Why p$frac{partial }{partial P} $ is the mean of binomial distribution function?












0














Why p$frac{partial }{partial P}[ {p+q}]^n $ is the mean of binomial distribution function? I know the mean should be $sum np(n) $ but why p$frac{partial }{partial P} [{p+q}]^n$ of the binomial distribution is the mean. And i dont see that to be used in Poisson or Gaussian distribution. So why is this the mean ?






statistics probability-distributions binomial-distribution







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  • Is this some sort of bizarre typo? The binomial $B(n,p)$ will have mean $np$.
    – Randall
    Jan 1 at 2:49










  • @Randall does it loik right now ?
    – user187604
    Jan 1 at 3:13
















0














Why p$frac{partial }{partial P}[ {p+q}]^n $ is the mean of binomial distribution function? I know the mean should be $sum np(n) $ but why p$frac{partial }{partial P} [{p+q}]^n$ of the binomial distribution is the mean. And i dont see that to be used in Poisson or Gaussian distribution. So why is this the mean ?






statistics probability-distributions binomial-distribution







share|cite|improve this question
























  • Is this some sort of bizarre typo? The binomial $B(n,p)$ will have mean $np$.
    – Randall
    Jan 1 at 2:49










  • @Randall does it loik right now ?
    – user187604
    Jan 1 at 3:13














0












0








0







Why p$frac{partial }{partial P}[ {p+q}]^n $ is the mean of binomial distribution function? I know the mean should be $sum np(n) $ but why p$frac{partial }{partial P} [{p+q}]^n$ of the binomial distribution is the mean. And i dont see that to be used in Poisson or Gaussian distribution. So why is this the mean ?






statistics probability-distributions binomial-distribution







share|cite|improve this question















Why p$frac{partial }{partial P}[ {p+q}]^n $ is the mean of binomial distribution function? I know the mean should be $sum np(n) $ but why p$frac{partial }{partial P} [{p+q}]^n$ of the binomial distribution is the mean. And i dont see that to be used in Poisson or Gaussian distribution. So why is this the mean ?






statistics probability-distributions binomial-distribution




statistics probability-distributions binomial-distribution






share|cite|improve this question















share|cite|improve this question













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edited Jan 1 at 3:11







user187604

















asked Jan 1 at 2:15









user187604user187604

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27811












  • Is this some sort of bizarre typo? The binomial $B(n,p)$ will have mean $np$.
    – Randall
    Jan 1 at 2:49










  • @Randall does it loik right now ?
    – user187604
    Jan 1 at 3:13


















  • Is this some sort of bizarre typo? The binomial $B(n,p)$ will have mean $np$.
    – Randall
    Jan 1 at 2:49










  • @Randall does it loik right now ?
    – user187604
    Jan 1 at 3:13
















Is this some sort of bizarre typo? The binomial $B(n,p)$ will have mean $np$.
– Randall
Jan 1 at 2:49




Is this some sort of bizarre typo? The binomial $B(n,p)$ will have mean $np$.
– Randall
Jan 1 at 2:49












@Randall does it loik right now ?
– user187604
Jan 1 at 3:13




@Randall does it loik right now ?
– user187604
Jan 1 at 3:13










1 Answer
1






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oldest

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2














Using the binomial expansion,



$$
sum_{i=1}^n ibinom{n}{i}p^iq^{n-i}=ptimessum_{i=1}^nbinom{n}{i}frac{partial}{partial p}left(p^iq^{n-i}right)=ptimes frac{partial}{partial p}(p+q)^n
$$






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  • Thanks for the answer.
    – user187604
    Jan 1 at 3:17











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Using the binomial expansion,



$$
sum_{i=1}^n ibinom{n}{i}p^iq^{n-i}=ptimessum_{i=1}^nbinom{n}{i}frac{partial}{partial p}left(p^iq^{n-i}right)=ptimes frac{partial}{partial p}(p+q)^n
$$






share|cite|improve this answer





















  • Thanks for the answer.
    – user187604
    Jan 1 at 3:17
















2














Using the binomial expansion,



$$
sum_{i=1}^n ibinom{n}{i}p^iq^{n-i}=ptimessum_{i=1}^nbinom{n}{i}frac{partial}{partial p}left(p^iq^{n-i}right)=ptimes frac{partial}{partial p}(p+q)^n
$$






share|cite|improve this answer





















  • Thanks for the answer.
    – user187604
    Jan 1 at 3:17














2












2








2






Using the binomial expansion,



$$
sum_{i=1}^n ibinom{n}{i}p^iq^{n-i}=ptimessum_{i=1}^nbinom{n}{i}frac{partial}{partial p}left(p^iq^{n-i}right)=ptimes frac{partial}{partial p}(p+q)^n
$$






share|cite|improve this answer












Using the binomial expansion,



$$
sum_{i=1}^n ibinom{n}{i}p^iq^{n-i}=ptimessum_{i=1}^nbinom{n}{i}frac{partial}{partial p}left(p^iq^{n-i}right)=ptimes frac{partial}{partial p}(p+q)^n
$$







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share|cite|improve this answer










answered Jan 1 at 3:14









d.k.o.d.k.o.

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8,622528












  • Thanks for the answer.
    – user187604
    Jan 1 at 3:17


















  • Thanks for the answer.
    – user187604
    Jan 1 at 3:17
















Thanks for the answer.
– user187604
Jan 1 at 3:17




Thanks for the answer.
– user187604
Jan 1 at 3:17


















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