how to prove Bayes theorem?
$begingroup$
how to prove Bayes theorem? (Develop from conditional probability )
2 criterias
-Total of law is probability
-multiplication rule
Can someone show me the mathematical proof & steps for this please?
conditional-probability
asked Dec 5 '18 at 16:12
Amber Amber
11
$endgroup$
add a comment |
$begingroup$
how to prove Bayes theorem? (Develop from conditional probability )
2 criterias
-Total of law is probability
-multiplication rule
Can someone show me the mathematical proof & steps for this please?
conditional-probability
asked Dec 5 '18 at 16:12
Amber Amber
11
$endgroup$
1
$begingroup$
There's a perfectly good proof on Wikipedia. Are there steps there that you don't follow and would like more explanation on?
$endgroup$
– Nick Peterson
Dec 5 '18 at 16:16
1
$begingroup$
There isn't much to prove at all...
$endgroup$
– Federico
Dec 5 '18 at 16:17
add a comment |
$begingroup$
how to prove Bayes theorem? (Develop from conditional probability )
2 criterias
-Total of law is probability
-multiplication rule
Can someone show me the mathematical proof & steps for this please?
conditional-probability
asked Dec 5 '18 at 16:12
Amber Amber
11
$endgroup$
how to prove Bayes theorem? (Develop from conditional probability )
2 criterias
-Total of law is probability
-multiplication rule
Can someone show me the mathematical proof & steps for this please?
conditional-probability
conditional-probability
asked Dec 5 '18 at 16:12
Amber Amber
11
asked Dec 5 '18 at 16:12
Amber Amber
11
asked Dec 5 '18 at 16:12
Amber Amber
11
asked Dec 5 '18 at 16:12
Amber Amber
11
asked Dec 5 '18 at 16:12
Amber Amber
11
11
1
$begingroup$
There's a perfectly good proof on Wikipedia. Are there steps there that you don't follow and would like more explanation on?
$endgroup$
– Nick Peterson
Dec 5 '18 at 16:16
1
$begingroup$
There isn't much to prove at all...
$endgroup$
– Federico
Dec 5 '18 at 16:17
add a comment |
1
$begingroup$
There's a perfectly good proof on Wikipedia. Are there steps there that you don't follow and would like more explanation on?
$endgroup$
– Nick Peterson
Dec 5 '18 at 16:16
1
$begingroup$
There isn't much to prove at all...
$endgroup$
– Federico
Dec 5 '18 at 16:17
1
1
$begingroup$
There's a perfectly good proof on Wikipedia. Are there steps there that you don't follow and would like more explanation on?
$endgroup$
– Nick Peterson
Dec 5 '18 at 16:16
$begingroup$
There's a perfectly good proof on Wikipedia. Are there steps there that you don't follow and would like more explanation on?
$endgroup$
– Nick Peterson
Dec 5 '18 at 16:16
1
1
$begingroup$
There isn't much to prove at all...
$endgroup$
– Federico
Dec 5 '18 at 16:17
$begingroup$
There isn't much to prove at all...
$endgroup$
– Federico
Dec 5 '18 at 16:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The conditional probability is defined as
$$
P(A|B) = frac{P(Acap B)}{P(B)},
$$
hence
$$
P(A|B) P(B) = P(B|A) P(A) .
$$
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
$endgroup$
add a comment |
$begingroup$
Hints
- Write $P(A cap B)$ two different ways using conditional probability
- these are therefore equal by Euclid's first axiom
- manipulate into Bayes' theorem (assuming none of the probabilities are $0$)
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The conditional probability is defined as
$$
P(A|B) = frac{P(Acap B)}{P(B)},
$$
hence
$$
P(A|B) P(B) = P(B|A) P(A) .
$$
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
$endgroup$
add a comment |
$begingroup$
The conditional probability is defined as
$$
P(A|B) = frac{P(Acap B)}{P(B)},
$$
hence
$$
P(A|B) P(B) = P(B|A) P(A) .
$$
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
$endgroup$
add a comment |
$begingroup$
The conditional probability is defined as
$$
P(A|B) = frac{P(Acap B)}{P(B)},
$$
hence
$$
P(A|B) P(B) = P(B|A) P(A) .
$$
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
$endgroup$
The conditional probability is defined as
$$
P(A|B) = frac{P(Acap B)}{P(B)},
$$
hence
$$
P(A|B) P(B) = P(B|A) P(A) .
$$
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
answered Dec 5 '18 at 16:19
FedericoFederico
5,014514
5,014514
add a comment |
add a comment |
$begingroup$
Hints
- Write $P(A cap B)$ two different ways using conditional probability
- these are therefore equal by Euclid's first axiom
- manipulate into Bayes' theorem (assuming none of the probabilities are $0$)
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
$endgroup$
add a comment |
$begingroup$
Hints
- Write $P(A cap B)$ two different ways using conditional probability
- these are therefore equal by Euclid's first axiom
- manipulate into Bayes' theorem (assuming none of the probabilities are $0$)
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
$endgroup$
add a comment |
$begingroup$
Hints
- Write $P(A cap B)$ two different ways using conditional probability
- these are therefore equal by Euclid's first axiom
- manipulate into Bayes' theorem (assuming none of the probabilities are $0$)
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
$endgroup$
Hints
- Write $P(A cap B)$ two different ways using conditional probability
- these are therefore equal by Euclid's first axiom
- manipulate into Bayes' theorem (assuming none of the probabilities are $0$)
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
answered Dec 5 '18 at 16:18
HenryHenry
99.7k479165
99.7k479165
add a comment |
add a comment |
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1
$begingroup$
There's a perfectly good proof on Wikipedia. Are there steps there that you don't follow and would like more explanation on?
$endgroup$
– Nick Peterson
Dec 5 '18 at 16:16
1
$begingroup$
There isn't much to prove at all...
$endgroup$
– Federico
Dec 5 '18 at 16:17