Finitely Additive and Countably Additive Property of Probability Function, $mathbb{P}$.












2












$begingroup$


In Grimmett and Stirzaker's Probability and Random Processes (section 1.3), for two disjoint events $A$ and $B$, we have that



$mathbb{P} (A cup B) = mathbb{P}(A) + mathbb{P}(B)$



From this statement, the authors 'jump' and state that $mathbb{P}$ should be finitely additive, and further along in the text, they 'jump' again and state that $mathbb{P}$ should be countably additive.



My questions:




  1. Why is $mathbb{P}$ only finitely additive? Isn't it possible to keep adding disjoint events ad infinitum : $mathbb{P} (A_1 cup A_2 cup ldots) = mathbb{P}(A_1) + mathbb{P}(A_2) + ldots$

  2. What is the difference between finitely additive and countably additive? I know that finitely additive just means I have a fixed number of events I need to add up but I am not sure of what countably additive means.


I have browsed the (many) other posts around this topic but they start discussing measure theory which I haven't studied yet in my course (I'm at undergrad level).






probability-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Finitely additive means that if you have any finite set of events $A_1,ldots,A_n$ which are pairwise independent (i.e. $A_icap A_j=varnothing$ for $ineq j$) then $mathbb(A_1cupldotscup A_n)=mathbb P(A_1)+ldots+mathbb P(A_n)$. Countable additivity means that we can do that with any sequence of events (which are pairwise disjoint), $langle A_nmid ninBbb Nrangle$, and not just finite sequences.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:38












  • $begingroup$
    So, $mathbb{P}$ being countably additive answers my first question because, as you state, that property is not just for finite sequences.
    $endgroup$
    – Mika'il
    Apr 20 '14 at 6:47












  • $begingroup$
    Yes, it answers part of your questions. Except the important(?) part, why is it just finitely additive. I don't know about that.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:51






  • 1




    $begingroup$
    Looking at your comments here it seems that the focus of interest is finite probability spaces. In that case, there's no harm in requiring only finite additivity, since there are not infinitely many distinct (let alone pairwise disjoint) events.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 7:39
















2












$begingroup$


In Grimmett and Stirzaker's Probability and Random Processes (section 1.3), for two disjoint events $A$ and $B$, we have that



$mathbb{P} (A cup B) = mathbb{P}(A) + mathbb{P}(B)$



From this statement, the authors 'jump' and state that $mathbb{P}$ should be finitely additive, and further along in the text, they 'jump' again and state that $mathbb{P}$ should be countably additive.



My questions:




  1. Why is $mathbb{P}$ only finitely additive? Isn't it possible to keep adding disjoint events ad infinitum : $mathbb{P} (A_1 cup A_2 cup ldots) = mathbb{P}(A_1) + mathbb{P}(A_2) + ldots$

  2. What is the difference between finitely additive and countably additive? I know that finitely additive just means I have a fixed number of events I need to add up but I am not sure of what countably additive means.


I have browsed the (many) other posts around this topic but they start discussing measure theory which I haven't studied yet in my course (I'm at undergrad level).






probability-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Finitely additive means that if you have any finite set of events $A_1,ldots,A_n$ which are pairwise independent (i.e. $A_icap A_j=varnothing$ for $ineq j$) then $mathbb(A_1cupldotscup A_n)=mathbb P(A_1)+ldots+mathbb P(A_n)$. Countable additivity means that we can do that with any sequence of events (which are pairwise disjoint), $langle A_nmid ninBbb Nrangle$, and not just finite sequences.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:38












  • $begingroup$
    So, $mathbb{P}$ being countably additive answers my first question because, as you state, that property is not just for finite sequences.
    $endgroup$
    – Mika'il
    Apr 20 '14 at 6:47












  • $begingroup$
    Yes, it answers part of your questions. Except the important(?) part, why is it just finitely additive. I don't know about that.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:51






  • 1




    $begingroup$
    Looking at your comments here it seems that the focus of interest is finite probability spaces. In that case, there's no harm in requiring only finite additivity, since there are not infinitely many distinct (let alone pairwise disjoint) events.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 7:39














2












2








2


1



$begingroup$


In Grimmett and Stirzaker's Probability and Random Processes (section 1.3), for two disjoint events $A$ and $B$, we have that



$mathbb{P} (A cup B) = mathbb{P}(A) + mathbb{P}(B)$



From this statement, the authors 'jump' and state that $mathbb{P}$ should be finitely additive, and further along in the text, they 'jump' again and state that $mathbb{P}$ should be countably additive.



My questions:




  1. Why is $mathbb{P}$ only finitely additive? Isn't it possible to keep adding disjoint events ad infinitum : $mathbb{P} (A_1 cup A_2 cup ldots) = mathbb{P}(A_1) + mathbb{P}(A_2) + ldots$

  2. What is the difference between finitely additive and countably additive? I know that finitely additive just means I have a fixed number of events I need to add up but I am not sure of what countably additive means.


I have browsed the (many) other posts around this topic but they start discussing measure theory which I haven't studied yet in my course (I'm at undergrad level).






probability-theory







share|cite|improve this question











$endgroup$




In Grimmett and Stirzaker's Probability and Random Processes (section 1.3), for two disjoint events $A$ and $B$, we have that



$mathbb{P} (A cup B) = mathbb{P}(A) + mathbb{P}(B)$



From this statement, the authors 'jump' and state that $mathbb{P}$ should be finitely additive, and further along in the text, they 'jump' again and state that $mathbb{P}$ should be countably additive.



My questions:




  1. Why is $mathbb{P}$ only finitely additive? Isn't it possible to keep adding disjoint events ad infinitum : $mathbb{P} (A_1 cup A_2 cup ldots) = mathbb{P}(A_1) + mathbb{P}(A_2) + ldots$

  2. What is the difference between finitely additive and countably additive? I know that finitely additive just means I have a fixed number of events I need to add up but I am not sure of what countably additive means.


I have browsed the (many) other posts around this topic but they start discussing measure theory which I haven't studied yet in my course (I'm at undergrad level).






probability-theory




probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 20 '14 at 6:37









Asaf Karagila♦

305k33435766




305k33435766










asked Apr 20 '14 at 6:33









Mika'ilMika'il

569424




569424












  • $begingroup$
    Finitely additive means that if you have any finite set of events $A_1,ldots,A_n$ which are pairwise independent (i.e. $A_icap A_j=varnothing$ for $ineq j$) then $mathbb(A_1cupldotscup A_n)=mathbb P(A_1)+ldots+mathbb P(A_n)$. Countable additivity means that we can do that with any sequence of events (which are pairwise disjoint), $langle A_nmid ninBbb Nrangle$, and not just finite sequences.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:38












  • $begingroup$
    So, $mathbb{P}$ being countably additive answers my first question because, as you state, that property is not just for finite sequences.
    $endgroup$
    – Mika'il
    Apr 20 '14 at 6:47












  • $begingroup$
    Yes, it answers part of your questions. Except the important(?) part, why is it just finitely additive. I don't know about that.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:51






  • 1




    $begingroup$
    Looking at your comments here it seems that the focus of interest is finite probability spaces. In that case, there's no harm in requiring only finite additivity, since there are not infinitely many distinct (let alone pairwise disjoint) events.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 7:39


















  • $begingroup$
    Finitely additive means that if you have any finite set of events $A_1,ldots,A_n$ which are pairwise independent (i.e. $A_icap A_j=varnothing$ for $ineq j$) then $mathbb(A_1cupldotscup A_n)=mathbb P(A_1)+ldots+mathbb P(A_n)$. Countable additivity means that we can do that with any sequence of events (which are pairwise disjoint), $langle A_nmid ninBbb Nrangle$, and not just finite sequences.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:38












  • $begingroup$
    So, $mathbb{P}$ being countably additive answers my first question because, as you state, that property is not just for finite sequences.
    $endgroup$
    – Mika'il
    Apr 20 '14 at 6:47












  • $begingroup$
    Yes, it answers part of your questions. Except the important(?) part, why is it just finitely additive. I don't know about that.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 6:51






  • 1




    $begingroup$
    Looking at your comments here it seems that the focus of interest is finite probability spaces. In that case, there's no harm in requiring only finite additivity, since there are not infinitely many distinct (let alone pairwise disjoint) events.
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '14 at 7:39
















$begingroup$
Finitely additive means that if you have any finite set of events $A_1,ldots,A_n$ which are pairwise independent (i.e. $A_icap A_j=varnothing$ for $ineq j$) then $mathbb(A_1cupldotscup A_n)=mathbb P(A_1)+ldots+mathbb P(A_n)$. Countable additivity means that we can do that with any sequence of events (which are pairwise disjoint), $langle A_nmid ninBbb Nrangle$, and not just finite sequences.
$endgroup$
– Asaf Karagila♦
Apr 20 '14 at 6:38






$begingroup$
Finitely additive means that if you have any finite set of events $A_1,ldots,A_n$ which are pairwise independent (i.e. $A_icap A_j=varnothing$ for $ineq j$) then $mathbb(A_1cupldotscup A_n)=mathbb P(A_1)+ldots+mathbb P(A_n)$. Countable additivity means that we can do that with any sequence of events (which are pairwise disjoint), $langle A_nmid ninBbb Nrangle$, and not just finite sequences.
$endgroup$
– Asaf Karagila♦
Apr 20 '14 at 6:38














$begingroup$
So, $mathbb{P}$ being countably additive answers my first question because, as you state, that property is not just for finite sequences.
$endgroup$
– Mika'il
Apr 20 '14 at 6:47






$begingroup$
So, $mathbb{P}$ being countably additive answers my first question because, as you state, that property is not just for finite sequences.
$endgroup$
– Mika'il
Apr 20 '14 at 6:47














$begingroup$
Yes, it answers part of your questions. Except the important(?) part, why is it just finitely additive. I don't know about that.
$endgroup$
– Asaf Karagila♦
Apr 20 '14 at 6:51




$begingroup$
Yes, it answers part of your questions. Except the important(?) part, why is it just finitely additive. I don't know about that.
$endgroup$
– Asaf Karagila♦
Apr 20 '14 at 6:51




1




1




$begingroup$
Looking at your comments here it seems that the focus of interest is finite probability spaces. In that case, there's no harm in requiring only finite additivity, since there are not infinitely many distinct (let alone pairwise disjoint) events.
$endgroup$
– Asaf Karagila♦
Apr 20 '14 at 7:39




$begingroup$
Looking at your comments here it seems that the focus of interest is finite probability spaces. In that case, there's no harm in requiring only finite additivity, since there are not infinitely many distinct (let alone pairwise disjoint) events.
$endgroup$
– Asaf Karagila♦
Apr 20 '14 at 7:39










1 Answer
1






active

oldest

votes


















0












$begingroup$

If $P(A cup B) = P(A) + P(B)$ whenever $A,B$ are disjoints sets, then



for $A_1 ldots A_n$ disjoints sets, one can consider
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1 cup B) $ where $B = A_2 cup ldots cup A_n$ is disjoint from $A$.



From the first property: $P(A_1 cup B) = P(A_1) + P(B)$



Proceed in the same manner $n$ times (or apply induction) to obtain that
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1) + P( A_2 ) + ldots P (A_n)$



And that is finite aditivity.



Finite additivity does not imply countable aditivity. For instance consider on $(mathbb{N} , mathcal{P}(mathbb{N}),)$ be such that
$$P(A) = begin{cases}0 & text{ if } A^c text{ is finite}\
1 & text{ otherwise }end{cases}$$



P is finite aditive but is not countably aditive






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    The final example here is not correct. For this function the probabilities of the even numbers, odd numbers, and natural numbers are all $1$, violating finite additivity. Instead, take $P(A)=1$ if $A^C$ is finite, $0$ otherwise (see math.stackexchange.com/questions/186280/… ).
    $endgroup$
    – Kevin P. Costello
    Oct 3 '17 at 17:58













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

If $P(A cup B) = P(A) + P(B)$ whenever $A,B$ are disjoints sets, then



for $A_1 ldots A_n$ disjoints sets, one can consider
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1 cup B) $ where $B = A_2 cup ldots cup A_n$ is disjoint from $A$.



From the first property: $P(A_1 cup B) = P(A_1) + P(B)$



Proceed in the same manner $n$ times (or apply induction) to obtain that
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1) + P( A_2 ) + ldots P (A_n)$



And that is finite aditivity.



Finite additivity does not imply countable aditivity. For instance consider on $(mathbb{N} , mathcal{P}(mathbb{N}),)$ be such that
$$P(A) = begin{cases}0 & text{ if } A^c text{ is finite}\
1 & text{ otherwise }end{cases}$$



P is finite aditive but is not countably aditive






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    The final example here is not correct. For this function the probabilities of the even numbers, odd numbers, and natural numbers are all $1$, violating finite additivity. Instead, take $P(A)=1$ if $A^C$ is finite, $0$ otherwise (see math.stackexchange.com/questions/186280/… ).
    $endgroup$
    – Kevin P. Costello
    Oct 3 '17 at 17:58


















0












$begingroup$

If $P(A cup B) = P(A) + P(B)$ whenever $A,B$ are disjoints sets, then



for $A_1 ldots A_n$ disjoints sets, one can consider
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1 cup B) $ where $B = A_2 cup ldots cup A_n$ is disjoint from $A$.



From the first property: $P(A_1 cup B) = P(A_1) + P(B)$



Proceed in the same manner $n$ times (or apply induction) to obtain that
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1) + P( A_2 ) + ldots P (A_n)$



And that is finite aditivity.



Finite additivity does not imply countable aditivity. For instance consider on $(mathbb{N} , mathcal{P}(mathbb{N}),)$ be such that
$$P(A) = begin{cases}0 & text{ if } A^c text{ is finite}\
1 & text{ otherwise }end{cases}$$



P is finite aditive but is not countably aditive






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    The final example here is not correct. For this function the probabilities of the even numbers, odd numbers, and natural numbers are all $1$, violating finite additivity. Instead, take $P(A)=1$ if $A^C$ is finite, $0$ otherwise (see math.stackexchange.com/questions/186280/… ).
    $endgroup$
    – Kevin P. Costello
    Oct 3 '17 at 17:58
















0












0








0





$begingroup$

If $P(A cup B) = P(A) + P(B)$ whenever $A,B$ are disjoints sets, then



for $A_1 ldots A_n$ disjoints sets, one can consider
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1 cup B) $ where $B = A_2 cup ldots cup A_n$ is disjoint from $A$.



From the first property: $P(A_1 cup B) = P(A_1) + P(B)$



Proceed in the same manner $n$ times (or apply induction) to obtain that
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1) + P( A_2 ) + ldots P (A_n)$



And that is finite aditivity.



Finite additivity does not imply countable aditivity. For instance consider on $(mathbb{N} , mathcal{P}(mathbb{N}),)$ be such that
$$P(A) = begin{cases}0 & text{ if } A^c text{ is finite}\
1 & text{ otherwise }end{cases}$$



P is finite aditive but is not countably aditive






share|cite|improve this answer











$endgroup$



If $P(A cup B) = P(A) + P(B)$ whenever $A,B$ are disjoints sets, then



for $A_1 ldots A_n$ disjoints sets, one can consider
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1 cup B) $ where $B = A_2 cup ldots cup A_n$ is disjoint from $A$.



From the first property: $P(A_1 cup B) = P(A_1) + P(B)$



Proceed in the same manner $n$ times (or apply induction) to obtain that
$P(A_1 cup A_2 cup ldots cup A_n) = P(A_1) + P( A_2 ) + ldots P (A_n)$



And that is finite aditivity.



Finite additivity does not imply countable aditivity. For instance consider on $(mathbb{N} , mathcal{P}(mathbb{N}),)$ be such that
$$P(A) = begin{cases}0 & text{ if } A^c text{ is finite}\
1 & text{ otherwise }end{cases}$$



P is finite aditive but is not countably aditive







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 4 '17 at 6:49

























answered Jun 23 '15 at 19:53









Conrado CostaConrado Costa

4,432932




4,432932








  • 2




    $begingroup$
    The final example here is not correct. For this function the probabilities of the even numbers, odd numbers, and natural numbers are all $1$, violating finite additivity. Instead, take $P(A)=1$ if $A^C$ is finite, $0$ otherwise (see math.stackexchange.com/questions/186280/… ).
    $endgroup$
    – Kevin P. Costello
    Oct 3 '17 at 17:58
















  • 2




    $begingroup$
    The final example here is not correct. For this function the probabilities of the even numbers, odd numbers, and natural numbers are all $1$, violating finite additivity. Instead, take $P(A)=1$ if $A^C$ is finite, $0$ otherwise (see math.stackexchange.com/questions/186280/… ).
    $endgroup$
    – Kevin P. Costello
    Oct 3 '17 at 17:58










2




2




$begingroup$
The final example here is not correct. For this function the probabilities of the even numbers, odd numbers, and natural numbers are all $1$, violating finite additivity. Instead, take $P(A)=1$ if $A^C$ is finite, $0$ otherwise (see math.stackexchange.com/questions/186280/… ).
$endgroup$
– Kevin P. Costello
Oct 3 '17 at 17:58






$begingroup$
The final example here is not correct. For this function the probabilities of the even numbers, odd numbers, and natural numbers are all $1$, violating finite additivity. Instead, take $P(A)=1$ if $A^C$ is finite, $0$ otherwise (see math.stackexchange.com/questions/186280/… ).
$endgroup$
– Kevin P. Costello
Oct 3 '17 at 17:58




















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