Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$












2















Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$.




It is suggested that I can transform it into
begin{align}
mathbb{E}[X(X-1)(X-2)(X-3)]
&=sum_{k=0}^n k(k-1)(k-2)mathbb{P}{X=k}\
&=sum_{k=3}^{n+3} (k-3)(k-4)(k-5)mathbb{P}{X=k-3}\
&=sum_{k=0}^n i(i-1)(i-2)mathbb{P}{X=i}
end{align}
But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one.



Please help.






summation combinations binomial-distribution expected-value







share|cite|improve this question




















  • 1




    Related: math.stackexchange.com/questions/1476676/…
    – heropup
    Nov 21 '18 at 6:24
















2















Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$.




It is suggested that I can transform it into
begin{align}
mathbb{E}[X(X-1)(X-2)(X-3)]
&=sum_{k=0}^n k(k-1)(k-2)mathbb{P}{X=k}\
&=sum_{k=3}^{n+3} (k-3)(k-4)(k-5)mathbb{P}{X=k-3}\
&=sum_{k=0}^n i(i-1)(i-2)mathbb{P}{X=i}
end{align}
But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one.



Please help.






summation combinations binomial-distribution expected-value







share|cite|improve this question




















  • 1




    Related: math.stackexchange.com/questions/1476676/…
    – heropup
    Nov 21 '18 at 6:24














2












2








2








Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$.




It is suggested that I can transform it into
begin{align}
mathbb{E}[X(X-1)(X-2)(X-3)]
&=sum_{k=0}^n k(k-1)(k-2)mathbb{P}{X=k}\
&=sum_{k=3}^{n+3} (k-3)(k-4)(k-5)mathbb{P}{X=k-3}\
&=sum_{k=0}^n i(i-1)(i-2)mathbb{P}{X=i}
end{align}
But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one.



Please help.






summation combinations binomial-distribution expected-value







share|cite|improve this question
















Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$.




It is suggested that I can transform it into
begin{align}
mathbb{E}[X(X-1)(X-2)(X-3)]
&=sum_{k=0}^n k(k-1)(k-2)mathbb{P}{X=k}\
&=sum_{k=3}^{n+3} (k-3)(k-4)(k-5)mathbb{P}{X=k-3}\
&=sum_{k=0}^n i(i-1)(i-2)mathbb{P}{X=i}
end{align}
But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one.



Please help.






summation combinations binomial-distribution expected-value




summation combinations 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 Nov 21 '18 at 6:18









Clement C.

49.7k33886




49.7k33886










asked Nov 21 '18 at 5:45









Michael Lee

385




385








  • 1




    Related: math.stackexchange.com/questions/1476676/…
    – heropup
    Nov 21 '18 at 6:24














  • 1




    Related: math.stackexchange.com/questions/1476676/…
    – heropup
    Nov 21 '18 at 6:24








1




1




Related: math.stackexchange.com/questions/1476676/…
– heropup
Nov 21 '18 at 6:24




Related: math.stackexchange.com/questions/1476676/…
– heropup
Nov 21 '18 at 6:24










4 Answers
4






active

oldest

votes


















0














Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is:
$$begin{align*}
mathbb{E}[X(X-1)(X-2)(X-3)]
&= sum_{k=0}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
&= sum_{k=4}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
&= sum_{k=4}^n k(k-1)(k-2)(k-3)binom{n}{k}p^k(1-p)^{n-k}\
&= sum_{k=4}^n k(k-1)(k-2)(k-3)frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\
&= sum_{k=4}^n frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\
&= sum_{k=4}^n frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\
&= n(n-1)(n-2)(n-3)p^4sum_{k=4}^n binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\
&= n(n-1)(n-2)(n-3)p^4sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}\
&= boxed{n(n-1)(n-2)(n-3)p^4}
end{align*}$$
since $sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.






share|cite|improve this answer





















  • Thank you for our help.
    – Michael Lee
    Nov 22 '18 at 2:48










  • @MichaelLee You're welcome!
    – Clement C.
    Nov 22 '18 at 2:56



















2














One way is to use characteristic functions. The characteristic function of $B(n,p)$ is $phi (t)=(pe^{it}+(1-p))^{n}$. The moments of $X$ are given by $i^{n}EX^{n}=phi^{(n)}(0)$. You can compute the first 4 moments of $X$ using this and then use the expanded form of $X(X_1)(X-2)(X-3)$.






share|cite|improve this answer





















  • This is somewhat overkill (assumes somewhat more advanced knowledge), and does not really follow the suggestion/hint at all, however.
    – Clement C.
    Nov 21 '18 at 6:13










  • @ClementC. The other two answers here use moment generating function instead of characteristic function. I am not sure if characteristic function is much more advanced than moment generating function. If OP says he is not familiar with characteristic functions I will gladly delete my answer.
    – Kavi Rama Murthy
    Nov 21 '18 at 7:46










  • No, your answer is worth having. Just feeling that all the characteristic functions/MGF approaches are a bit "not elementary" for this question.
    – Clement C.
    Nov 21 '18 at 7:49










  • Hi guys I am OP, well this is indeed a year 1 engineering math course so I really have no idea about these things. I guess I will go study on the idea of your answer later.
    – Michael Lee
    Nov 22 '18 at 2:47



















0














Hint: Probability Generating Function (see https://en.wikipedia.org/wiki/Probability-generating_function)



Look at the properties section, the part where it talks about the kth factorial moment.






share|cite|improve this answer





















  • proofwiki.org/wiki/…
    – KnowsNothing
    Nov 21 '18 at 5:57



















0














Look at $f(s)= E [ s^X]$.



1) Observe that



$$f(s) = (sp + (1-p))^n=(1+ (s-1)p)^n =sum_{k=0}^n binom{n}{k}p^k (s-1)^k=sum_{k=0}^n frac{n!}{(n-k)!} frac{(s-1)^k}{k!},$$
the RHS identified as the Taylor series of $f$ about $s=1$.



2) Differentiating under the expectation (which is just a finite sum here), we obtain $$f'(s) = E[ X s^{X-1}] ,~f''(s) = E[X (X-1) s^{X-2}],dots.$$



You're looking for $f^{(4)}(1)$.






share|cite|improve this answer





















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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is:
    $$begin{align*}
    mathbb{E}[X(X-1)(X-2)(X-3)]
    &= sum_{k=0}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)binom{n}{k}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\
    &= n(n-1)(n-2)(n-3)p^4sum_{k=4}^n binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\
    &= n(n-1)(n-2)(n-3)p^4sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}\
    &= boxed{n(n-1)(n-2)(n-3)p^4}
    end{align*}$$
    since $sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.






    share|cite|improve this answer





















    • Thank you for our help.
      – Michael Lee
      Nov 22 '18 at 2:48










    • @MichaelLee You're welcome!
      – Clement C.
      Nov 22 '18 at 2:56
















    0














    Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is:
    $$begin{align*}
    mathbb{E}[X(X-1)(X-2)(X-3)]
    &= sum_{k=0}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)binom{n}{k}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\
    &= n(n-1)(n-2)(n-3)p^4sum_{k=4}^n binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\
    &= n(n-1)(n-2)(n-3)p^4sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}\
    &= boxed{n(n-1)(n-2)(n-3)p^4}
    end{align*}$$
    since $sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.






    share|cite|improve this answer





















    • Thank you for our help.
      – Michael Lee
      Nov 22 '18 at 2:48










    • @MichaelLee You're welcome!
      – Clement C.
      Nov 22 '18 at 2:56














    0












    0








    0






    Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is:
    $$begin{align*}
    mathbb{E}[X(X-1)(X-2)(X-3)]
    &= sum_{k=0}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)binom{n}{k}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\
    &= n(n-1)(n-2)(n-3)p^4sum_{k=4}^n binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\
    &= n(n-1)(n-2)(n-3)p^4sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}\
    &= boxed{n(n-1)(n-2)(n-3)p^4}
    end{align*}$$
    since $sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.






    share|cite|improve this answer












    Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is:
    $$begin{align*}
    mathbb{E}[X(X-1)(X-2)(X-3)]
    &= sum_{k=0}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)mathbb{P}{X=k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)binom{n}{k}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n k(k-1)(k-2)(k-3)frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\
    &= sum_{k=4}^n frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\
    &= n(n-1)(n-2)(n-3)p^4sum_{k=4}^n binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\
    &= n(n-1)(n-2)(n-3)p^4sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}\
    &= boxed{n(n-1)(n-2)(n-3)p^4}
    end{align*}$$
    since $sum_{ell=0}^n binom{n-4}{ell}p^{ell}(1-p)^{(n-4)-ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 '18 at 6:06









    Clement C.

    49.7k33886




    49.7k33886












    • Thank you for our help.
      – Michael Lee
      Nov 22 '18 at 2:48










    • @MichaelLee You're welcome!
      – Clement C.
      Nov 22 '18 at 2:56


















    • Thank you for our help.
      – Michael Lee
      Nov 22 '18 at 2:48










    • @MichaelLee You're welcome!
      – Clement C.
      Nov 22 '18 at 2:56
















    Thank you for our help.
    – Michael Lee
    Nov 22 '18 at 2:48




    Thank you for our help.
    – Michael Lee
    Nov 22 '18 at 2:48












    @MichaelLee You're welcome!
    – Clement C.
    Nov 22 '18 at 2:56




    @MichaelLee You're welcome!
    – Clement C.
    Nov 22 '18 at 2:56











    2














    One way is to use characteristic functions. The characteristic function of $B(n,p)$ is $phi (t)=(pe^{it}+(1-p))^{n}$. The moments of $X$ are given by $i^{n}EX^{n}=phi^{(n)}(0)$. You can compute the first 4 moments of $X$ using this and then use the expanded form of $X(X_1)(X-2)(X-3)$.






    share|cite|improve this answer





















    • This is somewhat overkill (assumes somewhat more advanced knowledge), and does not really follow the suggestion/hint at all, however.
      – Clement C.
      Nov 21 '18 at 6:13










    • @ClementC. The other two answers here use moment generating function instead of characteristic function. I am not sure if characteristic function is much more advanced than moment generating function. If OP says he is not familiar with characteristic functions I will gladly delete my answer.
      – Kavi Rama Murthy
      Nov 21 '18 at 7:46










    • No, your answer is worth having. Just feeling that all the characteristic functions/MGF approaches are a bit "not elementary" for this question.
      – Clement C.
      Nov 21 '18 at 7:49










    • Hi guys I am OP, well this is indeed a year 1 engineering math course so I really have no idea about these things. I guess I will go study on the idea of your answer later.
      – Michael Lee
      Nov 22 '18 at 2:47
















    2














    One way is to use characteristic functions. The characteristic function of $B(n,p)$ is $phi (t)=(pe^{it}+(1-p))^{n}$. The moments of $X$ are given by $i^{n}EX^{n}=phi^{(n)}(0)$. You can compute the first 4 moments of $X$ using this and then use the expanded form of $X(X_1)(X-2)(X-3)$.






    share|cite|improve this answer





















    • This is somewhat overkill (assumes somewhat more advanced knowledge), and does not really follow the suggestion/hint at all, however.
      – Clement C.
      Nov 21 '18 at 6:13










    • @ClementC. The other two answers here use moment generating function instead of characteristic function. I am not sure if characteristic function is much more advanced than moment generating function. If OP says he is not familiar with characteristic functions I will gladly delete my answer.
      – Kavi Rama Murthy
      Nov 21 '18 at 7:46










    • No, your answer is worth having. Just feeling that all the characteristic functions/MGF approaches are a bit "not elementary" for this question.
      – Clement C.
      Nov 21 '18 at 7:49










    • Hi guys I am OP, well this is indeed a year 1 engineering math course so I really have no idea about these things. I guess I will go study on the idea of your answer later.
      – Michael Lee
      Nov 22 '18 at 2:47














    2












    2








    2






    One way is to use characteristic functions. The characteristic function of $B(n,p)$ is $phi (t)=(pe^{it}+(1-p))^{n}$. The moments of $X$ are given by $i^{n}EX^{n}=phi^{(n)}(0)$. You can compute the first 4 moments of $X$ using this and then use the expanded form of $X(X_1)(X-2)(X-3)$.






    share|cite|improve this answer












    One way is to use characteristic functions. The characteristic function of $B(n,p)$ is $phi (t)=(pe^{it}+(1-p))^{n}$. The moments of $X$ are given by $i^{n}EX^{n}=phi^{(n)}(0)$. You can compute the first 4 moments of $X$ using this and then use the expanded form of $X(X_1)(X-2)(X-3)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 '18 at 5:51









    Kavi Rama Murthy

    50.4k31854




    50.4k31854












    • This is somewhat overkill (assumes somewhat more advanced knowledge), and does not really follow the suggestion/hint at all, however.
      – Clement C.
      Nov 21 '18 at 6:13










    • @ClementC. The other two answers here use moment generating function instead of characteristic function. I am not sure if characteristic function is much more advanced than moment generating function. If OP says he is not familiar with characteristic functions I will gladly delete my answer.
      – Kavi Rama Murthy
      Nov 21 '18 at 7:46










    • No, your answer is worth having. Just feeling that all the characteristic functions/MGF approaches are a bit "not elementary" for this question.
      – Clement C.
      Nov 21 '18 at 7:49










    • Hi guys I am OP, well this is indeed a year 1 engineering math course so I really have no idea about these things. I guess I will go study on the idea of your answer later.
      – Michael Lee
      Nov 22 '18 at 2:47


















    • This is somewhat overkill (assumes somewhat more advanced knowledge), and does not really follow the suggestion/hint at all, however.
      – Clement C.
      Nov 21 '18 at 6:13










    • @ClementC. The other two answers here use moment generating function instead of characteristic function. I am not sure if characteristic function is much more advanced than moment generating function. If OP says he is not familiar with characteristic functions I will gladly delete my answer.
      – Kavi Rama Murthy
      Nov 21 '18 at 7:46










    • No, your answer is worth having. Just feeling that all the characteristic functions/MGF approaches are a bit "not elementary" for this question.
      – Clement C.
      Nov 21 '18 at 7:49










    • Hi guys I am OP, well this is indeed a year 1 engineering math course so I really have no idea about these things. I guess I will go study on the idea of your answer later.
      – Michael Lee
      Nov 22 '18 at 2:47
















    This is somewhat overkill (assumes somewhat more advanced knowledge), and does not really follow the suggestion/hint at all, however.
    – Clement C.
    Nov 21 '18 at 6:13




    This is somewhat overkill (assumes somewhat more advanced knowledge), and does not really follow the suggestion/hint at all, however.
    – Clement C.
    Nov 21 '18 at 6:13












    @ClementC. The other two answers here use moment generating function instead of characteristic function. I am not sure if characteristic function is much more advanced than moment generating function. If OP says he is not familiar with characteristic functions I will gladly delete my answer.
    – Kavi Rama Murthy
    Nov 21 '18 at 7:46




    @ClementC. The other two answers here use moment generating function instead of characteristic function. I am not sure if characteristic function is much more advanced than moment generating function. If OP says he is not familiar with characteristic functions I will gladly delete my answer.
    – Kavi Rama Murthy
    Nov 21 '18 at 7:46












    No, your answer is worth having. Just feeling that all the characteristic functions/MGF approaches are a bit "not elementary" for this question.
    – Clement C.
    Nov 21 '18 at 7:49




    No, your answer is worth having. Just feeling that all the characteristic functions/MGF approaches are a bit "not elementary" for this question.
    – Clement C.
    Nov 21 '18 at 7:49












    Hi guys I am OP, well this is indeed a year 1 engineering math course so I really have no idea about these things. I guess I will go study on the idea of your answer later.
    – Michael Lee
    Nov 22 '18 at 2:47




    Hi guys I am OP, well this is indeed a year 1 engineering math course so I really have no idea about these things. I guess I will go study on the idea of your answer later.
    – Michael Lee
    Nov 22 '18 at 2:47











    0














    Hint: Probability Generating Function (see https://en.wikipedia.org/wiki/Probability-generating_function)



    Look at the properties section, the part where it talks about the kth factorial moment.






    share|cite|improve this answer





















    • proofwiki.org/wiki/…
      – KnowsNothing
      Nov 21 '18 at 5:57
















    0














    Hint: Probability Generating Function (see https://en.wikipedia.org/wiki/Probability-generating_function)



    Look at the properties section, the part where it talks about the kth factorial moment.






    share|cite|improve this answer





















    • proofwiki.org/wiki/…
      – KnowsNothing
      Nov 21 '18 at 5:57














    0












    0








    0






    Hint: Probability Generating Function (see https://en.wikipedia.org/wiki/Probability-generating_function)



    Look at the properties section, the part where it talks about the kth factorial moment.






    share|cite|improve this answer












    Hint: Probability Generating Function (see https://en.wikipedia.org/wiki/Probability-generating_function)



    Look at the properties section, the part where it talks about the kth factorial moment.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 '18 at 5:51









    KnowsNothing

    355




    355












    • proofwiki.org/wiki/…
      – KnowsNothing
      Nov 21 '18 at 5:57


















    • proofwiki.org/wiki/…
      – KnowsNothing
      Nov 21 '18 at 5:57
















    proofwiki.org/wiki/…
    – KnowsNothing
    Nov 21 '18 at 5:57




    proofwiki.org/wiki/…
    – KnowsNothing
    Nov 21 '18 at 5:57











    0














    Look at $f(s)= E [ s^X]$.



    1) Observe that



    $$f(s) = (sp + (1-p))^n=(1+ (s-1)p)^n =sum_{k=0}^n binom{n}{k}p^k (s-1)^k=sum_{k=0}^n frac{n!}{(n-k)!} frac{(s-1)^k}{k!},$$
    the RHS identified as the Taylor series of $f$ about $s=1$.



    2) Differentiating under the expectation (which is just a finite sum here), we obtain $$f'(s) = E[ X s^{X-1}] ,~f''(s) = E[X (X-1) s^{X-2}],dots.$$



    You're looking for $f^{(4)}(1)$.






    share|cite|improve this answer


























      0














      Look at $f(s)= E [ s^X]$.



      1) Observe that



      $$f(s) = (sp + (1-p))^n=(1+ (s-1)p)^n =sum_{k=0}^n binom{n}{k}p^k (s-1)^k=sum_{k=0}^n frac{n!}{(n-k)!} frac{(s-1)^k}{k!},$$
      the RHS identified as the Taylor series of $f$ about $s=1$.



      2) Differentiating under the expectation (which is just a finite sum here), we obtain $$f'(s) = E[ X s^{X-1}] ,~f''(s) = E[X (X-1) s^{X-2}],dots.$$



      You're looking for $f^{(4)}(1)$.






      share|cite|improve this answer
























        0












        0








        0






        Look at $f(s)= E [ s^X]$.



        1) Observe that



        $$f(s) = (sp + (1-p))^n=(1+ (s-1)p)^n =sum_{k=0}^n binom{n}{k}p^k (s-1)^k=sum_{k=0}^n frac{n!}{(n-k)!} frac{(s-1)^k}{k!},$$
        the RHS identified as the Taylor series of $f$ about $s=1$.



        2) Differentiating under the expectation (which is just a finite sum here), we obtain $$f'(s) = E[ X s^{X-1}] ,~f''(s) = E[X (X-1) s^{X-2}],dots.$$



        You're looking for $f^{(4)}(1)$.






        share|cite|improve this answer












        Look at $f(s)= E [ s^X]$.



        1) Observe that



        $$f(s) = (sp + (1-p))^n=(1+ (s-1)p)^n =sum_{k=0}^n binom{n}{k}p^k (s-1)^k=sum_{k=0}^n frac{n!}{(n-k)!} frac{(s-1)^k}{k!},$$
        the RHS identified as the Taylor series of $f$ about $s=1$.



        2) Differentiating under the expectation (which is just a finite sum here), we obtain $$f'(s) = E[ X s^{X-1}] ,~f''(s) = E[X (X-1) s^{X-2}],dots.$$



        You're looking for $f^{(4)}(1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 '18 at 6:01









        Fnacool

        4,976511




        4,976511






























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