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Analysis of limit on summation of probability

1

$begingroup$

This question relates to proving a random variable N on $(mathbfOmega ,mathfrakF,mathbbP)$ which has image 0,1,2,... has expectation $sum_k=0^infty mathbbP(N> k)$ if the sum exists.

So far my proof is as follows, let $f$ and $F$ be the probability mass function and cumulative distribution function respectively.

The expectation of N is
$$beginalignsum_k=left 0,1,2,... right kmathbbP(N=k)
&=lim_nrightarrow inftysum_k=0^n kmathbbP(N=k)
\&=lim_nrightarrow inftysum_k=0^n kf(k)
\&=lim_nrightarrow inftysum_k=1^n k[F(k)-F(k-1)] text (k=0 term is 0)endalign$$
The sum simplifies to: $$beginalign
lim_nrightarrow inftysum_k=1^n (F(n)-F(k-1))
&=lim_nrightarrow inftysum_k=1^n mathbbP(k-1<N<n)
\&=lim_nrightarrow inftysum_k=0^n mathbbP(k<N<n)endalign$$

My question is as follows: in my final line, does the nature of the analysis of the sum allow me to take the limit within it so it becomes $sum_k=0^infty lim_nrightarrow inftymathbbP(k<N<n)$, which is indeed $sum_k=0^infty mathbbP(N> k)$ the required result? Or is there a more approproate alternative route that I can take?

probability expected-value

share|cite|improve this question

edited Apr 1 at 2:25

Carol Barnes

asked Apr 1 at 1:06

Carol BarnesCarol Barnes

62

$endgroup$

add a comment | 

1

$begingroup$

This question relates to proving a random variable N on $(mathbfOmega ,mathfrakF,mathbbP)$ which has image 0,1,2,... has expectation $sum_k=0^infty mathbbP(N> k)$ if the sum exists.

So far my proof is as follows, let $f$ and $F$ be the probability mass function and cumulative distribution function respectively.

The expectation of N is
$$beginalignsum_k=left 0,1,2,... right kmathbbP(N=k)
&=lim_nrightarrow inftysum_k=0^n kmathbbP(N=k)
\&=lim_nrightarrow inftysum_k=0^n kf(k)
\&=lim_nrightarrow inftysum_k=1^n k[F(k)-F(k-1)] text (k=0 term is 0)endalign$$
The sum simplifies to: $$beginalign
lim_nrightarrow inftysum_k=1^n (F(n)-F(k-1))
&=lim_nrightarrow inftysum_k=1^n mathbbP(k-1<N<n)
\&=lim_nrightarrow inftysum_k=0^n mathbbP(k<N<n)endalign$$

My question is as follows: in my final line, does the nature of the analysis of the sum allow me to take the limit within it so it becomes $sum_k=0^infty lim_nrightarrow inftymathbbP(k<N<n)$, which is indeed $sum_k=0^infty mathbbP(N> k)$ the required result? Or is there a more approproate alternative route that I can take?

probability expected-value

share|cite|improve this question

edited Apr 1 at 2:25

Carol Barnes

asked Apr 1 at 1:06

Carol BarnesCarol Barnes

62

$endgroup$

add a comment | 

$begingroup$

This question relates to proving a random variable N on $(mathbfOmega ,mathfrakF,mathbbP)$ which has image 0,1,2,... has expectation $sum_k=0^infty mathbbP(N> k)$ if the sum exists.

So far my proof is as follows, let $f$ and $F$ be the probability mass function and cumulative distribution function respectively.

The expectation of N is
$$beginalignsum_k=left 0,1,2,... right kmathbbP(N=k)
&=lim_nrightarrow inftysum_k=0^n kmathbbP(N=k)
\&=lim_nrightarrow inftysum_k=0^n kf(k)
\&=lim_nrightarrow inftysum_k=1^n k[F(k)-F(k-1)] text (k=0 term is 0)endalign$$
The sum simplifies to: $$beginalign
lim_nrightarrow inftysum_k=1^n (F(n)-F(k-1))
&=lim_nrightarrow inftysum_k=1^n mathbbP(k-1<N<n)
\&=lim_nrightarrow inftysum_k=0^n mathbbP(k<N<n)endalign$$

My question is as follows: in my final line, does the nature of the analysis of the sum allow me to take the limit within it so it becomes $sum_k=0^infty lim_nrightarrow inftymathbbP(k<N<n)$, which is indeed $sum_k=0^infty mathbbP(N> k)$ the required result? Or is there a more approproate alternative route that I can take?

probability expected-value

share|cite|improve this question

edited Apr 1 at 2:25

Carol Barnes

asked Apr 1 at 1:06

Carol BarnesCarol Barnes

62

$endgroup$

share|cite|improve this question

edited Apr 1 at 2:25

Carol Barnes

asked Apr 1 at 1:06

Carol BarnesCarol Barnes

62

share|cite|improve this question

edited Apr 1 at 2:25

Carol Barnes

asked Apr 1 at 1:06

Carol BarnesCarol Barnes

62

asked Apr 1 at 1:06

Carol BarnesCarol Barnes

62

asked Apr 1 at 1:06

Carol BarnesCarol Barnes

62

1 Answer
1

active

oldest

votes

0

$begingroup$

Yes, it is legitimate to take the limit inside. You can apply Monotone Convergence Theorem noting that $I_k leq nI_N<n$ is non-negative and increasing.

share|cite|improve this answer

answered Apr 1 at 5:59

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

add a comment | 

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0

$begingroup$

Yes, it is legitimate to take the limit inside. You can apply Monotone Convergence Theorem noting that $I_k leq nI_N<n$ is non-negative and increasing.

share|cite|improve this answer

answered Apr 1 at 5:59

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

add a comment | 

0

$begingroup$

Yes, it is legitimate to take the limit inside. You can apply Monotone Convergence Theorem noting that $I_k leq nI_N<n$ is non-negative and increasing.

share|cite|improve this answer

answered Apr 1 at 5:59

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

add a comment | 

$begingroup$

Yes, it is legitimate to take the limit inside. You can apply Monotone Convergence Theorem noting that $I_k leq nI_N<n$ is non-negative and increasing.

share|cite|improve this answer

answered Apr 1 at 5:59

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

share|cite|improve this answer

answered Apr 1 at 5:59

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

answered Apr 1 at 5:59

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

answered Apr 1 at 5:59

Kavi Rama MurthyKavi Rama Murthy

74.9k53270