Why does $P(X+ Y =k) = sum_{i = 0}^k P(X+ Y = k, X = i)$ hold?
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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
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
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add a comment |
$begingroup$
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
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
$endgroup$
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 $Xin {0, , 1, , ..., k}$
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– Jakobian
Jan 7 at 21:24
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Law of Total Probability.
$endgroup$
– Clarinetist
Jan 7 at 21:25
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@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
add a comment |
$begingroup$
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
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
$endgroup$
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
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
asked Jan 7 at 21:22
ParsevalParseval
3,0741719
3,0741719
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
This is perfect! Thank you ver much!
$endgroup$
– Parseval
Jan 7 at 22:36
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
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.
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
This is perfect! Thank you ver much!
$endgroup$
– Parseval
Jan 7 at 22:36
add a comment |
$begingroup$
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.
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
This is perfect! Thank you ver much!
$endgroup$
– Parseval
Jan 7 at 22:36
add a comment |
$begingroup$
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.
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
$endgroup$
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.
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
answered Jan 7 at 21:48
kimchi loverkimchi lover
11.8k31229
11.8k31229
$begingroup$
This is perfect! Thank you ver much!
$endgroup$
– Parseval
Jan 7 at 22:36
add a comment |
$begingroup$
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
add a comment |
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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