Mixing
Total probability of Conditional Probability
$begingroup$
Suppose that we have a conditional probability $P(x | y)$ and a partition $x'$ of the sample space. How can we apply the law of total probability to conclude that
$$P(x|y) = sumlimits_{x'}^{} P((x|y)cap x')?$$
probability
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
asked May 14 '15 at 10:35
XingdongXingdong
97811524
$endgroup$
add a comment |
$begingroup$
Suppose that we have a conditional probability $P(x | y)$ and a partition $x'$ of the sample space. How can we apply the law of total probability to conclude that
$$P(x|y) = sumlimits_{x'}^{} P((x|y)cap x')?$$
probability
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
asked May 14 '15 at 10:35
XingdongXingdong
97811524
$endgroup$
add a comment |
$begingroup$
Suppose that we have a conditional probability $P(x | y)$ and a partition $x'$ of the sample space. How can we apply the law of total probability to conclude that
$$P(x|y) = sumlimits_{x'}^{} P((x|y)cap x')?$$
probability
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
asked May 14 '15 at 10:35
XingdongXingdong
97811524
$endgroup$
Suppose that we have a conditional probability $P(x | y)$ and a partition $x'$ of the sample space. How can we apply the law of total probability to conclude that
$$P(x|y) = sumlimits_{x'}^{} P((x|y)cap x')?$$
probability
probability
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
asked May 14 '15 at 10:35
XingdongXingdong
97811524
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
asked May 14 '15 at 10:35
XingdongXingdong
97811524
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
edited May 14 '15 at 10:59
Tomek Kania
12.3k11946
12.3k11946
asked May 14 '15 at 10:35
XingdongXingdong
97811524
asked May 14 '15 at 10:35
XingdongXingdong
97811524
asked May 14 '15 at 10:35
XingdongXingdong
97811524
97811524
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is rather straightforward, use the fact that for any events $A$ and $B$
$$P( A cap B )= P(B)P(A|B).$$
Thus $P((x|y) cap x^{prime}) = P((x|y)|x^{prime})P(x^{prime})$, so by the law of total probability
$$P(x|y) = sum_{x^{prime} }P((x|y)|x^{prime})P(x^{prime})=sum_{x^{prime} } P((x|y) cap x^{prime}).$$
Caution: Using $x^{prime}$ as the elements from the partition is not the best notation, $prime$ is often used to denote the complement of an event.
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
$endgroup$
$begingroup$
$P((x|y)|x') = P(x| y, x')$, by the way, is this true ?
$endgroup$
– Xingdong
May 14 '15 at 11:22
$begingroup$
$P(x|y, x^{prime}) = frac{P(x cap y|x^{prime} )}{P(y | x^{prime} )}$
$endgroup$
– m_gnacik
May 14 '15 at 12:39
$begingroup$
So is there any alternative or equivalent representation of $P((x|y)|x')$ ?
$endgroup$
– Xingdong
May 14 '15 at 13:49
$begingroup$
It depends what would you like to get; you could apply Bayes' theorem and obtain $$ P( (x|y)|x^{prime}) = frac{P(x^{prime}|(x|y))P(x|y)}{P(x^{prime})}.$$ If you happy with my answer you can accept it. Cheers.
$endgroup$
– m_gnacik
May 14 '15 at 14:20
add a comment |
$begingroup$
In addition, I think this answer also provides a good view of what happens here.
Proof of Total Probability Theorem for Conditional Probability
I find it rather confusing to use condition with
$$sum_{z}P((x|y)|z)P(z) \text{z here in place of x' in question}$$
for the event space, $Omega$ for $Z$ here is not so clear. The space for $Z$, or the value of $P(Z)$ here should be defined over $Omega_y cap Omega_Z$.
This makes the overall answer be either:
$$P(x|y) = sum^{Omega_ycap Omega_Z}_zP(x|ycap z)P(z) = sum_zP(xcap z|y)$$
Also, as mentioned by Graham Kemp in the comment:
∣ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function.
answered Jan 30 at 8:37
DogemoreDogemore
31
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
It is rather straightforward, use the fact that for any events $A$ and $B$
$$P( A cap B )= P(B)P(A|B).$$
Thus $P((x|y) cap x^{prime}) = P((x|y)|x^{prime})P(x^{prime})$, so by the law of total probability
$$P(x|y) = sum_{x^{prime} }P((x|y)|x^{prime})P(x^{prime})=sum_{x^{prime} } P((x|y) cap x^{prime}).$$
Caution: Using $x^{prime}$ as the elements from the partition is not the best notation, $prime$ is often used to denote the complement of an event.
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
$endgroup$
$begingroup$
$P((x|y)|x') = P(x| y, x')$, by the way, is this true ?
$endgroup$
– Xingdong
May 14 '15 at 11:22
$begingroup$
$P(x|y, x^{prime}) = frac{P(x cap y|x^{prime} )}{P(y | x^{prime} )}$
$endgroup$
– m_gnacik
May 14 '15 at 12:39
$begingroup$
So is there any alternative or equivalent representation of $P((x|y)|x')$ ?
$endgroup$
– Xingdong
May 14 '15 at 13:49
$begingroup$
It depends what would you like to get; you could apply Bayes' theorem and obtain $$ P( (x|y)|x^{prime}) = frac{P(x^{prime}|(x|y))P(x|y)}{P(x^{prime})}.$$ If you happy with my answer you can accept it. Cheers.
$endgroup$
– m_gnacik
May 14 '15 at 14:20
add a comment |
$begingroup$
It is rather straightforward, use the fact that for any events $A$ and $B$
$$P( A cap B )= P(B)P(A|B).$$
Thus $P((x|y) cap x^{prime}) = P((x|y)|x^{prime})P(x^{prime})$, so by the law of total probability
$$P(x|y) = sum_{x^{prime} }P((x|y)|x^{prime})P(x^{prime})=sum_{x^{prime} } P((x|y) cap x^{prime}).$$
Caution: Using $x^{prime}$ as the elements from the partition is not the best notation, $prime$ is often used to denote the complement of an event.
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
$endgroup$
$begingroup$
$P((x|y)|x') = P(x| y, x')$, by the way, is this true ?
$endgroup$
– Xingdong
May 14 '15 at 11:22
$begingroup$
$P(x|y, x^{prime}) = frac{P(x cap y|x^{prime} )}{P(y | x^{prime} )}$
$endgroup$
– m_gnacik
May 14 '15 at 12:39
$begingroup$
So is there any alternative or equivalent representation of $P((x|y)|x')$ ?
$endgroup$
– Xingdong
May 14 '15 at 13:49
$begingroup$
It depends what would you like to get; you could apply Bayes' theorem and obtain $$ P( (x|y)|x^{prime}) = frac{P(x^{prime}|(x|y))P(x|y)}{P(x^{prime})}.$$ If you happy with my answer you can accept it. Cheers.
$endgroup$
– m_gnacik
May 14 '15 at 14:20
add a comment |
$begingroup$
It is rather straightforward, use the fact that for any events $A$ and $B$
$$P( A cap B )= P(B)P(A|B).$$
Thus $P((x|y) cap x^{prime}) = P((x|y)|x^{prime})P(x^{prime})$, so by the law of total probability
$$P(x|y) = sum_{x^{prime} }P((x|y)|x^{prime})P(x^{prime})=sum_{x^{prime} } P((x|y) cap x^{prime}).$$
Caution: Using $x^{prime}$ as the elements from the partition is not the best notation, $prime$ is often used to denote the complement of an event.
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
$endgroup$
It is rather straightforward, use the fact that for any events $A$ and $B$
$$P( A cap B )= P(B)P(A|B).$$
Thus $P((x|y) cap x^{prime}) = P((x|y)|x^{prime})P(x^{prime})$, so by the law of total probability
$$P(x|y) = sum_{x^{prime} }P((x|y)|x^{prime})P(x^{prime})=sum_{x^{prime} } P((x|y) cap x^{prime}).$$
Caution: Using $x^{prime}$ as the elements from the partition is not the best notation, $prime$ is often used to denote the complement of an event.
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
edited May 14 '15 at 11:01
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
answered May 14 '15 at 10:54
m_gnacikm_gnacik
2,4381120
2,4381120
$begingroup$
$P((x|y)|x') = P(x| y, x')$, by the way, is this true ?
$endgroup$
– Xingdong
May 14 '15 at 11:22
$begingroup$
$P(x|y, x^{prime}) = frac{P(x cap y|x^{prime} )}{P(y | x^{prime} )}$
$endgroup$
– m_gnacik
May 14 '15 at 12:39
$begingroup$
So is there any alternative or equivalent representation of $P((x|y)|x')$ ?
$endgroup$
– Xingdong
May 14 '15 at 13:49
$begingroup$
It depends what would you like to get; you could apply Bayes' theorem and obtain $$ P( (x|y)|x^{prime}) = frac{P(x^{prime}|(x|y))P(x|y)}{P(x^{prime})}.$$ If you happy with my answer you can accept it. Cheers.
$endgroup$
– m_gnacik
May 14 '15 at 14:20
add a comment |
$begingroup$
$P((x|y)|x') = P(x| y, x')$, by the way, is this true ?
$endgroup$
– Xingdong
May 14 '15 at 11:22
$begingroup$
$P(x|y, x^{prime}) = frac{P(x cap y|x^{prime} )}{P(y | x^{prime} )}$
$endgroup$
– m_gnacik
May 14 '15 at 12:39
$begingroup$
So is there any alternative or equivalent representation of $P((x|y)|x')$ ?
$endgroup$
– Xingdong
May 14 '15 at 13:49
$begingroup$
It depends what would you like to get; you could apply Bayes' theorem and obtain $$ P( (x|y)|x^{prime}) = frac{P(x^{prime}|(x|y))P(x|y)}{P(x^{prime})}.$$ If you happy with my answer you can accept it. Cheers.
$endgroup$
– m_gnacik
May 14 '15 at 14:20
$begingroup$
$P((x|y)|x') = P(x| y, x')$, by the way, is this true ?
$endgroup$
– Xingdong
May 14 '15 at 11:22
$begingroup$
$P((x|y)|x') = P(x| y, x')$, by the way, is this true ?
$endgroup$
– Xingdong
May 14 '15 at 11:22
$begingroup$
$P(x|y, x^{prime}) = frac{P(x cap y|x^{prime} )}{P(y | x^{prime} )}$
$endgroup$
– m_gnacik
May 14 '15 at 12:39
$begingroup$
$P(x|y, x^{prime}) = frac{P(x cap y|x^{prime} )}{P(y | x^{prime} )}$
$endgroup$
– m_gnacik
May 14 '15 at 12:39
$begingroup$
So is there any alternative or equivalent representation of $P((x|y)|x')$ ?
$endgroup$
– Xingdong
May 14 '15 at 13:49
$begingroup$
So is there any alternative or equivalent representation of $P((x|y)|x')$ ?
$endgroup$
– Xingdong
May 14 '15 at 13:49
$begingroup$
It depends what would you like to get; you could apply Bayes' theorem and obtain $$ P( (x|y)|x^{prime}) = frac{P(x^{prime}|(x|y))P(x|y)}{P(x^{prime})}.$$ If you happy with my answer you can accept it. Cheers.
$endgroup$
– m_gnacik
May 14 '15 at 14:20
$begingroup$
It depends what would you like to get; you could apply Bayes' theorem and obtain $$ P( (x|y)|x^{prime}) = frac{P(x^{prime}|(x|y))P(x|y)}{P(x^{prime})}.$$ If you happy with my answer you can accept it. Cheers.
$endgroup$
– m_gnacik
May 14 '15 at 14:20
add a comment |
$begingroup$
In addition, I think this answer also provides a good view of what happens here.
Proof of Total Probability Theorem for Conditional Probability
I find it rather confusing to use condition with
$$sum_{z}P((x|y)|z)P(z) \text{z here in place of x' in question}$$
for the event space, $Omega$ for $Z$ here is not so clear. The space for $Z$, or the value of $P(Z)$ here should be defined over $Omega_y cap Omega_Z$.
This makes the overall answer be either:
$$P(x|y) = sum^{Omega_ycap Omega_Z}_zP(x|ycap z)P(z) = sum_zP(xcap z|y)$$
Also, as mentioned by Graham Kemp in the comment:
∣ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function.
answered Jan 30 at 8:37
DogemoreDogemore
31
$endgroup$
add a comment |
$begingroup$
In addition, I think this answer also provides a good view of what happens here.
Proof of Total Probability Theorem for Conditional Probability
I find it rather confusing to use condition with
$$sum_{z}P((x|y)|z)P(z) \text{z here in place of x' in question}$$
for the event space, $Omega$ for $Z$ here is not so clear. The space for $Z$, or the value of $P(Z)$ here should be defined over $Omega_y cap Omega_Z$.
This makes the overall answer be either:
$$P(x|y) = sum^{Omega_ycap Omega_Z}_zP(x|ycap z)P(z) = sum_zP(xcap z|y)$$
Also, as mentioned by Graham Kemp in the comment:
∣ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function.
answered Jan 30 at 8:37
DogemoreDogemore
31
$endgroup$
add a comment |
$begingroup$
In addition, I think this answer also provides a good view of what happens here.
Proof of Total Probability Theorem for Conditional Probability
I find it rather confusing to use condition with
$$sum_{z}P((x|y)|z)P(z) \text{z here in place of x' in question}$$
for the event space, $Omega$ for $Z$ here is not so clear. The space for $Z$, or the value of $P(Z)$ here should be defined over $Omega_y cap Omega_Z$.
This makes the overall answer be either:
$$P(x|y) = sum^{Omega_ycap Omega_Z}_zP(x|ycap z)P(z) = sum_zP(xcap z|y)$$
Also, as mentioned by Graham Kemp in the comment:
∣ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function.
answered Jan 30 at 8:37
DogemoreDogemore
31
$endgroup$
In addition, I think this answer also provides a good view of what happens here.
Proof of Total Probability Theorem for Conditional Probability
I find it rather confusing to use condition with
$$sum_{z}P((x|y)|z)P(z) \text{z here in place of x' in question}$$
for the event space, $Omega$ for $Z$ here is not so clear. The space for $Z$, or the value of $P(Z)$ here should be defined over $Omega_y cap Omega_Z$.
This makes the overall answer be either:
$$P(x|y) = sum^{Omega_ycap Omega_Z}_zP(x|ycap z)P(z) = sum_zP(xcap z|y)$$
Also, as mentioned by Graham Kemp in the comment:
∣ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function.
answered Jan 30 at 8:37
DogemoreDogemore
31
edited Jan 30 at 9:03
answered Jan 30 at 8:37
DogemoreDogemore
31
answered Jan 30 at 8:37
DogemoreDogemore
31
answered Jan 30 at 8:37
DogemoreDogemore
31
31
add a comment |
add a comment |
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