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Harmonic series in probability mass function problem

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0

$begingroup$

Suppose $X$ is a discrete random variable with possible values $1, 2, 3,dots.$ Further, suppose the p.m.f is $$cleft(frac1x-frac1x+1right)enspacetexts.t. $c > 0$$$
Find c and $E[X].$

Idea:

We have $$1=sum_x=1^inftycleft(frac1x-frac1x+1right)=cleft(sum_x=1^inftyfrac1x-sum_x=1^inftyfrac1x+1right)$$
But since $sum_x=1^inftyfrac1x$ is a harmonic series, diverges. Thus, there is no value for $c$.

Since it diverges, $E[X]$ does not exist.

Questions:

Is it possible for c not to exist? Did I do a mistake?

Update:

$$1=sum_x=1^inftycleft(frac1x-frac1x+1right)=csum_x=1^inftyleft(frac1x-frac1x+1right)$$
by telescoping series we have
$$1=ccdot 1$$

So, our p.m.f is $$left(frac1x-frac1x+1right)$$
But, $$E[X]=1cdotfrac12+2cdotfrac16+3cdotfrac112+4cdot frac120+dots=frac12+frac13+frac14+frac15+dots$$
But that diverges. So, $E[X]$ doesn't exist.

probability sequences-and-series

share|cite|improve this question

edited Mar 4 at 6:36

Tomás Palamás

asked Mar 4 at 6:07

Tomás PalamásTomás Palamás

365211

$endgroup$

add a comment | 

0

$begingroup$

Suppose $X$ is a discrete random variable with possible values $1, 2, 3,dots.$ Further, suppose the p.m.f is $$cleft(frac1x-frac1x+1right)enspacetexts.t. $c > 0$$$
Find c and $E[X].$

Idea:

We have $$1=sum_x=1^inftycleft(frac1x-frac1x+1right)=cleft(sum_x=1^inftyfrac1x-sum_x=1^inftyfrac1x+1right)$$
But since $sum_x=1^inftyfrac1x$ is a harmonic series, diverges. Thus, there is no value for $c$.

Since it diverges, $E[X]$ does not exist.

Questions:

Is it possible for c not to exist? Did I do a mistake?

Update:

$$1=sum_x=1^inftycleft(frac1x-frac1x+1right)=csum_x=1^inftyleft(frac1x-frac1x+1right)$$
by telescoping series we have
$$1=ccdot 1$$

So, our p.m.f is $$left(frac1x-frac1x+1right)$$
But, $$E[X]=1cdotfrac12+2cdotfrac16+3cdotfrac112+4cdot frac120+dots=frac12+frac13+frac14+frac15+dots$$
But that diverges. So, $E[X]$ doesn't exist.

probability sequences-and-series

share|cite|improve this question

edited Mar 4 at 6:36

Tomás Palamás

asked Mar 4 at 6:07

Tomás PalamásTomás Palamás

365211

$endgroup$

add a comment | 

$begingroup$

Suppose $X$ is a discrete random variable with possible values $1, 2, 3,dots.$ Further, suppose the p.m.f is $$cleft(frac1x-frac1x+1right)enspacetexts.t. $c > 0$$$
Find c and $E[X].$

Idea:

We have $$1=sum_x=1^inftycleft(frac1x-frac1x+1right)=cleft(sum_x=1^inftyfrac1x-sum_x=1^inftyfrac1x+1right)$$
But since $sum_x=1^inftyfrac1x$ is a harmonic series, diverges. Thus, there is no value for $c$.

Since it diverges, $E[X]$ does not exist.

Questions:

Is it possible for c not to exist? Did I do a mistake?

Update:

$$1=sum_x=1^inftycleft(frac1x-frac1x+1right)=csum_x=1^inftyleft(frac1x-frac1x+1right)$$
by telescoping series we have
$$1=ccdot 1$$

So, our p.m.f is $$left(frac1x-frac1x+1right)$$
But, $$E[X]=1cdotfrac12+2cdotfrac16+3cdotfrac112+4cdot frac120+dots=frac12+frac13+frac14+frac15+dots$$
But that diverges. So, $E[X]$ doesn't exist.

probability sequences-and-series

share|cite|improve this question

edited Mar 4 at 6:36

Tomás Palamás

asked Mar 4 at 6:07

Tomás PalamásTomás Palamás

365211

$endgroup$

share|cite|improve this question

edited Mar 4 at 6:36

Tomás Palamás

asked Mar 4 at 6:07

Tomás PalamásTomás Palamás

365211

share|cite|improve this question

edited Mar 4 at 6:36

Tomás Palamás

asked Mar 4 at 6:07

Tomás PalamásTomás Palamás

365211

asked Mar 4 at 6:07

Tomás PalamásTomás Palamás

365211

asked Mar 4 at 6:07

Tomás PalamásTomás Palamás

365211

2 Answers
2

active

oldest

votes

0

$begingroup$

Hint: Evaluate the series up to a finite $N$ first, then take the limit as $Nto infty$. The series up to a finite $N$ will be a telescoping series, i.e. most of the terms will cancel, making it easy to evaluate.

share|cite|improve this answer

answered Mar 4 at 6:10

Minus One-TwelfthMinus One-Twelfth

2,823413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see it, I forgot about that series! I will post the solution in a bit but I am seeing that $c=1$?
  $endgroup$
  – Tomás Palamás
  Mar 4 at 6:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $sumlimits_k=1^Nleft(frac1k - frac1k+1right) = 1 - frac1N+1to 1$, you are right. :)
  $endgroup$
  – Minus One-Twelfth
  Mar 4 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I updated the OP. Is it possible for the expected value not to exist?
  $endgroup$
  – Tomás Palamás
  Mar 4 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is possible.
  $endgroup$
  – Minus One-Twelfth
  Mar 4 at 7:46
  
  

  
  
  
  

add a comment | 

0

$begingroup$

You can write $$sum a_n-b_n=sum a_n-sum b_n$$only if at least one of $sum a_n$ or $sum b_n$ is bounded. In this case$$sum_x=1^infty1over x-1over x+1=left(1-1over 2right)\+left(1over 2-1over 3right)\+left(1over 3-1over 4right)\+left(1over 4-1over 5right)\+cdots\=1$$therefore $$c=1$$ and the rest is easy.

share|cite|improve this answer

answered Mar 16 at 21:50

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

add a comment | 

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0

$begingroup$

Hint: Evaluate the series up to a finite $N$ first, then take the limit as $Nto infty$. The series up to a finite $N$ will be a telescoping series, i.e. most of the terms will cancel, making it easy to evaluate.

share|cite|improve this answer

answered Mar 4 at 6:10

Minus One-TwelfthMinus One-Twelfth

2,823413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see it, I forgot about that series! I will post the solution in a bit but I am seeing that $c=1$?
  $endgroup$
  – Tomás Palamás
  Mar 4 at 6:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $sumlimits_k=1^Nleft(frac1k - frac1k+1right) = 1 - frac1N+1to 1$, you are right. :)
  $endgroup$
  – Minus One-Twelfth
  Mar 4 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I updated the OP. Is it possible for the expected value not to exist?
  $endgroup$
  – Tomás Palamás
  Mar 4 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is possible.
  $endgroup$
  – Minus One-Twelfth
  Mar 4 at 7:46
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Evaluate the series up to a finite $N$ first, then take the limit as $Nto infty$. The series up to a finite $N$ will be a telescoping series, i.e. most of the terms will cancel, making it easy to evaluate.

share|cite|improve this answer

answered Mar 4 at 6:10

Minus One-TwelfthMinus One-Twelfth

2,823413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see it, I forgot about that series! I will post the solution in a bit but I am seeing that $c=1$?
  $endgroup$
  – Tomás Palamás
  Mar 4 at 6:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $sumlimits_k=1^Nleft(frac1k - frac1k+1right) = 1 - frac1N+1to 1$, you are right. :)
  $endgroup$
  – Minus One-Twelfth
  Mar 4 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I updated the OP. Is it possible for the expected value not to exist?
  $endgroup$
  – Tomás Palamás
  Mar 4 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is possible.
  $endgroup$
  – Minus One-Twelfth
  Mar 4 at 7:46
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Evaluate the series up to a finite $N$ first, then take the limit as $Nto infty$. The series up to a finite $N$ will be a telescoping series, i.e. most of the terms will cancel, making it easy to evaluate.

share|cite|improve this answer

answered Mar 4 at 6:10

Minus One-TwelfthMinus One-Twelfth

2,823413

$endgroup$

share|cite|improve this answer

answered Mar 4 at 6:10

Minus One-TwelfthMinus One-Twelfth

2,823413

answered Mar 4 at 6:10

Minus One-TwelfthMinus One-Twelfth

2,823413

answered Mar 4 at 6:10

Minus One-TwelfthMinus One-Twelfth

2,823413

0

$begingroup$

You can write $$sum a_n-b_n=sum a_n-sum b_n$$only if at least one of $sum a_n$ or $sum b_n$ is bounded. In this case$$sum_x=1^infty1over x-1over x+1=left(1-1over 2right)\+left(1over 2-1over 3right)\+left(1over 3-1over 4right)\+left(1over 4-1over 5right)\+cdots\=1$$therefore $$c=1$$ and the rest is easy.

share|cite|improve this answer

answered Mar 16 at 21:50

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

add a comment | 

0

$begingroup$

You can write $$sum a_n-b_n=sum a_n-sum b_n$$only if at least one of $sum a_n$ or $sum b_n$ is bounded. In this case$$sum_x=1^infty1over x-1over x+1=left(1-1over 2right)\+left(1over 2-1over 3right)\+left(1over 3-1over 4right)\+left(1over 4-1over 5right)\+cdots\=1$$therefore $$c=1$$ and the rest is easy.

share|cite|improve this answer

answered Mar 16 at 21:50

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

add a comment | 

$begingroup$

You can write $$sum a_n-b_n=sum a_n-sum b_n$$only if at least one of $sum a_n$ or $sum b_n$ is bounded. In this case$$sum_x=1^infty1over x-1over x+1=left(1-1over 2right)\+left(1over 2-1over 3right)\+left(1over 3-1over 4right)\+left(1over 4-1over 5right)\+cdots\=1$$therefore $$c=1$$ and the rest is easy.

share|cite|improve this answer

answered Mar 16 at 21:50

Mostafa AyazMostafa Ayaz

17.6k31039

$endgroup$

share|cite|improve this answer

answered Mar 16 at 21:50

Mostafa AyazMostafa Ayaz

17.6k31039

answered Mar 16 at 21:50

Mostafa AyazMostafa Ayaz

17.6k31039

answered Mar 16 at 21:50

Mostafa AyazMostafa Ayaz

17.6k31039