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Show that $sum_r=0^n binomnr r p^r q^n-r = np$, given $p+q=1$

Expected Value of a Binomial distribution?Sum $(1-x)^n$ $sum_r=1^n$ $r$ $nchoose r$ $(fracx1-x)^r$Finishing proof of identity $sum_k=b^n binomnk binomkb = 2^n-b binomnb$Show that $sumlimits_k=0^nbinom2n2k^!2-sumlimits_k=0^n-1binom2n2k+1^!2=(-1)^nbinom2nn$How can you show that $binom n7=sum_k=7^n binom k-1 6$?Proving combinatorially $sum_k=0^n k binom n k ^2=nbinom2n-1n-1$Binomial Coefficient Identity Involving SummationHow to show $sum_k=0^nbinomn+kkfrac12^k=2^n$How to prove $sum_k=0^n binomk+1k = binomn+2n$prove $sum_r=1^nbinomnrbinomn-1r-1=binom2n-1n-1$Prove $sum_r=0^n binomnr binomn+rr (-2)^r =(-1)^nsum_r=0^n binomnr^2 2^r$Combinatorial proof of $sum_k=1^n k^2 =binomn+13 + binomn+23$

1

1

$begingroup$

How can I prove

$$sum_r=0^n binomnr r p^r q^n-r = np,$$
given that $p+q=1$?

I think it is about applying
$$(1+x)^n=sum_r=0^n binomnr x^r.$$
Although I am quite unsure about the approach, I think it might give a way.

combinatorics binomial-theorem

share|cite|improve this question

edited Mar 22 at 9:20

Later

674

asked Mar 22 at 8:59

Tony1970Tony1970

205

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is equivalent to proving the expected value of a Binomial distribution. Discussion/proofs can be found here: math.stackexchange.com/questions/226237/….
  $endgroup$
  – Minus One-Twelfth
  Mar 22 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth I must have missed
  $endgroup$
  – Tony1970
  Mar 22 at 9:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is closely related to Sum $(1-x)^n$ $sum_r=1^n$ $r$ $nchoose r$ $(fracx1-x)^r$, if not equivalent.
  $endgroup$
  – robjohn♦
  Mar 22 at 10:34
  

  
  
  
  

add a comment | 

1

1

$begingroup$

How can I prove

$$sum_r=0^n binomnr r p^r q^n-r = np,$$
given that $p+q=1$?

I think it is about applying
$$(1+x)^n=sum_r=0^n binomnr x^r.$$
Although I am quite unsure about the approach, I think it might give a way.

combinatorics binomial-theorem

share|cite|improve this question

edited Mar 22 at 9:20

Later

674

asked Mar 22 at 8:59

Tony1970Tony1970

205

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is equivalent to proving the expected value of a Binomial distribution. Discussion/proofs can be found here: math.stackexchange.com/questions/226237/….
  $endgroup$
  – Minus One-Twelfth
  Mar 22 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth I must have missed
  $endgroup$
  – Tony1970
  Mar 22 at 9:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is closely related to Sum $(1-x)^n$ $sum_r=1^n$ $r$ $nchoose r$ $(fracx1-x)^r$, if not equivalent.
  $endgroup$
  – robjohn♦
  Mar 22 at 10:34
  

  
  
  
  

add a comment | 

$begingroup$

How can I prove

$$sum_r=0^n binomnr r p^r q^n-r = np,$$
given that $p+q=1$?

I think it is about applying
$$(1+x)^n=sum_r=0^n binomnr x^r.$$
Although I am quite unsure about the approach, I think it might give a way.

combinatorics binomial-theorem

share|cite|improve this question

edited Mar 22 at 9:20

Later

674

asked Mar 22 at 8:59

Tony1970Tony1970

205

$endgroup$

share|cite|improve this question

edited Mar 22 at 9:20

Later

674

asked Mar 22 at 8:59

Tony1970Tony1970

205

share|cite|improve this question

edited Mar 22 at 9:20

Later

674

asked Mar 22 at 8:59

Tony1970Tony1970

205

edited Mar 22 at 9:20

Later

674

edited Mar 22 at 9:20

Later

674

asked Mar 22 at 8:59

Tony1970Tony1970

205

asked Mar 22 at 8:59

Tony1970Tony1970

205

3 Answers
3

active

oldest

votes

4

$begingroup$

For $nge rge1,$ $$rbinom nr=nbinomn-1r-1$$

$$impliessum_r=0^n binomnr r p^r q^n-r =npsum_r=1^nbinomn-1r-1p^r-1q^n-1-(r-1)=np(p+q)^n-1$$

share|cite|improve this answer

edited Mar 22 at 9:09

answered Mar 22 at 9:02

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you meant $$frac rcolorrednbinom nr=binomn-1r-1$$
  $endgroup$
  – Darkrai
  Mar 22 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Darkrai, Thanks for the observation. The position of $=$ was messed
  $endgroup$
  – lab bhattacharjee
  Mar 22 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasLesgourgues, These days, I'm too absent minded
  $endgroup$
  – lab bhattacharjee
  Mar 22 at 9:09
  

  
  
  
  

add a comment | 

4

$begingroup$

Another approach uses calculus. Since $sum_rge 0binomnrx^r=(1+x)^n$, $$sum_rge 0binomnrrx^r=xfracddx(1+x)^n=nx(1+x)^n-1.$$Set $x=fracpq$ then multiply by $q^n$, giving $$sum_rbinomnrrp^rq^n-r=npq^n-1left(1+tfracpqright)^n-1=np(p+q)^n-1=np.$$

share|cite|improve this answer

answered Mar 22 at 9:07

J.G.J.G.

32.9k23250

$endgroup$

add a comment | 

1

$begingroup$

Here is a probabilistic approach:

• 
  $sum_r=0^n binomnr r p^r q^n-r = E[X]$, where


• 
  $X$ follows a binomial distribution $B(p,n)$, which means


• 
  $X = sum_i=1^nX_i$ with $X_i$ i.i.d. where $X_i in 0,1$ and $P(X_i=1) = p$
  


• $Rightarrow E[X_i] = 1cdot p + 0 cdot (1-p) = p$

Now, it follows
$$E[X] = E left[sum_i=1^nX_iright] = sum_i=1^n E[X_i] =sum_i=1^n p = np$$

share|cite|improve this answer

answered Mar 22 at 9:43

trancelocationtrancelocation

13.6k1829

$endgroup$

add a comment | 

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4

$begingroup$

For $nge rge1,$ $$rbinom nr=nbinomn-1r-1$$

$$impliessum_r=0^n binomnr r p^r q^n-r =npsum_r=1^nbinomn-1r-1p^r-1q^n-1-(r-1)=np(p+q)^n-1$$

share|cite|improve this answer

edited Mar 22 at 9:09

answered Mar 22 at 9:02

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you meant $$frac rcolorrednbinom nr=binomn-1r-1$$
  $endgroup$
  – Darkrai
  Mar 22 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Darkrai, Thanks for the observation. The position of $=$ was messed
  $endgroup$
  – lab bhattacharjee
  Mar 22 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasLesgourgues, These days, I'm too absent minded
  $endgroup$
  – lab bhattacharjee
  Mar 22 at 9:09
  

  
  
  
  

add a comment | 

4

$begingroup$

For $nge rge1,$ $$rbinom nr=nbinomn-1r-1$$

$$impliessum_r=0^n binomnr r p^r q^n-r =npsum_r=1^nbinomn-1r-1p^r-1q^n-1-(r-1)=np(p+q)^n-1$$

share|cite|improve this answer

edited Mar 22 at 9:09

answered Mar 22 at 9:02

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you meant $$frac rcolorrednbinom nr=binomn-1r-1$$
  $endgroup$
  – Darkrai
  Mar 22 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Darkrai, Thanks for the observation. The position of $=$ was messed
  $endgroup$
  – lab bhattacharjee
  Mar 22 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasLesgourgues, These days, I'm too absent minded
  $endgroup$
  – lab bhattacharjee
  Mar 22 at 9:09
  

  
  
  
  

add a comment | 

$begingroup$

For $nge rge1,$ $$rbinom nr=nbinomn-1r-1$$

$$impliessum_r=0^n binomnr r p^r q^n-r =npsum_r=1^nbinomn-1r-1p^r-1q^n-1-(r-1)=np(p+q)^n-1$$

share|cite|improve this answer

edited Mar 22 at 9:09

answered Mar 22 at 9:02

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

share|cite|improve this answer

edited Mar 22 at 9:09

answered Mar 22 at 9:02

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 22 at 9:02

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 22 at 9:02

lab bhattacharjeelab bhattacharjee

228k15158279

4

$begingroup$

Another approach uses calculus. Since $sum_rge 0binomnrx^r=(1+x)^n$, $$sum_rge 0binomnrrx^r=xfracddx(1+x)^n=nx(1+x)^n-1.$$Set $x=fracpq$ then multiply by $q^n$, giving $$sum_rbinomnrrp^rq^n-r=npq^n-1left(1+tfracpqright)^n-1=np(p+q)^n-1=np.$$

share|cite|improve this answer

answered Mar 22 at 9:07

J.G.J.G.

32.9k23250

$endgroup$

add a comment | 

4

$begingroup$

Another approach uses calculus. Since $sum_rge 0binomnrx^r=(1+x)^n$, $$sum_rge 0binomnrrx^r=xfracddx(1+x)^n=nx(1+x)^n-1.$$Set $x=fracpq$ then multiply by $q^n$, giving $$sum_rbinomnrrp^rq^n-r=npq^n-1left(1+tfracpqright)^n-1=np(p+q)^n-1=np.$$

share|cite|improve this answer

answered Mar 22 at 9:07

J.G.J.G.

32.9k23250

$endgroup$

add a comment | 

$begingroup$

Another approach uses calculus. Since $sum_rge 0binomnrx^r=(1+x)^n$, $$sum_rge 0binomnrrx^r=xfracddx(1+x)^n=nx(1+x)^n-1.$$Set $x=fracpq$ then multiply by $q^n$, giving $$sum_rbinomnrrp^rq^n-r=npq^n-1left(1+tfracpqright)^n-1=np(p+q)^n-1=np.$$

share|cite|improve this answer

answered Mar 22 at 9:07

J.G.J.G.

32.9k23250

$endgroup$

share|cite|improve this answer

answered Mar 22 at 9:07

J.G.J.G.

32.9k23250

answered Mar 22 at 9:07

J.G.J.G.

32.9k23250

answered Mar 22 at 9:07

J.G.J.G.

32.9k23250

1

$begingroup$

Here is a probabilistic approach:

• 
  $sum_r=0^n binomnr r p^r q^n-r = E[X]$, where


• 
  $X$ follows a binomial distribution $B(p,n)$, which means


• 
  $X = sum_i=1^nX_i$ with $X_i$ i.i.d. where $X_i in 0,1$ and $P(X_i=1) = p$
  


• $Rightarrow E[X_i] = 1cdot p + 0 cdot (1-p) = p$

Now, it follows
$$E[X] = E left[sum_i=1^nX_iright] = sum_i=1^n E[X_i] =sum_i=1^n p = np$$

share|cite|improve this answer

answered Mar 22 at 9:43

trancelocationtrancelocation

13.6k1829

$endgroup$

add a comment | 

1

$begingroup$

Here is a probabilistic approach:

• 
  $sum_r=0^n binomnr r p^r q^n-r = E[X]$, where


• 
  $X$ follows a binomial distribution $B(p,n)$, which means


• 
  $X = sum_i=1^nX_i$ with $X_i$ i.i.d. where $X_i in 0,1$ and $P(X_i=1) = p$
  


• $Rightarrow E[X_i] = 1cdot p + 0 cdot (1-p) = p$

Now, it follows
$$E[X] = E left[sum_i=1^nX_iright] = sum_i=1^n E[X_i] =sum_i=1^n p = np$$

share|cite|improve this answer

answered Mar 22 at 9:43

trancelocationtrancelocation

13.6k1829

$endgroup$

add a comment | 

$begingroup$

Here is a probabilistic approach:

• 
  $sum_r=0^n binomnr r p^r q^n-r = E[X]$, where


• 
  $X$ follows a binomial distribution $B(p,n)$, which means


• 
  $X = sum_i=1^nX_i$ with $X_i$ i.i.d. where $X_i in 0,1$ and $P(X_i=1) = p$
  


• $Rightarrow E[X_i] = 1cdot p + 0 cdot (1-p) = p$

Now, it follows
$$E[X] = E left[sum_i=1^nX_iright] = sum_i=1^n E[X_i] =sum_i=1^n p = np$$

share|cite|improve this answer

answered Mar 22 at 9:43

trancelocationtrancelocation

13.6k1829

$endgroup$

share|cite|improve this answer

answered Mar 22 at 9:43

trancelocationtrancelocation

13.6k1829

answered Mar 22 at 9:43

trancelocationtrancelocation

13.6k1829

answered Mar 22 at 9:43

trancelocationtrancelocation

13.6k1829