Why does $P(X+ Y =k) = sum_{i = 0}^k P(X+ Y = k, X = i)$ hold?












0












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Can someone explain, intuitively or otherwise, why the following is true:



$$P(X+ Y =k) = sum_{i = 0}^k P(X+ Y = k, X = i)tag 1$$



How can the probability of $X+Y=k$ be the same as the probability of $X+Y=k$ and $X=i$ summed over $i=0,1,...,k?$ I can't seem to grasp this, apparently basic concept and it discourages me a lot.






probability probability-theory







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  • 2




    $begingroup$
    ${omegain Omega : X(omega)+Y(omega) = k} = bigcup_{i=0}^k {omegainOmega : X(omega)+Y(omega)=k, , X(omega)=i } $, and the sum is disjoint. That's of course if $Xin {0, , 1, , ..., k}$
    $endgroup$
    – Jakobian
    Jan 7 at 21:24












  • $begingroup$
    Law of Total Probability.
    $endgroup$
    – Clarinetist
    Jan 7 at 21:25










  • $begingroup$
    @Clarinetist I know it is the LOTP with the partitions and stuff but I can't understand it. Apologies for not making it clear in my question.
    $endgroup$
    – Parseval
    Jan 7 at 21:26
















0












$begingroup$


Can someone explain, intuitively or otherwise, why the following is true:



$$P(X+ Y =k) = sum_{i = 0}^k P(X+ Y = k, X = i)tag 1$$



How can the probability of $X+Y=k$ be the same as the probability of $X+Y=k$ and $X=i$ summed over $i=0,1,...,k?$ I can't seem to grasp this, apparently basic concept and it discourages me a lot.






probability probability-theory







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    ${omegain Omega : X(omega)+Y(omega) = k} = bigcup_{i=0}^k {omegainOmega : X(omega)+Y(omega)=k, , X(omega)=i } $, and the sum is disjoint. That's of course if $Xin {0, , 1, , ..., k}$
    $endgroup$
    – Jakobian
    Jan 7 at 21:24












  • $begingroup$
    Law of Total Probability.
    $endgroup$
    – Clarinetist
    Jan 7 at 21:25










  • $begingroup$
    @Clarinetist I know it is the LOTP with the partitions and stuff but I can't understand it. Apologies for not making it clear in my question.
    $endgroup$
    – Parseval
    Jan 7 at 21:26














0












0








0





$begingroup$


Can someone explain, intuitively or otherwise, why the following is true:



$$P(X+ Y =k) = sum_{i = 0}^k P(X+ Y = k, X = i)tag 1$$



How can the probability of $X+Y=k$ be the same as the probability of $X+Y=k$ and $X=i$ summed over $i=0,1,...,k?$ I can't seem to grasp this, apparently basic concept and it discourages me a lot.






probability probability-theory







share|cite|improve this question









$endgroup$




Can someone explain, intuitively or otherwise, why the following is true:



$$P(X+ Y =k) = sum_{i = 0}^k P(X+ Y = k, X = i)tag 1$$



How can the probability of $X+Y=k$ be the same as the probability of $X+Y=k$ and $X=i$ summed over $i=0,1,...,k?$ I can't seem to grasp this, apparently basic concept and it discourages me a lot.






probability probability-theory




probability probability-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 21:22









ParsevalParseval

2,7901718




2,7901718








  • 2




    $begingroup$
    ${omegain Omega : X(omega)+Y(omega) = k} = bigcup_{i=0}^k {omegainOmega : X(omega)+Y(omega)=k, , X(omega)=i } $, and the sum is disjoint. That's of course if $Xin {0, , 1, , ..., k}$
    $endgroup$
    – Jakobian
    Jan 7 at 21:24












  • $begingroup$
    Law of Total Probability.
    $endgroup$
    – Clarinetist
    Jan 7 at 21:25










  • $begingroup$
    @Clarinetist I know it is the LOTP with the partitions and stuff but I can't understand it. Apologies for not making it clear in my question.
    $endgroup$
    – Parseval
    Jan 7 at 21:26














  • 2




    $begingroup$
    ${omegain Omega : X(omega)+Y(omega) = k} = bigcup_{i=0}^k {omegainOmega : X(omega)+Y(omega)=k, , X(omega)=i } $, and the sum is disjoint. That's of course if $Xin {0, , 1, , ..., k}$
    $endgroup$
    – Jakobian
    Jan 7 at 21:24












  • $begingroup$
    Law of Total Probability.
    $endgroup$
    – Clarinetist
    Jan 7 at 21:25










  • $begingroup$
    @Clarinetist I know it is the LOTP with the partitions and stuff but I can't understand it. Apologies for not making it clear in my question.
    $endgroup$
    – Parseval
    Jan 7 at 21:26








2




2




$begingroup$
${omegain Omega : X(omega)+Y(omega) = k} = bigcup_{i=0}^k {omegainOmega : X(omega)+Y(omega)=k, , X(omega)=i } $, and the sum is disjoint. That's of course if $Xin {0, , 1, , ..., k}$
$endgroup$
– Jakobian
Jan 7 at 21:24






$begingroup$
${omegain Omega : X(omega)+Y(omega) = k} = bigcup_{i=0}^k {omegainOmega : X(omega)+Y(omega)=k, , X(omega)=i } $, and the sum is disjoint. That's of course if $Xin {0, , 1, , ..., k}$
$endgroup$
– Jakobian
Jan 7 at 21:24














$begingroup$
Law of Total Probability.
$endgroup$
– Clarinetist
Jan 7 at 21:25




$begingroup$
Law of Total Probability.
$endgroup$
– Clarinetist
Jan 7 at 21:25












$begingroup$
@Clarinetist I know it is the LOTP with the partitions and stuff but I can't understand it. Apologies for not making it clear in my question.
$endgroup$
– Parseval
Jan 7 at 21:26




$begingroup$
@Clarinetist I know it is the LOTP with the partitions and stuff but I can't understand it. Apologies for not making it clear in my question.
$endgroup$
– Parseval
Jan 7 at 21:26










1 Answer
1






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1












$begingroup$

Let me take a try.



Your formula holds only if $X$ can take on non-negative integer values only.



Then it is just an algebraic way of saying, for instance (taking $k=3$)




the chance I have 3 kids is equal to

the sum of the chance I have 3 girls and 0 boys plus the chance I have 2 girls and 1 boy plus the chance I have 1 girl and 2 boys plus the chance I have 0 girls and 3 boys.




Which itself is a probabilistic version of




If I have 3 children, then exactly one of the following is true: I have 0 sons, I have 1 son, I have 2 sons, I have 3 sons.




Here $X$ and $Y$ stand for the number of male and female children I have, and the non-negative integer condition obviously holds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is perfect! Thank you ver much!
    $endgroup$
    – Parseval
    Jan 7 at 22:36











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let me take a try.



Your formula holds only if $X$ can take on non-negative integer values only.



Then it is just an algebraic way of saying, for instance (taking $k=3$)




the chance I have 3 kids is equal to

the sum of the chance I have 3 girls and 0 boys plus the chance I have 2 girls and 1 boy plus the chance I have 1 girl and 2 boys plus the chance I have 0 girls and 3 boys.




Which itself is a probabilistic version of




If I have 3 children, then exactly one of the following is true: I have 0 sons, I have 1 son, I have 2 sons, I have 3 sons.




Here $X$ and $Y$ stand for the number of male and female children I have, and the non-negative integer condition obviously holds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is perfect! Thank you ver much!
    $endgroup$
    – Parseval
    Jan 7 at 22:36
















1












$begingroup$

Let me take a try.



Your formula holds only if $X$ can take on non-negative integer values only.



Then it is just an algebraic way of saying, for instance (taking $k=3$)




the chance I have 3 kids is equal to

the sum of the chance I have 3 girls and 0 boys plus the chance I have 2 girls and 1 boy plus the chance I have 1 girl and 2 boys plus the chance I have 0 girls and 3 boys.




Which itself is a probabilistic version of




If I have 3 children, then exactly one of the following is true: I have 0 sons, I have 1 son, I have 2 sons, I have 3 sons.




Here $X$ and $Y$ stand for the number of male and female children I have, and the non-negative integer condition obviously holds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is perfect! Thank you ver much!
    $endgroup$
    – Parseval
    Jan 7 at 22:36














1












1








1





$begingroup$

Let me take a try.



Your formula holds only if $X$ can take on non-negative integer values only.



Then it is just an algebraic way of saying, for instance (taking $k=3$)




the chance I have 3 kids is equal to

the sum of the chance I have 3 girls and 0 boys plus the chance I have 2 girls and 1 boy plus the chance I have 1 girl and 2 boys plus the chance I have 0 girls and 3 boys.




Which itself is a probabilistic version of




If I have 3 children, then exactly one of the following is true: I have 0 sons, I have 1 son, I have 2 sons, I have 3 sons.




Here $X$ and $Y$ stand for the number of male and female children I have, and the non-negative integer condition obviously holds.






share|cite|improve this answer









$endgroup$



Let me take a try.



Your formula holds only if $X$ can take on non-negative integer values only.



Then it is just an algebraic way of saying, for instance (taking $k=3$)




the chance I have 3 kids is equal to

the sum of the chance I have 3 girls and 0 boys plus the chance I have 2 girls and 1 boy plus the chance I have 1 girl and 2 boys plus the chance I have 0 girls and 3 boys.




Which itself is a probabilistic version of




If I have 3 children, then exactly one of the following is true: I have 0 sons, I have 1 son, I have 2 sons, I have 3 sons.




Here $X$ and $Y$ stand for the number of male and female children I have, and the non-negative integer condition obviously holds.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 21:48









kimchi loverkimchi lover

9,74631128




9,74631128












  • $begingroup$
    This is perfect! Thank you ver much!
    $endgroup$
    – Parseval
    Jan 7 at 22:36


















  • $begingroup$
    This is perfect! Thank you ver much!
    $endgroup$
    – Parseval
    Jan 7 at 22:36
















$begingroup$
This is perfect! Thank you ver much!
$endgroup$
– Parseval
Jan 7 at 22:36




$begingroup$
This is perfect! Thank you ver much!
$endgroup$
– Parseval
Jan 7 at 22:36


















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