X+1)^{n-1}).
      end{align*}
      $$



      Assuming the lines are numbered 1-6, I will try to explain and show where I'm stuck (the bolded parts):




      1. This is just by definition, but I don't understand why it is necessarily going from 0 to infinity, and why does it look for $P(X=i), 0leq ileq infty $ (hope my questions was clear)


      2. Since for $j=0$ it doesn't contribute to the sum, I can omit it and count from 1


      3. This line is not clear to me, why can I reduce the power and the denominator? doesn't make much sense to me

      4. Nothing special here

      5. We change the lower bound again to count from 0, why again I can simply add 1 to the power and denominator?

      6. Why is this conclusion correct?


      Thanks in advance!






      probability summation expected-value




      probability summation expected-value






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      edited Jan 18 at 9:15









      Yuval Filmus

      48.7k471145




      48.7k471145










      asked Jan 18 at 8:17









      superuser123superuser123

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      47828






















          2 Answers
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          $begingroup$

          $1$ - the expected value of a function $f(x)$ of a discrete random variable $X$ is $sumlimits_x f(x), mathbb P(X=x)$. Here $f(x)=x^n$. For a Poisson random variable with $mathbb P(X=x) = e^{-lambda}dfrac{lambda^x}{x!}$, the possible values of $X$ are the non-negative integers, so the sum is over the non-negative integers



          $3$ - By dividing both the numerator and the denominator by non-zero $j$ you can say $dfrac{j^n}{j!} = dfrac{j^{n-1}}{(j-1)!}$



          $5$ - if you do the substitution $i=j-1$, then $j^{n-1}=(i+1)^{n-1}$ and $lambda^{j-1}=lambda^i$ and $(j-1)!=i!$ and $j=1 implies i=0$



          $6$ - this is the reverse of the first point: you have inside the brackets $sumlimits_x g(x), mathbb P(X=x)$ so this is the expectation of $g(x)=(x+1)^{n-1}$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$


            1. This is like for any set of numbers the sum of the numbers is taken when finding their mean. This comes from the definition of expected value.

            2. Sure.

            3. You're simply cancelling the $j$ from $j^n$ with the $j$ from $frac{1}{j!}$. That is, $j^n = j^{n-1}j$ and $j! = j(j-1)!$.

            4. Okay.

            5. This is a basic change of index. The basic form for this is $sum_{k=2}^{infty}k = sum_{j=1}^{infty}(j+1)$.

            6. We substitute line 5 into the original form to get this result.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

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              active

              oldest

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              active

              oldest

              votes









              2












              $begingroup$

              $1$ - the expected value of a function $f(x)$ of a discrete random variable $X$ is $sumlimits_x f(x), mathbb P(X=x)$. Here $f(x)=x^n$. For a Poisson random variable with $mathbb P(X=x) = e^{-lambda}dfrac{lambda^x}{x!}$, the possible values of $X$ are the non-negative integers, so the sum is over the non-negative integers



              $3$ - By dividing both the numerator and the denominator by non-zero $j$ you can say $dfrac{j^n}{j!} = dfrac{j^{n-1}}{(j-1)!}$



              $5$ - if you do the substitution $i=j-1$, then $j^{n-1}=(i+1)^{n-1}$ and $lambda^{j-1}=lambda^i$ and $(j-1)!=i!$ and $j=1 implies i=0$



              $6$ - this is the reverse of the first point: you have inside the brackets $sumlimits_x g(x), mathbb P(X=x)$ so this is the expectation of $g(x)=(x+1)^{n-1}$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                $1$ - the expected value of a function $f(x)$ of a discrete random variable $X$ is $sumlimits_x f(x), mathbb P(X=x)$. Here $f(x)=x^n$. For a Poisson random variable with $mathbb P(X=x) = e^{-lambda}dfrac{lambda^x}{x!}$, the possible values of $X$ are the non-negative integers, so the sum is over the non-negative integers



                $3$ - By dividing both the numerator and the denominator by non-zero $j$ you can say $dfrac{j^n}{j!} = dfrac{j^{n-1}}{(j-1)!}$



                $5$ - if you do the substitution $i=j-1$, then $j^{n-1}=(i+1)^{n-1}$ and $lambda^{j-1}=lambda^i$ and $(j-1)!=i!$ and $j=1 implies i=0$



                $6$ - this is the reverse of the first point: you have inside the brackets $sumlimits_x g(x), mathbb P(X=x)$ so this is the expectation of $g(x)=(x+1)^{n-1}$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $1$ - the expected value of a function $f(x)$ of a discrete random variable $X$ is $sumlimits_x f(x), mathbb P(X=x)$. Here $f(x)=x^n$. For a Poisson random variable with $mathbb P(X=x) = e^{-lambda}dfrac{lambda^x}{x!}$, the possible values of $X$ are the non-negative integers, so the sum is over the non-negative integers



                  $3$ - By dividing both the numerator and the denominator by non-zero $j$ you can say $dfrac{j^n}{j!} = dfrac{j^{n-1}}{(j-1)!}$



                  $5$ - if you do the substitution $i=j-1$, then $j^{n-1}=(i+1)^{n-1}$ and $lambda^{j-1}=lambda^i$ and $(j-1)!=i!$ and $j=1 implies i=0$



                  $6$ - this is the reverse of the first point: you have inside the brackets $sumlimits_x g(x), mathbb P(X=x)$ so this is the expectation of $g(x)=(x+1)^{n-1}$






                  share|cite|improve this answer









                  $endgroup$



                  $1$ - the expected value of a function $f(x)$ of a discrete random variable $X$ is $sumlimits_x f(x), mathbb P(X=x)$. Here $f(x)=x^n$. For a Poisson random variable with $mathbb P(X=x) = e^{-lambda}dfrac{lambda^x}{x!}$, the possible values of $X$ are the non-negative integers, so the sum is over the non-negative integers



                  $3$ - By dividing both the numerator and the denominator by non-zero $j$ you can say $dfrac{j^n}{j!} = dfrac{j^{n-1}}{(j-1)!}$



                  $5$ - if you do the substitution $i=j-1$, then $j^{n-1}=(i+1)^{n-1}$ and $lambda^{j-1}=lambda^i$ and $(j-1)!=i!$ and $j=1 implies i=0$



                  $6$ - this is the reverse of the first point: you have inside the brackets $sumlimits_x g(x), mathbb P(X=x)$ so this is the expectation of $g(x)=(x+1)^{n-1}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 8:43









                  HenryHenry

                  100k481168




                  100k481168























                      1












                      $begingroup$


                      1. This is like for any set of numbers the sum of the numbers is taken when finding their mean. This comes from the definition of expected value.

                      2. Sure.

                      3. You're simply cancelling the $j$ from $j^n$ with the $j$ from $frac{1}{j!}$. That is, $j^n = j^{n-1}j$ and $j! = j(j-1)!$.

                      4. Okay.

                      5. This is a basic change of index. The basic form for this is $sum_{k=2}^{infty}k = sum_{j=1}^{infty}(j+1)$.

                      6. We substitute line 5 into the original form to get this result.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$


                        1. This is like for any set of numbers the sum of the numbers is taken when finding their mean. This comes from the definition of expected value.

                        2. Sure.

                        3. You're simply cancelling the $j$ from $j^n$ with the $j$ from $frac{1}{j!}$. That is, $j^n = j^{n-1}j$ and $j! = j(j-1)!$.

                        4. Okay.

                        5. This is a basic change of index. The basic form for this is $sum_{k=2}^{infty}k = sum_{j=1}^{infty}(j+1)$.

                        6. We substitute line 5 into the original form to get this result.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$


                          1. This is like for any set of numbers the sum of the numbers is taken when finding their mean. This comes from the definition of expected value.

                          2. Sure.

                          3. You're simply cancelling the $j$ from $j^n$ with the $j$ from $frac{1}{j!}$. That is, $j^n = j^{n-1}j$ and $j! = j(j-1)!$.

                          4. Okay.

                          5. This is a basic change of index. The basic form for this is $sum_{k=2}^{infty}k = sum_{j=1}^{infty}(j+1)$.

                          6. We substitute line 5 into the original form to get this result.






                          share|cite|improve this answer









                          $endgroup$




                          1. This is like for any set of numbers the sum of the numbers is taken when finding their mean. This comes from the definition of expected value.

                          2. Sure.

                          3. You're simply cancelling the $j$ from $j^n$ with the $j$ from $frac{1}{j!}$. That is, $j^n = j^{n-1}j$ and $j! = j(j-1)!$.

                          4. Okay.

                          5. This is a basic change of index. The basic form for this is $sum_{k=2}^{infty}k = sum_{j=1}^{infty}(j+1)$.

                          6. We substitute line 5 into the original form to get this result.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 18 at 8:47









                          DohlemanDohleman

                          395212




                          395212






























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