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How to derive this relation that I found intuitively?

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1

$begingroup$

By intuition, I found that the result of evaluating the following expression

$$ frac1M fracsum_N=0^M fracM!(M-N)!N! N e^cNsum_N=0^M fracM!(M-N)!N! e^cN $$

does not depend on the positive value of the integer $M$, i.e. it only depends on $cinmathbb R$.

I corroborated this with the help of a simple Python script. How to show analytically that this is true?

probability problem-solving

share|cite|improve this question

edited 1 hour ago

Shalop

9,63811130

asked 2 hours ago

user1420303user1420303

1086

$endgroup$

add a comment | 

1

$begingroup$

By intuition, I found that the result of evaluating the following expression

$$ frac1M fracsum_N=0^M fracM!(M-N)!N! N e^cNsum_N=0^M fracM!(M-N)!N! e^cN $$

does not depend on the positive value of the integer $M$, i.e. it only depends on $cinmathbb R$.

I corroborated this with the help of a simple Python script. How to show analytically that this is true?

probability problem-solving

share|cite|improve this question

edited 1 hour ago

Shalop

9,63811130

asked 2 hours ago

user1420303user1420303

1086

$endgroup$

add a comment | 

$begingroup$

By intuition, I found that the result of evaluating the following expression

$$ frac1M fracsum_N=0^M fracM!(M-N)!N! N e^cNsum_N=0^M fracM!(M-N)!N! e^cN $$

does not depend on the positive value of the integer $M$, i.e. it only depends on $cinmathbb R$.

I corroborated this with the help of a simple Python script. How to show analytically that this is true?

probability problem-solving

share|cite|improve this question

edited 1 hour ago

Shalop

9,63811130

asked 2 hours ago

user1420303user1420303

1086

$endgroup$

share|cite|improve this question

edited 1 hour ago

Shalop

9,63811130

asked 2 hours ago

user1420303user1420303

1086

share|cite|improve this question

edited 1 hour ago

Shalop

9,63811130

asked 2 hours ago

user1420303user1420303

1086

edited 1 hour ago

Shalop

9,63811130

edited 1 hour ago

Shalop

9,63811130

asked 2 hours ago

user1420303user1420303

1086

asked 2 hours ago

user1420303user1420303

1086

3 Answers
3

active

oldest

votes

4

$begingroup$

By the Binomial Theorem, the denominator is $(e^c+1)^M$. The numerator can also be manipulated to apply the Binomial Theorem:
beginalign*
sum_N=0^M fracM!(M-N)!N! Ne^cN &= sum_N=1^M fracMcdot (M-1)!((M-1)-(N-1))!(N-1)!e^c(N-1)cdot e^c\
&=Me^c sum_N=1^M binomM-1N-1e^c(N-1)\
&=Me^c sum_K=0^M-1 binomM-1Ke^cK\
&=Me^c (e^c+1)^M-1.
endalign*

So the whole expression simplifies to:
$$frace^ce^c+1.$$

share|cite|improve this answer

edited 2 hours ago

Vincent

3,33311231

answered 2 hours ago

kccukccu

13.1k11331

$endgroup$

add a comment | 

2

$begingroup$

By the binomial theorem we have
begineqnarray*
sum_N=0^M binomMN x^N =(1+x)^M.
endeqnarray*
Differentiate this and we have
begineqnarray*
fracddx sum_N=0^M binomMN x^N =sum_N=0^M binomMNN x^N-1 =M (1+x)^M-1.
endeqnarray*
So your expression becomes
begineqnarray*
frac1M frac sum_N=0^M N binomMN x^N sum_N=0^M binomMN x^N = fracx(1+x)
endeqnarray*
and all the $M$ dependency has cancelled.

share|cite|improve this answer

answered 2 hours ago

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

add a comment | 

0

$begingroup$

Let $phi(c) = Bbb E[e^cZ]$ denote the moment generating function of a standard bernoilli(1/2) random variable $Z$ taking values in $0,1$. Since the sum of $M$ independent copies gives the binomial distribution, the quantity in question is simply $$frac1M fracpartial_c (phi(c)^M)phi(c)^M = frac1M fracMphi(c)^M-1phi'(c)phi(c)^M = fracphi'(c)phi(c),$$ which only depends on $c$.

Note: I (perhaps unnecessarily) added the probability tag to this question.

share|cite|improve this answer

answered 1 hour ago

ShalopShalop

9,63811130

$endgroup$

add a comment | 

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4

$begingroup$

By the Binomial Theorem, the denominator is $(e^c+1)^M$. The numerator can also be manipulated to apply the Binomial Theorem:
beginalign*
sum_N=0^M fracM!(M-N)!N! Ne^cN &= sum_N=1^M fracMcdot (M-1)!((M-1)-(N-1))!(N-1)!e^c(N-1)cdot e^c\
&=Me^c sum_N=1^M binomM-1N-1e^c(N-1)\
&=Me^c sum_K=0^M-1 binomM-1Ke^cK\
&=Me^c (e^c+1)^M-1.
endalign*

So the whole expression simplifies to:
$$frace^ce^c+1.$$

share|cite|improve this answer

edited 2 hours ago

Vincent

3,33311231

answered 2 hours ago

kccukccu

13.1k11331

$endgroup$

add a comment | 

4

$begingroup$

By the Binomial Theorem, the denominator is $(e^c+1)^M$. The numerator can also be manipulated to apply the Binomial Theorem:
beginalign*
sum_N=0^M fracM!(M-N)!N! Ne^cN &= sum_N=1^M fracMcdot (M-1)!((M-1)-(N-1))!(N-1)!e^c(N-1)cdot e^c\
&=Me^c sum_N=1^M binomM-1N-1e^c(N-1)\
&=Me^c sum_K=0^M-1 binomM-1Ke^cK\
&=Me^c (e^c+1)^M-1.
endalign*

So the whole expression simplifies to:
$$frace^ce^c+1.$$

share|cite|improve this answer

edited 2 hours ago

Vincent

3,33311231

answered 2 hours ago

kccukccu

13.1k11331

$endgroup$

add a comment | 

$begingroup$

By the Binomial Theorem, the denominator is $(e^c+1)^M$. The numerator can also be manipulated to apply the Binomial Theorem:
beginalign*
sum_N=0^M fracM!(M-N)!N! Ne^cN &= sum_N=1^M fracMcdot (M-1)!((M-1)-(N-1))!(N-1)!e^c(N-1)cdot e^c\
&=Me^c sum_N=1^M binomM-1N-1e^c(N-1)\
&=Me^c sum_K=0^M-1 binomM-1Ke^cK\
&=Me^c (e^c+1)^M-1.
endalign*

So the whole expression simplifies to:
$$frace^ce^c+1.$$

share|cite|improve this answer

edited 2 hours ago

Vincent

3,33311231

answered 2 hours ago

kccukccu

13.1k11331

$endgroup$

share|cite|improve this answer

edited 2 hours ago

Vincent

3,33311231

answered 2 hours ago

kccukccu

13.1k11331

edited 2 hours ago

Vincent

3,33311231

edited 2 hours ago

Vincent

3,33311231

answered 2 hours ago

kccukccu

13.1k11331

answered 2 hours ago

kccukccu

13.1k11331

2

$begingroup$

By the binomial theorem we have
begineqnarray*
sum_N=0^M binomMN x^N =(1+x)^M.
endeqnarray*
Differentiate this and we have
begineqnarray*
fracddx sum_N=0^M binomMN x^N =sum_N=0^M binomMNN x^N-1 =M (1+x)^M-1.
endeqnarray*
So your expression becomes
begineqnarray*
frac1M frac sum_N=0^M N binomMN x^N sum_N=0^M binomMN x^N = fracx(1+x)
endeqnarray*
and all the $M$ dependency has cancelled.

share|cite|improve this answer

answered 2 hours ago

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

add a comment | 

2

$begingroup$

By the binomial theorem we have
begineqnarray*
sum_N=0^M binomMN x^N =(1+x)^M.
endeqnarray*
Differentiate this and we have
begineqnarray*
fracddx sum_N=0^M binomMN x^N =sum_N=0^M binomMNN x^N-1 =M (1+x)^M-1.
endeqnarray*
So your expression becomes
begineqnarray*
frac1M frac sum_N=0^M N binomMN x^N sum_N=0^M binomMN x^N = fracx(1+x)
endeqnarray*
and all the $M$ dependency has cancelled.

share|cite|improve this answer

answered 2 hours ago

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

add a comment | 

$begingroup$

By the binomial theorem we have
begineqnarray*
sum_N=0^M binomMN x^N =(1+x)^M.
endeqnarray*
Differentiate this and we have
begineqnarray*
fracddx sum_N=0^M binomMN x^N =sum_N=0^M binomMNN x^N-1 =M (1+x)^M-1.
endeqnarray*
So your expression becomes
begineqnarray*
frac1M frac sum_N=0^M N binomMN x^N sum_N=0^M binomMN x^N = fracx(1+x)
endeqnarray*
and all the $M$ dependency has cancelled.

share|cite|improve this answer

answered 2 hours ago

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Donald SplutterwitDonald Splutterwit

23.5k21448

answered 2 hours ago

Donald SplutterwitDonald Splutterwit

23.5k21448

answered 2 hours ago

Donald SplutterwitDonald Splutterwit

23.5k21448

0

$begingroup$

Let $phi(c) = Bbb E[e^cZ]$ denote the moment generating function of a standard bernoilli(1/2) random variable $Z$ taking values in $0,1$. Since the sum of $M$ independent copies gives the binomial distribution, the quantity in question is simply $$frac1M fracpartial_c (phi(c)^M)phi(c)^M = frac1M fracMphi(c)^M-1phi'(c)phi(c)^M = fracphi'(c)phi(c),$$ which only depends on $c$.

Note: I (perhaps unnecessarily) added the probability tag to this question.

share|cite|improve this answer

answered 1 hour ago

ShalopShalop

9,63811130

$endgroup$

add a comment | 

0

$begingroup$

Let $phi(c) = Bbb E[e^cZ]$ denote the moment generating function of a standard bernoilli(1/2) random variable $Z$ taking values in $0,1$. Since the sum of $M$ independent copies gives the binomial distribution, the quantity in question is simply $$frac1M fracpartial_c (phi(c)^M)phi(c)^M = frac1M fracMphi(c)^M-1phi'(c)phi(c)^M = fracphi'(c)phi(c),$$ which only depends on $c$.

Note: I (perhaps unnecessarily) added the probability tag to this question.

share|cite|improve this answer

answered 1 hour ago

ShalopShalop

9,63811130

$endgroup$

add a comment | 

$begingroup$

Let $phi(c) = Bbb E[e^cZ]$ denote the moment generating function of a standard bernoilli(1/2) random variable $Z$ taking values in $0,1$. Since the sum of $M$ independent copies gives the binomial distribution, the quantity in question is simply $$frac1M fracpartial_c (phi(c)^M)phi(c)^M = frac1M fracMphi(c)^M-1phi'(c)phi(c)^M = fracphi'(c)phi(c),$$ which only depends on $c$.

Note: I (perhaps unnecessarily) added the probability tag to this question.

share|cite|improve this answer

answered 1 hour ago

ShalopShalop

9,63811130

$endgroup$

share|cite|improve this answer

answered 1 hour ago

ShalopShalop

9,63811130

answered 1 hour ago

ShalopShalop

9,63811130

answered 1 hour ago

ShalopShalop

9,63811130