If $X$ has a binomial distribution with parameters $n$ and $p$, calculate $textbf{E}[(1+X)^{-1}]$ [duplicate]












0












$begingroup$



This question already has an answer here:




  • Find the expected value of $frac{1}{X+1}$ where $X$ is binomial

    4 answers




If $X$ has a binomial distribution with parameters $n$ and $p$, show that



begin{align*}
textbf{E}left(frac{1}{1+X}right) = frac{1-(1-p)^{n+1}}{(n+1)p}
end{align*}



MY ATTEMPT



Since $Xsimtext{Binomial}(n,p)$, we have
begin{align*}
textbf{E}left(frac{1}{1+X}right) & = sum_{x=0}^{n}{nchoose x}frac{p^{x}(1-p)^{n-x}}{1+x} = (1-p)^{n}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x}\\
& = frac{(1-p)^{n+1}}{p}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x+1}
end{align*}



This is as far as I can get. Could someone help me out? Thanks in advance!






probability binomial-distribution expected-value







share|cite|improve this question











$endgroup$



marked as duplicate by StubbornAtom, NCh, max_zorn, Andrew, Lord Shark the Unknown Jan 30 at 5:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Where are the combinatorial coefficients?
    $endgroup$
    – Will M.
    Jan 28 at 23:22










  • $begingroup$
    Apparently I missed them. I have edited it, though.
    $endgroup$
    – user1337
    Jan 28 at 23:27






  • 3




    $begingroup$
    $dbinom{n}{x} dfrac{1}{x+1} p^x (1-p)^{n-x}=dfrac{1}{p(n+1)}dbinom{n+1}{x+1}p^{x+1}(1-p)^{(n+1)-(x+1)}.$ Cheers
    $endgroup$
    – Will M.
    Jan 28 at 23:36










  • $begingroup$
    I suggest not using $x$ to denote a discrete variable. How about $k$ or $m$?
    $endgroup$
    – Aditya Dua
    Jan 29 at 0:08
















0












$begingroup$



This question already has an answer here:




  • Find the expected value of $frac{1}{X+1}$ where $X$ is binomial

    4 answers




If $X$ has a binomial distribution with parameters $n$ and $p$, show that



begin{align*}
textbf{E}left(frac{1}{1+X}right) = frac{1-(1-p)^{n+1}}{(n+1)p}
end{align*}



MY ATTEMPT



Since $Xsimtext{Binomial}(n,p)$, we have
begin{align*}
textbf{E}left(frac{1}{1+X}right) & = sum_{x=0}^{n}{nchoose x}frac{p^{x}(1-p)^{n-x}}{1+x} = (1-p)^{n}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x}\\
& = frac{(1-p)^{n+1}}{p}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x+1}
end{align*}



This is as far as I can get. Could someone help me out? Thanks in advance!






probability binomial-distribution expected-value







share|cite|improve this question











$endgroup$



marked as duplicate by StubbornAtom, NCh, max_zorn, Andrew, Lord Shark the Unknown Jan 30 at 5:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Where are the combinatorial coefficients?
    $endgroup$
    – Will M.
    Jan 28 at 23:22










  • $begingroup$
    Apparently I missed them. I have edited it, though.
    $endgroup$
    – user1337
    Jan 28 at 23:27






  • 3




    $begingroup$
    $dbinom{n}{x} dfrac{1}{x+1} p^x (1-p)^{n-x}=dfrac{1}{p(n+1)}dbinom{n+1}{x+1}p^{x+1}(1-p)^{(n+1)-(x+1)}.$ Cheers
    $endgroup$
    – Will M.
    Jan 28 at 23:36










  • $begingroup$
    I suggest not using $x$ to denote a discrete variable. How about $k$ or $m$?
    $endgroup$
    – Aditya Dua
    Jan 29 at 0:08














0












0








0


0



$begingroup$



This question already has an answer here:




  • Find the expected value of $frac{1}{X+1}$ where $X$ is binomial

    4 answers




If $X$ has a binomial distribution with parameters $n$ and $p$, show that



begin{align*}
textbf{E}left(frac{1}{1+X}right) = frac{1-(1-p)^{n+1}}{(n+1)p}
end{align*}



MY ATTEMPT



Since $Xsimtext{Binomial}(n,p)$, we have
begin{align*}
textbf{E}left(frac{1}{1+X}right) & = sum_{x=0}^{n}{nchoose x}frac{p^{x}(1-p)^{n-x}}{1+x} = (1-p)^{n}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x}\\
& = frac{(1-p)^{n+1}}{p}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x+1}
end{align*}



This is as far as I can get. Could someone help me out? Thanks in advance!






probability binomial-distribution expected-value







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Find the expected value of $frac{1}{X+1}$ where $X$ is binomial

    4 answers




If $X$ has a binomial distribution with parameters $n$ and $p$, show that



begin{align*}
textbf{E}left(frac{1}{1+X}right) = frac{1-(1-p)^{n+1}}{(n+1)p}
end{align*}



MY ATTEMPT



Since $Xsimtext{Binomial}(n,p)$, we have
begin{align*}
textbf{E}left(frac{1}{1+X}right) & = sum_{x=0}^{n}{nchoose x}frac{p^{x}(1-p)^{n-x}}{1+x} = (1-p)^{n}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x}\\
& = frac{(1-p)^{n+1}}{p}sum_{x=0}^{n}{nchoose x}frac{1}{1+x}left(frac{p}{1-p}right)^{x+1}
end{align*}



This is as far as I can get. Could someone help me out? Thanks in advance!





This question already has an answer here:




  • Find the expected value of $frac{1}{X+1}$ where $X$ is binomial

    4 answers







probability binomial-distribution expected-value




probability binomial-distribution expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 23:26







user1337

















asked Jan 28 at 23:14









user1337user1337

46510




46510




marked as duplicate by StubbornAtom, NCh, max_zorn, Andrew, Lord Shark the Unknown Jan 30 at 5:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by StubbornAtom, NCh, max_zorn, Andrew, Lord Shark the Unknown Jan 30 at 5:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Where are the combinatorial coefficients?
    $endgroup$
    – Will M.
    Jan 28 at 23:22










  • $begingroup$
    Apparently I missed them. I have edited it, though.
    $endgroup$
    – user1337
    Jan 28 at 23:27






  • 3




    $begingroup$
    $dbinom{n}{x} dfrac{1}{x+1} p^x (1-p)^{n-x}=dfrac{1}{p(n+1)}dbinom{n+1}{x+1}p^{x+1}(1-p)^{(n+1)-(x+1)}.$ Cheers
    $endgroup$
    – Will M.
    Jan 28 at 23:36










  • $begingroup$
    I suggest not using $x$ to denote a discrete variable. How about $k$ or $m$?
    $endgroup$
    – Aditya Dua
    Jan 29 at 0:08


















  • $begingroup$
    Where are the combinatorial coefficients?
    $endgroup$
    – Will M.
    Jan 28 at 23:22










  • $begingroup$
    Apparently I missed them. I have edited it, though.
    $endgroup$
    – user1337
    Jan 28 at 23:27






  • 3




    $begingroup$
    $dbinom{n}{x} dfrac{1}{x+1} p^x (1-p)^{n-x}=dfrac{1}{p(n+1)}dbinom{n+1}{x+1}p^{x+1}(1-p)^{(n+1)-(x+1)}.$ Cheers
    $endgroup$
    – Will M.
    Jan 28 at 23:36










  • $begingroup$
    I suggest not using $x$ to denote a discrete variable. How about $k$ or $m$?
    $endgroup$
    – Aditya Dua
    Jan 29 at 0:08
















$begingroup$
Where are the combinatorial coefficients?
$endgroup$
– Will M.
Jan 28 at 23:22




$begingroup$
Where are the combinatorial coefficients?
$endgroup$
– Will M.
Jan 28 at 23:22












$begingroup$
Apparently I missed them. I have edited it, though.
$endgroup$
– user1337
Jan 28 at 23:27




$begingroup$
Apparently I missed them. I have edited it, though.
$endgroup$
– user1337
Jan 28 at 23:27




3




3




$begingroup$
$dbinom{n}{x} dfrac{1}{x+1} p^x (1-p)^{n-x}=dfrac{1}{p(n+1)}dbinom{n+1}{x+1}p^{x+1}(1-p)^{(n+1)-(x+1)}.$ Cheers
$endgroup$
– Will M.
Jan 28 at 23:36




$begingroup$
$dbinom{n}{x} dfrac{1}{x+1} p^x (1-p)^{n-x}=dfrac{1}{p(n+1)}dbinom{n+1}{x+1}p^{x+1}(1-p)^{(n+1)-(x+1)}.$ Cheers
$endgroup$
– Will M.
Jan 28 at 23:36












$begingroup$
I suggest not using $x$ to denote a discrete variable. How about $k$ or $m$?
$endgroup$
– Aditya Dua
Jan 29 at 0:08




$begingroup$
I suggest not using $x$ to denote a discrete variable. How about $k$ or $m$?
$endgroup$
– Aditya Dua
Jan 29 at 0:08










2 Answers
2






active

oldest

votes


















1












$begingroup$

You can also sidestep the summation by using the moment generating function (via the Laplace transform). For any $X>-1$,



$$frac{1}{1+X} = int_0^{+infty} e^{-(1+X)t}dt,$$



so, taking the expectation,



$$mathbb{E} left(frac{1}{1+X}right) = mathbb{E} left( int_0^{+infty} e^{-(1+X)t}dtright) = int_0^{+infty} e^{-t} mathbb{E} left( e^{-tX}right) dt.$$



The moment generating function of the binomial distribution is $mathbb{E} left( e^{-tX}right) = (1-p+pe^{-t})^n$. Insert this value in the integral, solve the integral (there is an explicit antiderivative), and get the result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Dear Thomine, thank you very much for your contribution! But we haven't studied moment generating functions so far. Could you provide an answer based on another approach?
    $endgroup$
    – user1337
    Jan 28 at 23:54





















1












$begingroup$

$X sim B(n,p)$



$$mathbb{E}left( {1 over 1+X}right) = sum_{k=0}^n {n choose k}{1 over 1+k} p^k(1-p)^{n-k} = (1-p)^nsum_{k=0}^n {n choose k}{1 over k+1} left( {p over 1-p}right)^k$$



We know that
$$(1+y)^n = sum_{k=0}^n {n choose k} y^k$$



Integrating both sides with respect to $y$ from 0 to $x$ gives



$${(1+x)^{n+1} over n+1} -{1 over n+1} = sum_{k=0}^n {n choose k} {x^{k+1} over k+1}$$



Setting $x = {p over 1-p}$ gives



$${1 over n+1} left( {1 over (1-p)^{n+1}} -1 right) = {p over 1-p}sum_{k=0}^n {n choose k} {1 over k+1} left( {p over 1-p}right)^k$$



Multiplying the left side by ${1-p over p}$ and further by $(1-p)^n$ should give us our final answer.



$$mathbb{E}left( {1 over 1+X}right) = {1-p over p(n+1)} left( {1 over 1-p} - (1-p)^nright) = {1 - (1-p)^{n+1} over p(n+1)}$$






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You can also sidestep the summation by using the moment generating function (via the Laplace transform). For any $X>-1$,



    $$frac{1}{1+X} = int_0^{+infty} e^{-(1+X)t}dt,$$



    so, taking the expectation,



    $$mathbb{E} left(frac{1}{1+X}right) = mathbb{E} left( int_0^{+infty} e^{-(1+X)t}dtright) = int_0^{+infty} e^{-t} mathbb{E} left( e^{-tX}right) dt.$$



    The moment generating function of the binomial distribution is $mathbb{E} left( e^{-tX}right) = (1-p+pe^{-t})^n$. Insert this value in the integral, solve the integral (there is an explicit antiderivative), and get the result.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Dear Thomine, thank you very much for your contribution! But we haven't studied moment generating functions so far. Could you provide an answer based on another approach?
      $endgroup$
      – user1337
      Jan 28 at 23:54


















    1












    $begingroup$

    You can also sidestep the summation by using the moment generating function (via the Laplace transform). For any $X>-1$,



    $$frac{1}{1+X} = int_0^{+infty} e^{-(1+X)t}dt,$$



    so, taking the expectation,



    $$mathbb{E} left(frac{1}{1+X}right) = mathbb{E} left( int_0^{+infty} e^{-(1+X)t}dtright) = int_0^{+infty} e^{-t} mathbb{E} left( e^{-tX}right) dt.$$



    The moment generating function of the binomial distribution is $mathbb{E} left( e^{-tX}right) = (1-p+pe^{-t})^n$. Insert this value in the integral, solve the integral (there is an explicit antiderivative), and get the result.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Dear Thomine, thank you very much for your contribution! But we haven't studied moment generating functions so far. Could you provide an answer based on another approach?
      $endgroup$
      – user1337
      Jan 28 at 23:54
















    1












    1








    1





    $begingroup$

    You can also sidestep the summation by using the moment generating function (via the Laplace transform). For any $X>-1$,



    $$frac{1}{1+X} = int_0^{+infty} e^{-(1+X)t}dt,$$



    so, taking the expectation,



    $$mathbb{E} left(frac{1}{1+X}right) = mathbb{E} left( int_0^{+infty} e^{-(1+X)t}dtright) = int_0^{+infty} e^{-t} mathbb{E} left( e^{-tX}right) dt.$$



    The moment generating function of the binomial distribution is $mathbb{E} left( e^{-tX}right) = (1-p+pe^{-t})^n$. Insert this value in the integral, solve the integral (there is an explicit antiderivative), and get the result.






    share|cite|improve this answer









    $endgroup$



    You can also sidestep the summation by using the moment generating function (via the Laplace transform). For any $X>-1$,



    $$frac{1}{1+X} = int_0^{+infty} e^{-(1+X)t}dt,$$



    so, taking the expectation,



    $$mathbb{E} left(frac{1}{1+X}right) = mathbb{E} left( int_0^{+infty} e^{-(1+X)t}dtright) = int_0^{+infty} e^{-t} mathbb{E} left( e^{-tX}right) dt.$$



    The moment generating function of the binomial distribution is $mathbb{E} left( e^{-tX}right) = (1-p+pe^{-t})^n$. Insert this value in the integral, solve the integral (there is an explicit antiderivative), and get the result.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 28 at 23:47









    D. ThomineD. Thomine

    7,7991538




    7,7991538












    • $begingroup$
      Dear Thomine, thank you very much for your contribution! But we haven't studied moment generating functions so far. Could you provide an answer based on another approach?
      $endgroup$
      – user1337
      Jan 28 at 23:54




















    • $begingroup$
      Dear Thomine, thank you very much for your contribution! But we haven't studied moment generating functions so far. Could you provide an answer based on another approach?
      $endgroup$
      – user1337
      Jan 28 at 23:54


















    $begingroup$
    Dear Thomine, thank you very much for your contribution! But we haven't studied moment generating functions so far. Could you provide an answer based on another approach?
    $endgroup$
    – user1337
    Jan 28 at 23:54






    $begingroup$
    Dear Thomine, thank you very much for your contribution! But we haven't studied moment generating functions so far. Could you provide an answer based on another approach?
    $endgroup$
    – user1337
    Jan 28 at 23:54













    1












    $begingroup$

    $X sim B(n,p)$



    $$mathbb{E}left( {1 over 1+X}right) = sum_{k=0}^n {n choose k}{1 over 1+k} p^k(1-p)^{n-k} = (1-p)^nsum_{k=0}^n {n choose k}{1 over k+1} left( {p over 1-p}right)^k$$



    We know that
    $$(1+y)^n = sum_{k=0}^n {n choose k} y^k$$



    Integrating both sides with respect to $y$ from 0 to $x$ gives



    $${(1+x)^{n+1} over n+1} -{1 over n+1} = sum_{k=0}^n {n choose k} {x^{k+1} over k+1}$$



    Setting $x = {p over 1-p}$ gives



    $${1 over n+1} left( {1 over (1-p)^{n+1}} -1 right) = {p over 1-p}sum_{k=0}^n {n choose k} {1 over k+1} left( {p over 1-p}right)^k$$



    Multiplying the left side by ${1-p over p}$ and further by $(1-p)^n$ should give us our final answer.



    $$mathbb{E}left( {1 over 1+X}right) = {1-p over p(n+1)} left( {1 over 1-p} - (1-p)^nright) = {1 - (1-p)^{n+1} over p(n+1)}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $X sim B(n,p)$



      $$mathbb{E}left( {1 over 1+X}right) = sum_{k=0}^n {n choose k}{1 over 1+k} p^k(1-p)^{n-k} = (1-p)^nsum_{k=0}^n {n choose k}{1 over k+1} left( {p over 1-p}right)^k$$



      We know that
      $$(1+y)^n = sum_{k=0}^n {n choose k} y^k$$



      Integrating both sides with respect to $y$ from 0 to $x$ gives



      $${(1+x)^{n+1} over n+1} -{1 over n+1} = sum_{k=0}^n {n choose k} {x^{k+1} over k+1}$$



      Setting $x = {p over 1-p}$ gives



      $${1 over n+1} left( {1 over (1-p)^{n+1}} -1 right) = {p over 1-p}sum_{k=0}^n {n choose k} {1 over k+1} left( {p over 1-p}right)^k$$



      Multiplying the left side by ${1-p over p}$ and further by $(1-p)^n$ should give us our final answer.



      $$mathbb{E}left( {1 over 1+X}right) = {1-p over p(n+1)} left( {1 over 1-p} - (1-p)^nright) = {1 - (1-p)^{n+1} over p(n+1)}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $X sim B(n,p)$



        $$mathbb{E}left( {1 over 1+X}right) = sum_{k=0}^n {n choose k}{1 over 1+k} p^k(1-p)^{n-k} = (1-p)^nsum_{k=0}^n {n choose k}{1 over k+1} left( {p over 1-p}right)^k$$



        We know that
        $$(1+y)^n = sum_{k=0}^n {n choose k} y^k$$



        Integrating both sides with respect to $y$ from 0 to $x$ gives



        $${(1+x)^{n+1} over n+1} -{1 over n+1} = sum_{k=0}^n {n choose k} {x^{k+1} over k+1}$$



        Setting $x = {p over 1-p}$ gives



        $${1 over n+1} left( {1 over (1-p)^{n+1}} -1 right) = {p over 1-p}sum_{k=0}^n {n choose k} {1 over k+1} left( {p over 1-p}right)^k$$



        Multiplying the left side by ${1-p over p}$ and further by $(1-p)^n$ should give us our final answer.



        $$mathbb{E}left( {1 over 1+X}right) = {1-p over p(n+1)} left( {1 over 1-p} - (1-p)^nright) = {1 - (1-p)^{n+1} over p(n+1)}$$






        share|cite|improve this answer









        $endgroup$



        $X sim B(n,p)$



        $$mathbb{E}left( {1 over 1+X}right) = sum_{k=0}^n {n choose k}{1 over 1+k} p^k(1-p)^{n-k} = (1-p)^nsum_{k=0}^n {n choose k}{1 over k+1} left( {p over 1-p}right)^k$$



        We know that
        $$(1+y)^n = sum_{k=0}^n {n choose k} y^k$$



        Integrating both sides with respect to $y$ from 0 to $x$ gives



        $${(1+x)^{n+1} over n+1} -{1 over n+1} = sum_{k=0}^n {n choose k} {x^{k+1} over k+1}$$



        Setting $x = {p over 1-p}$ gives



        $${1 over n+1} left( {1 over (1-p)^{n+1}} -1 right) = {p over 1-p}sum_{k=0}^n {n choose k} {1 over k+1} left( {p over 1-p}right)^k$$



        Multiplying the left side by ${1-p over p}$ and further by $(1-p)^n$ should give us our final answer.



        $$mathbb{E}left( {1 over 1+X}right) = {1-p over p(n+1)} left( {1 over 1-p} - (1-p)^nright) = {1 - (1-p)^{n+1} over p(n+1)}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 29 at 7:07









        Aditya DuaAditya Dua

        1,15418




        1,15418















            Popular posts from this blog

            MongoDB - Not Authorized To Execute Command

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
            Read more

            How to fix TextFormField cause rebuild widget in Flutter

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
            Read more

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Image
            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
            Read more