Understanding marginalization in Bishop's Pattern Recognition and Machine Learning
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On page 374 in Bishop's PRML, he does the following:
If
$$
p(a,b,c) = p(a)p(c|a)p(b|c)
$$
then
$$
p(a,b) = p(a)sum p(c|a)p(b|c) = p(a)p(b|a)
$$
I don't understand how he does the final step, going from the marginalization sum to the very last expression. Could someone derive this step by step?
Many thanks
probability-theory
asked Dec 3 '18 at 9:48
SandiSandi
255112
$endgroup$
add a comment |
$begingroup$
On page 374 in Bishop's PRML, he does the following:
If
$$
p(a,b,c) = p(a)p(c|a)p(b|c)
$$
then
$$
p(a,b) = p(a)sum p(c|a)p(b|c) = p(a)p(b|a)
$$
I don't understand how he does the final step, going from the marginalization sum to the very last expression. Could someone derive this step by step?
Many thanks
probability-theory
asked Dec 3 '18 at 9:48
SandiSandi
255112
$endgroup$
1
$begingroup$
If you ignore the middle it is straight from definition of bedingd wahrscheinlichkeit. (conditional probability.)
$endgroup$
– mathreadler
Dec 3 '18 at 9:54
$begingroup$
Oh, yeah, I missed that. But why then does Bishop bother with the marginalization in the middle?
$endgroup$
– Sandi
Dec 3 '18 at 10:01
$begingroup$
I have a copy of Bishop's in a locker ~1000 kilometers from here. I really can't check the context on page 374 from over here.
$endgroup$
– mathreadler
Dec 3 '18 at 10:08
add a comment |
$begingroup$
On page 374 in Bishop's PRML, he does the following:
If
$$
p(a,b,c) = p(a)p(c|a)p(b|c)
$$
then
$$
p(a,b) = p(a)sum p(c|a)p(b|c) = p(a)p(b|a)
$$
I don't understand how he does the final step, going from the marginalization sum to the very last expression. Could someone derive this step by step?
Many thanks
probability-theory
asked Dec 3 '18 at 9:48
SandiSandi
255112
$endgroup$
On page 374 in Bishop's PRML, he does the following:
If
$$
p(a,b,c) = p(a)p(c|a)p(b|c)
$$
then
$$
p(a,b) = p(a)sum p(c|a)p(b|c) = p(a)p(b|a)
$$
I don't understand how he does the final step, going from the marginalization sum to the very last expression. Could someone derive this step by step?
Many thanks
probability-theory
probability-theory
asked Dec 3 '18 at 9:48
SandiSandi
255112
asked Dec 3 '18 at 9:48
SandiSandi
255112
asked Dec 3 '18 at 9:48
SandiSandi
255112
asked Dec 3 '18 at 9:48
SandiSandi
255112
asked Dec 3 '18 at 9:48
SandiSandi
255112
255112
1
$begingroup$
If you ignore the middle it is straight from definition of bedingd wahrscheinlichkeit. (conditional probability.)
$endgroup$
– mathreadler
Dec 3 '18 at 9:54
$begingroup$
Oh, yeah, I missed that. But why then does Bishop bother with the marginalization in the middle?
$endgroup$
– Sandi
Dec 3 '18 at 10:01
$begingroup$
I have a copy of Bishop's in a locker ~1000 kilometers from here. I really can't check the context on page 374 from over here.
$endgroup$
– mathreadler
Dec 3 '18 at 10:08
add a comment |
1
$begingroup$
If you ignore the middle it is straight from definition of bedingd wahrscheinlichkeit. (conditional probability.)
$endgroup$
– mathreadler
Dec 3 '18 at 9:54
$begingroup$
Oh, yeah, I missed that. But why then does Bishop bother with the marginalization in the middle?
$endgroup$
– Sandi
Dec 3 '18 at 10:01
$begingroup$
I have a copy of Bishop's in a locker ~1000 kilometers from here. I really can't check the context on page 374 from over here.
$endgroup$
– mathreadler
Dec 3 '18 at 10:08
1
1
$begingroup$
If you ignore the middle it is straight from definition of bedingd wahrscheinlichkeit. (conditional probability.)
$endgroup$
– mathreadler
Dec 3 '18 at 9:54
$begingroup$
If you ignore the middle it is straight from definition of bedingd wahrscheinlichkeit. (conditional probability.)
$endgroup$
– mathreadler
Dec 3 '18 at 9:54
$begingroup$
Oh, yeah, I missed that. But why then does Bishop bother with the marginalization in the middle?
$endgroup$
– Sandi
Dec 3 '18 at 10:01
$begingroup$
Oh, yeah, I missed that. But why then does Bishop bother with the marginalization in the middle?
$endgroup$
– Sandi
Dec 3 '18 at 10:01
$begingroup$
I have a copy of Bishop's in a locker ~1000 kilometers from here. I really can't check the context on page 374 from over here.
$endgroup$
– mathreadler
Dec 3 '18 at 10:08
$begingroup$
I have a copy of Bishop's in a locker ~1000 kilometers from here. I really can't check the context on page 374 from over here.
$endgroup$
– mathreadler
Dec 3 '18 at 10:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Definition of conditional probability
$$p(b|a) = frac{p(a,b)}{p(a)}$$
Just use first and last expression and rewrite it and you see it become the same as above definition. The wikipedia just uses a set theoretic expression for above.
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
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add a comment |
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1 Answer
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$begingroup$
Definition of conditional probability
$$p(b|a) = frac{p(a,b)}{p(a)}$$
Just use first and last expression and rewrite it and you see it become the same as above definition. The wikipedia just uses a set theoretic expression for above.
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
$endgroup$
add a comment |
$begingroup$
Definition of conditional probability
$$p(b|a) = frac{p(a,b)}{p(a)}$$
Just use first and last expression and rewrite it and you see it become the same as above definition. The wikipedia just uses a set theoretic expression for above.
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
$endgroup$
add a comment |
$begingroup$
Definition of conditional probability
$$p(b|a) = frac{p(a,b)}{p(a)}$$
Just use first and last expression and rewrite it and you see it become the same as above definition. The wikipedia just uses a set theoretic expression for above.
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
$endgroup$
Definition of conditional probability
$$p(b|a) = frac{p(a,b)}{p(a)}$$
Just use first and last expression and rewrite it and you see it become the same as above definition. The wikipedia just uses a set theoretic expression for above.
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
answered Dec 3 '18 at 9:56
mathreadlermathreadler
14.8k72160
14.8k72160
add a comment |
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1
$begingroup$
If you ignore the middle it is straight from definition of bedingd wahrscheinlichkeit. (conditional probability.)
$endgroup$
– mathreadler
Dec 3 '18 at 9:54
$begingroup$
Oh, yeah, I missed that. But why then does Bishop bother with the marginalization in the middle?
$endgroup$
– Sandi
Dec 3 '18 at 10:01
$begingroup$
I have a copy of Bishop's in a locker ~1000 kilometers from here. I really can't check the context on page 374 from over here.
$endgroup$
– mathreadler
Dec 3 '18 at 10:08