Mixing
Order of Summation in proof, Grimmett and Stirzaker 3.11.13.a
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
The question asks to prove:
$E(X) = sum_n=0^inftyP(X>n)$
$E(X) = sum_m=0^inftymP(X=m) = sum_m=0^inftysum_n=0^m-1P(X=m) = sum_n=0^inftysum_m=n+1^inftyP(X=m) = sum_n=0^inftyP(X>n)$
My only problem is understanding how the new limits when the summations are changed going from the 3rd to the 4th expression.
For example $P(x=0)$ is included in $sum_m=0^inftysum_n=0^m-1P(X=m)$ but it doesn't seem to be included in $sum_n=0^inftysum_m=n+1^inftyP(X=m)$?
Can someone explain this?
probability
asked Jul 16 at 19:19
Bazman
365211
add a comment |Â
up vote
1
down vote
favorite
The question asks to prove:
$E(X) = sum_n=0^inftyP(X>n)$
$E(X) = sum_m=0^inftymP(X=m) = sum_m=0^inftysum_n=0^m-1P(X=m) = sum_n=0^inftysum_m=n+1^inftyP(X=m) = sum_n=0^inftyP(X>n)$
My only problem is understanding how the new limits when the summations are changed going from the 3rd to the 4th expression.
For example $P(x=0)$ is included in $sum_m=0^inftysum_n=0^m-1P(X=m)$ but it doesn't seem to be included in $sum_n=0^inftysum_m=n+1^inftyP(X=m)$?
Can someone explain this?
probability
asked Jul 16 at 19:19
Bazman
365211
I don’t think $P(X=0)$ is included in the third expression.
– Theoretical Economist
Jul 16 at 19:24
1
In the third equality, they have used $m=sum_n=0^m-11$. Clearly, we have $0le nle m-1$ which also means $mge n+1, nge0$. Loosely speaking, the sums can be interchanged as long as they are convergent.
– StubbornAtom
Jul 16 at 19:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The question asks to prove:
$E(X) = sum_n=0^inftyP(X>n)$
$E(X) = sum_m=0^inftymP(X=m) = sum_m=0^inftysum_n=0^m-1P(X=m) = sum_n=0^inftysum_m=n+1^inftyP(X=m) = sum_n=0^inftyP(X>n)$
My only problem is understanding how the new limits when the summations are changed going from the 3rd to the 4th expression.
For example $P(x=0)$ is included in $sum_m=0^inftysum_n=0^m-1P(X=m)$ but it doesn't seem to be included in $sum_n=0^inftysum_m=n+1^inftyP(X=m)$?
Can someone explain this?
probability
asked Jul 16 at 19:19
Bazman
365211
The question asks to prove:
$E(X) = sum_n=0^inftyP(X>n)$
$E(X) = sum_m=0^inftymP(X=m) = sum_m=0^inftysum_n=0^m-1P(X=m) = sum_n=0^inftysum_m=n+1^inftyP(X=m) = sum_n=0^inftyP(X>n)$
My only problem is understanding how the new limits when the summations are changed going from the 3rd to the 4th expression.
For example $P(x=0)$ is included in $sum_m=0^inftysum_n=0^m-1P(X=m)$ but it doesn't seem to be included in $sum_n=0^inftysum_m=n+1^inftyP(X=m)$?
Can someone explain this?
probability
asked Jul 16 at 19:19
Bazman
365211
asked Jul 16 at 19:19
Bazman
365211
asked Jul 16 at 19:19
Bazman
365211
asked Jul 16 at 19:19
Bazman
365211
365211
I don’t think $P(X=0)$ is included in the third expression.
– Theoretical Economist
Jul 16 at 19:24
1
In the third equality, they have used $m=sum_n=0^m-11$. Clearly, we have $0le nle m-1$ which also means $mge n+1, nge0$. Loosely speaking, the sums can be interchanged as long as they are convergent.
– StubbornAtom
Jul 16 at 19:30
add a comment |Â
I don’t think $P(X=0)$ is included in the third expression.
– Theoretical Economist
Jul 16 at 19:24
1
In the third equality, they have used $m=sum_n=0^m-11$. Clearly, we have $0le nle m-1$ which also means $mge n+1, nge0$. Loosely speaking, the sums can be interchanged as long as they are convergent.
– StubbornAtom
Jul 16 at 19:30
I don’t think $P(X=0)$ is included in the third expression.
– Theoretical Economist
Jul 16 at 19:24
I don’t think $P(X=0)$ is included in the third expression.
– Theoretical Economist
Jul 16 at 19:24
1
1
In the third equality, they have used $m=sum_n=0^m-11$. Clearly, we have $0le nle m-1$ which also means $mge n+1, nge0$. Loosely speaking, the sums can be interchanged as long as they are convergent.
– StubbornAtom
Jul 16 at 19:30
In the third equality, they have used $m=sum_n=0^m-11$. Clearly, we have $0le nle m-1$ which also means $mge n+1, nge0$. Loosely speaking, the sums can be interchanged as long as they are convergent.
– StubbornAtom
Jul 16 at 19:30
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Use the indicator
$$mathbf1_(n leqslant m-1) = mathbf1_(m geqslant n+1) = begincases1, & n leqslant m-1\0, & n > m-1 endcases$$
Thus,
$$sum_m=0^infty sum_n=0^m-1 P(X = m) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(n leqslant m-1) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(m geqslant n+1)\ = sum_n=0^infty sum_m=0^infty P(X = m) mathbf1_(m geqslant n+1) \ = sum_n=0^infty sum_m=n+1^infty P(X = m) $$
We can interchange the sums in the second line by Tonelli's theorem since the terms are non-negative.
answered Jul 16 at 20:12
RRL
43.7k42260
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Use the indicator
$$mathbf1_(n leqslant m-1) = mathbf1_(m geqslant n+1) = begincases1, & n leqslant m-1\0, & n > m-1 endcases$$
Thus,
$$sum_m=0^infty sum_n=0^m-1 P(X = m) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(n leqslant m-1) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(m geqslant n+1)\ = sum_n=0^infty sum_m=0^infty P(X = m) mathbf1_(m geqslant n+1) \ = sum_n=0^infty sum_m=n+1^infty P(X = m) $$
We can interchange the sums in the second line by Tonelli's theorem since the terms are non-negative.
answered Jul 16 at 20:12
RRL
43.7k42260
add a comment |Â
up vote
1
down vote
accepted
Use the indicator
$$mathbf1_(n leqslant m-1) = mathbf1_(m geqslant n+1) = begincases1, & n leqslant m-1\0, & n > m-1 endcases$$
Thus,
$$sum_m=0^infty sum_n=0^m-1 P(X = m) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(n leqslant m-1) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(m geqslant n+1)\ = sum_n=0^infty sum_m=0^infty P(X = m) mathbf1_(m geqslant n+1) \ = sum_n=0^infty sum_m=n+1^infty P(X = m) $$
We can interchange the sums in the second line by Tonelli's theorem since the terms are non-negative.
answered Jul 16 at 20:12
RRL
43.7k42260
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Use the indicator
$$mathbf1_(n leqslant m-1) = mathbf1_(m geqslant n+1) = begincases1, & n leqslant m-1\0, & n > m-1 endcases$$
Thus,
$$sum_m=0^infty sum_n=0^m-1 P(X = m) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(n leqslant m-1) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(m geqslant n+1)\ = sum_n=0^infty sum_m=0^infty P(X = m) mathbf1_(m geqslant n+1) \ = sum_n=0^infty sum_m=n+1^infty P(X = m) $$
We can interchange the sums in the second line by Tonelli's theorem since the terms are non-negative.
answered Jul 16 at 20:12
RRL
43.7k42260
Use the indicator
$$mathbf1_(n leqslant m-1) = mathbf1_(m geqslant n+1) = begincases1, & n leqslant m-1\0, & n > m-1 endcases$$
Thus,
$$sum_m=0^infty sum_n=0^m-1 P(X = m) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(n leqslant m-1) = sum_m=0^infty sum_n=0^infty P(X = m) mathbf1_(m geqslant n+1)\ = sum_n=0^infty sum_m=0^infty P(X = m) mathbf1_(m geqslant n+1) \ = sum_n=0^infty sum_m=n+1^infty P(X = m) $$
We can interchange the sums in the second line by Tonelli's theorem since the terms are non-negative.
answered Jul 16 at 20:12
RRL
43.7k42260
answered Jul 16 at 20:12
RRL
43.7k42260
answered Jul 16 at 20:12
RRL
43.7k42260
answered Jul 16 at 20:12
RRL
43.7k42260
43.7k42260
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853750%2forder-of-summation-in-proof-grimmett-and-stirzaker-3-11-13-a%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I don’t think $P(X=0)$ is included in the third expression.
– Theoretical Economist
Jul 16 at 19:24
1
In the third equality, they have used $m=sum_n=0^m-11$. Clearly, we have $0le nle m-1$ which also means $mge n+1, nge0$. Loosely speaking, the sums can be interchanged as long as they are convergent.
– StubbornAtom
Jul 16 at 19:30