What is the intuition or proof behind the conditional Bayes' theorem?
$begingroup$
In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:
$$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$
What is the intuition or mathematical proof behind this?
statistics proof-writing bayes-theorem
edited Jan 19 at 1:25
nbro
2,48863475
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
$endgroup$
add a comment |
$begingroup$
In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:
$$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$
What is the intuition or mathematical proof behind this?
statistics proof-writing bayes-theorem
edited Jan 19 at 1:25
nbro
2,48863475
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
$endgroup$
add a comment |
$begingroup$
In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:
$$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$
What is the intuition or mathematical proof behind this?
statistics proof-writing bayes-theorem
edited Jan 19 at 1:25
nbro
2,48863475
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
$endgroup$
In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:
$$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$
What is the intuition or mathematical proof behind this?
statistics proof-writing bayes-theorem
statistics proof-writing bayes-theorem
edited Jan 19 at 1:25
nbro
2,48863475
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
edited Jan 19 at 1:25
nbro
2,48863475
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
edited Jan 19 at 1:25
nbro
2,48863475
edited Jan 19 at 1:25
nbro
2,48863475
edited Jan 19 at 1:25
nbro
2,48863475
2,48863475
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
asked Aug 11 '14 at 23:26
ZhuluZhulu
3711612
3711612
add a comment |
add a comment |
1 Answer
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$begingroup$
It's directly equivalent to:
$$begin{align}
Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
\[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
end{align}$$
But with conditioning when given $C$:
$$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
\[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
end{align}$$
How do you know equation 3 is true?
$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
\[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
\[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
end{align}$
Also, can you explain how you derived the denominator in equation 4?
By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.
$$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
\ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
end{align}$$
I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).
Yes. Thuswise:
$$begin{align}
Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
\[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
\[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
\[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
\[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
end{align}$$
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
It's directly equivalent to:
$$begin{align}
Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
\[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
end{align}$$
But with conditioning when given $C$:
$$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
\[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
end{align}$$
How do you know equation 3 is true?
$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
\[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
\[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
end{align}$
Also, can you explain how you derived the denominator in equation 4?
By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.
$$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
\ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
end{align}$$
I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).
Yes. Thuswise:
$$begin{align}
Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
\[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
\[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
\[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
\[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
end{align}$$
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
It's directly equivalent to:
$$begin{align}
Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
\[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
end{align}$$
But with conditioning when given $C$:
$$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
\[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
end{align}$$
How do you know equation 3 is true?
$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
\[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
\[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
end{align}$
Also, can you explain how you derived the denominator in equation 4?
By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.
$$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
\ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
end{align}$$
I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).
Yes. Thuswise:
$$begin{align}
Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
\[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
\[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
\[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
\[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
end{align}$$
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
It's directly equivalent to:
$$begin{align}
Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
\[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
end{align}$$
But with conditioning when given $C$:
$$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
\[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
end{align}$$
How do you know equation 3 is true?
$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
\[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
\[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
end{align}$
Also, can you explain how you derived the denominator in equation 4?
By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.
$$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
\ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
end{align}$$
I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).
Yes. Thuswise:
$$begin{align}
Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
\[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
\[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
\[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
\[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
end{align}$$
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
$endgroup$
It's directly equivalent to:
$$begin{align}
Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
\[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
end{align}$$
But with conditioning when given $C$:
$$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
\[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
end{align}$$
How do you know equation 3 is true?
$begin{align}
Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
\[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
\[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
end{align}$
Also, can you explain how you derived the denominator in equation 4?
By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.
$$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
\ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
end{align}$$
I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).
Yes. Thuswise:
$$begin{align}
Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
\[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
\[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
\[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
\[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
end{align}$$
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
edited Aug 12 '14 at 2:43
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
answered Aug 11 '14 at 23:57
Graham KempGraham Kemp
88.1k43579
88.1k43579
add a comment |
add a comment |
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