Help simplifying Bayes' theorem for multiple conditions
$begingroup$
I'm trying to use Bayes' theorem to calculate the likelihood, in my example, of religion H being true considering evidence for and against it. I think I understand Bayes' theorem well enough to do this in two steps, calculating the odds that H is true considering evidence against it, and then taking that result as a new prior probability to factor in evidence for it.
I'm sure there's a way to combine those two steps into one, but the explanations I found were not that clear and used a notation that I'm not very familiar with.
So for example prior probability H is true due to the number of possible true religions P(H) is 0.001, the likelihood we'd find particular arguments against it if it were true P(Aa|H) is 0.01, the likelihood that we'd find particular arguments against it if it were not true P(Aa|H') is 0.6, the likelihood we'd find particular arguments for it if it were true P(Af|H) is 0.8, and the likelihood that we'd find particular arguments for it if it were not true P(Af|H') is 0.2. And in my example the arguments against it and arguments for it are both observed.
So then I do the math in two stages, first considering arguments against it:
P(H|Aa) = P(Aa|H) * P(H) / [P(Aa|H) * P(H) + P(Aa|H’) * P(H’)]
P(H|Aa) = 0.01 * 0.001 / [0.01 * 0.001 + 0.6 * 0.999]
P(H|Aa) = 0.000017
So now I take that and calculate considering arguments for it:
P(H|Af) = P(Af|H) * P(H) / [P(Af|H) * P(H) + P(Af|H’) * P(H’)]
P(H|Af) = 0.8 * 0.000017 / [0.8 * 0.000017 + 0.2 * 0.999983]
P(H|Af) = 0.000068
This gives me a 0.0068% probability that the religion is true given the above assumptions. (I hope I didn't mess up that math.)
But what would be a single equation to do both steps at once?
Edit: I think I may have found the simplification:
P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)]
P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999]
P(H|Aa, Af) = 0.000067
Which is the same result (slight difference probably due to my rounding in the intermediate step of the separate calculations). Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
probability bayes-theorem
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
$endgroup$
add a comment |
$begingroup$
I'm trying to use Bayes' theorem to calculate the likelihood, in my example, of religion H being true considering evidence for and against it. I think I understand Bayes' theorem well enough to do this in two steps, calculating the odds that H is true considering evidence against it, and then taking that result as a new prior probability to factor in evidence for it.
I'm sure there's a way to combine those two steps into one, but the explanations I found were not that clear and used a notation that I'm not very familiar with.
So for example prior probability H is true due to the number of possible true religions P(H) is 0.001, the likelihood we'd find particular arguments against it if it were true P(Aa|H) is 0.01, the likelihood that we'd find particular arguments against it if it were not true P(Aa|H') is 0.6, the likelihood we'd find particular arguments for it if it were true P(Af|H) is 0.8, and the likelihood that we'd find particular arguments for it if it were not true P(Af|H') is 0.2. And in my example the arguments against it and arguments for it are both observed.
So then I do the math in two stages, first considering arguments against it:
P(H|Aa) = P(Aa|H) * P(H) / [P(Aa|H) * P(H) + P(Aa|H’) * P(H’)]
P(H|Aa) = 0.01 * 0.001 / [0.01 * 0.001 + 0.6 * 0.999]
P(H|Aa) = 0.000017
So now I take that and calculate considering arguments for it:
P(H|Af) = P(Af|H) * P(H) / [P(Af|H) * P(H) + P(Af|H’) * P(H’)]
P(H|Af) = 0.8 * 0.000017 / [0.8 * 0.000017 + 0.2 * 0.999983]
P(H|Af) = 0.000068
This gives me a 0.0068% probability that the religion is true given the above assumptions. (I hope I didn't mess up that math.)
But what would be a single equation to do both steps at once?
Edit: I think I may have found the simplification:
P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)]
P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999]
P(H|Aa, Af) = 0.000067
Which is the same result (slight difference probably due to my rounding in the intermediate step of the separate calculations). Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
probability bayes-theorem
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
$endgroup$
add a comment |
$begingroup$
I'm trying to use Bayes' theorem to calculate the likelihood, in my example, of religion H being true considering evidence for and against it. I think I understand Bayes' theorem well enough to do this in two steps, calculating the odds that H is true considering evidence against it, and then taking that result as a new prior probability to factor in evidence for it.
I'm sure there's a way to combine those two steps into one, but the explanations I found were not that clear and used a notation that I'm not very familiar with.
So for example prior probability H is true due to the number of possible true religions P(H) is 0.001, the likelihood we'd find particular arguments against it if it were true P(Aa|H) is 0.01, the likelihood that we'd find particular arguments against it if it were not true P(Aa|H') is 0.6, the likelihood we'd find particular arguments for it if it were true P(Af|H) is 0.8, and the likelihood that we'd find particular arguments for it if it were not true P(Af|H') is 0.2. And in my example the arguments against it and arguments for it are both observed.
So then I do the math in two stages, first considering arguments against it:
P(H|Aa) = P(Aa|H) * P(H) / [P(Aa|H) * P(H) + P(Aa|H’) * P(H’)]
P(H|Aa) = 0.01 * 0.001 / [0.01 * 0.001 + 0.6 * 0.999]
P(H|Aa) = 0.000017
So now I take that and calculate considering arguments for it:
P(H|Af) = P(Af|H) * P(H) / [P(Af|H) * P(H) + P(Af|H’) * P(H’)]
P(H|Af) = 0.8 * 0.000017 / [0.8 * 0.000017 + 0.2 * 0.999983]
P(H|Af) = 0.000068
This gives me a 0.0068% probability that the religion is true given the above assumptions. (I hope I didn't mess up that math.)
But what would be a single equation to do both steps at once?
Edit: I think I may have found the simplification:
P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)]
P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999]
P(H|Aa, Af) = 0.000067
Which is the same result (slight difference probably due to my rounding in the intermediate step of the separate calculations). Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
probability bayes-theorem
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
$endgroup$
I'm trying to use Bayes' theorem to calculate the likelihood, in my example, of religion H being true considering evidence for and against it. I think I understand Bayes' theorem well enough to do this in two steps, calculating the odds that H is true considering evidence against it, and then taking that result as a new prior probability to factor in evidence for it.
I'm sure there's a way to combine those two steps into one, but the explanations I found were not that clear and used a notation that I'm not very familiar with.
So for example prior probability H is true due to the number of possible true religions P(H) is 0.001, the likelihood we'd find particular arguments against it if it were true P(Aa|H) is 0.01, the likelihood that we'd find particular arguments against it if it were not true P(Aa|H') is 0.6, the likelihood we'd find particular arguments for it if it were true P(Af|H) is 0.8, and the likelihood that we'd find particular arguments for it if it were not true P(Af|H') is 0.2. And in my example the arguments against it and arguments for it are both observed.
So then I do the math in two stages, first considering arguments against it:
P(H|Aa) = P(Aa|H) * P(H) / [P(Aa|H) * P(H) + P(Aa|H’) * P(H’)]
P(H|Aa) = 0.01 * 0.001 / [0.01 * 0.001 + 0.6 * 0.999]
P(H|Aa) = 0.000017
So now I take that and calculate considering arguments for it:
P(H|Af) = P(Af|H) * P(H) / [P(Af|H) * P(H) + P(Af|H’) * P(H’)]
P(H|Af) = 0.8 * 0.000017 / [0.8 * 0.000017 + 0.2 * 0.999983]
P(H|Af) = 0.000068
This gives me a 0.0068% probability that the religion is true given the above assumptions. (I hope I didn't mess up that math.)
But what would be a single equation to do both steps at once?
Edit: I think I may have found the simplification:
P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)]
P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999]
P(H|Aa, Af) = 0.000067
Which is the same result (slight difference probably due to my rounding in the intermediate step of the separate calculations). Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
probability bayes-theorem
probability bayes-theorem
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
edited Jul 25 '17 at 21:46
Aaliyah
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
asked Jul 25 '17 at 19:23
AaliyahAaliyah
537
537
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Maybe this general formula for multiple conditions helps?
$$P(A|B, C)= frac{P(B|A, C) * P(A|C)}{P(B|C)} $$
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
$endgroup$
$begingroup$
Can you explain that in terms of the arithmetic like my example above? How do you reach the values for P(B|A,C), P(A|C), P(B|C), and what do they represent?
$endgroup$
– Aaliyah
Jul 25 '17 at 20:29
$begingroup$
@Aaliyah It's a general formula, so you could say $A = H$, $B=Aa$, and $C= Af$. Something like $P(H|Aa,Af)$ would then be the probability that the religion is true given both the arguments against and the arguments for it (which is what you are looking for, right?). Now, the tricky one is $P(Aa|H,Af)$, which you can't really derive from the others ... indeed, I strongly suspect you did something wrong in your calculations and treated a conditional probability as a kind of event: note how you first calculate $P(H|Aa)$, and then plug in that value for $P(H)$ later ... as if $H = H|Aa$ ...
$endgroup$
– Bram28
Jul 25 '17 at 20:44
$begingroup$
Why can't I treat it as a kind of event? Isn't it adding new information? Whether I start with calculating with arguments against or I start with calculating for arguments for, I end up with the same ultimate result. (Well, I got 0.0067% instead of 0.0068%, but I will chalk the difference up to my rounding for intermediate steps.)
$endgroup$
– Aaliyah
Jul 25 '17 at 21:16
$begingroup$
I think I may have found the simplification: P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)] P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999] P(H|Aa, Af) = 0.000067 Which is the same result. Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
$endgroup$
– Aaliyah
Jul 25 '17 at 21:39
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2371536%2fhelp-simplifying-bayes-theorem-for-multiple-conditions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Maybe this general formula for multiple conditions helps?
$$P(A|B, C)= frac{P(B|A, C) * P(A|C)}{P(B|C)} $$
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
$endgroup$
$begingroup$
Can you explain that in terms of the arithmetic like my example above? How do you reach the values for P(B|A,C), P(A|C), P(B|C), and what do they represent?
$endgroup$
– Aaliyah
Jul 25 '17 at 20:29
$begingroup$
@Aaliyah It's a general formula, so you could say $A = H$, $B=Aa$, and $C= Af$. Something like $P(H|Aa,Af)$ would then be the probability that the religion is true given both the arguments against and the arguments for it (which is what you are looking for, right?). Now, the tricky one is $P(Aa|H,Af)$, which you can't really derive from the others ... indeed, I strongly suspect you did something wrong in your calculations and treated a conditional probability as a kind of event: note how you first calculate $P(H|Aa)$, and then plug in that value for $P(H)$ later ... as if $H = H|Aa$ ...
$endgroup$
– Bram28
Jul 25 '17 at 20:44
$begingroup$
Why can't I treat it as a kind of event? Isn't it adding new information? Whether I start with calculating with arguments against or I start with calculating for arguments for, I end up with the same ultimate result. (Well, I got 0.0067% instead of 0.0068%, but I will chalk the difference up to my rounding for intermediate steps.)
$endgroup$
– Aaliyah
Jul 25 '17 at 21:16
$begingroup$
I think I may have found the simplification: P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)] P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999] P(H|Aa, Af) = 0.000067 Which is the same result. Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
$endgroup$
– Aaliyah
Jul 25 '17 at 21:39
add a comment |
$begingroup$
Maybe this general formula for multiple conditions helps?
$$P(A|B, C)= frac{P(B|A, C) * P(A|C)}{P(B|C)} $$
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
$endgroup$
$begingroup$
Can you explain that in terms of the arithmetic like my example above? How do you reach the values for P(B|A,C), P(A|C), P(B|C), and what do they represent?
$endgroup$
– Aaliyah
Jul 25 '17 at 20:29
$begingroup$
@Aaliyah It's a general formula, so you could say $A = H$, $B=Aa$, and $C= Af$. Something like $P(H|Aa,Af)$ would then be the probability that the religion is true given both the arguments against and the arguments for it (which is what you are looking for, right?). Now, the tricky one is $P(Aa|H,Af)$, which you can't really derive from the others ... indeed, I strongly suspect you did something wrong in your calculations and treated a conditional probability as a kind of event: note how you first calculate $P(H|Aa)$, and then plug in that value for $P(H)$ later ... as if $H = H|Aa$ ...
$endgroup$
– Bram28
Jul 25 '17 at 20:44
$begingroup$
Why can't I treat it as a kind of event? Isn't it adding new information? Whether I start with calculating with arguments against or I start with calculating for arguments for, I end up with the same ultimate result. (Well, I got 0.0067% instead of 0.0068%, but I will chalk the difference up to my rounding for intermediate steps.)
$endgroup$
– Aaliyah
Jul 25 '17 at 21:16
$begingroup$
I think I may have found the simplification: P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)] P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999] P(H|Aa, Af) = 0.000067 Which is the same result. Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
$endgroup$
– Aaliyah
Jul 25 '17 at 21:39
add a comment |
$begingroup$
Maybe this general formula for multiple conditions helps?
$$P(A|B, C)= frac{P(B|A, C) * P(A|C)}{P(B|C)} $$
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
$endgroup$
Maybe this general formula for multiple conditions helps?
$$P(A|B, C)= frac{P(B|A, C) * P(A|C)}{P(B|C)} $$
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
edited Jul 25 '17 at 19:47
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
answered Jul 25 '17 at 19:30
Bram28Bram28
62.2k44793
62.2k44793
$begingroup$
Can you explain that in terms of the arithmetic like my example above? How do you reach the values for P(B|A,C), P(A|C), P(B|C), and what do they represent?
$endgroup$
– Aaliyah
Jul 25 '17 at 20:29
$begingroup$
@Aaliyah It's a general formula, so you could say $A = H$, $B=Aa$, and $C= Af$. Something like $P(H|Aa,Af)$ would then be the probability that the religion is true given both the arguments against and the arguments for it (which is what you are looking for, right?). Now, the tricky one is $P(Aa|H,Af)$, which you can't really derive from the others ... indeed, I strongly suspect you did something wrong in your calculations and treated a conditional probability as a kind of event: note how you first calculate $P(H|Aa)$, and then plug in that value for $P(H)$ later ... as if $H = H|Aa$ ...
$endgroup$
– Bram28
Jul 25 '17 at 20:44
$begingroup$
Why can't I treat it as a kind of event? Isn't it adding new information? Whether I start with calculating with arguments against or I start with calculating for arguments for, I end up with the same ultimate result. (Well, I got 0.0067% instead of 0.0068%, but I will chalk the difference up to my rounding for intermediate steps.)
$endgroup$
– Aaliyah
Jul 25 '17 at 21:16
$begingroup$
I think I may have found the simplification: P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)] P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999] P(H|Aa, Af) = 0.000067 Which is the same result. Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
$endgroup$
– Aaliyah
Jul 25 '17 at 21:39
add a comment |
$begingroup$
Can you explain that in terms of the arithmetic like my example above? How do you reach the values for P(B|A,C), P(A|C), P(B|C), and what do they represent?
$endgroup$
– Aaliyah
Jul 25 '17 at 20:29
$begingroup$
@Aaliyah It's a general formula, so you could say $A = H$, $B=Aa$, and $C= Af$. Something like $P(H|Aa,Af)$ would then be the probability that the religion is true given both the arguments against and the arguments for it (which is what you are looking for, right?). Now, the tricky one is $P(Aa|H,Af)$, which you can't really derive from the others ... indeed, I strongly suspect you did something wrong in your calculations and treated a conditional probability as a kind of event: note how you first calculate $P(H|Aa)$, and then plug in that value for $P(H)$ later ... as if $H = H|Aa$ ...
$endgroup$
– Bram28
Jul 25 '17 at 20:44
$begingroup$
Why can't I treat it as a kind of event? Isn't it adding new information? Whether I start with calculating with arguments against or I start with calculating for arguments for, I end up with the same ultimate result. (Well, I got 0.0067% instead of 0.0068%, but I will chalk the difference up to my rounding for intermediate steps.)
$endgroup$
– Aaliyah
Jul 25 '17 at 21:16
$begingroup$
I think I may have found the simplification: P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)] P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999] P(H|Aa, Af) = 0.000067 Which is the same result. Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
$endgroup$
– Aaliyah
Jul 25 '17 at 21:39
$begingroup$
Can you explain that in terms of the arithmetic like my example above? How do you reach the values for P(B|A,C), P(A|C), P(B|C), and what do they represent?
$endgroup$
– Aaliyah
Jul 25 '17 at 20:29
$begingroup$
Can you explain that in terms of the arithmetic like my example above? How do you reach the values for P(B|A,C), P(A|C), P(B|C), and what do they represent?
$endgroup$
– Aaliyah
Jul 25 '17 at 20:29
$begingroup$
@Aaliyah It's a general formula, so you could say $A = H$, $B=Aa$, and $C= Af$. Something like $P(H|Aa,Af)$ would then be the probability that the religion is true given both the arguments against and the arguments for it (which is what you are looking for, right?). Now, the tricky one is $P(Aa|H,Af)$, which you can't really derive from the others ... indeed, I strongly suspect you did something wrong in your calculations and treated a conditional probability as a kind of event: note how you first calculate $P(H|Aa)$, and then plug in that value for $P(H)$ later ... as if $H = H|Aa$ ...
$endgroup$
– Bram28
Jul 25 '17 at 20:44
$begingroup$
@Aaliyah It's a general formula, so you could say $A = H$, $B=Aa$, and $C= Af$. Something like $P(H|Aa,Af)$ would then be the probability that the religion is true given both the arguments against and the arguments for it (which is what you are looking for, right?). Now, the tricky one is $P(Aa|H,Af)$, which you can't really derive from the others ... indeed, I strongly suspect you did something wrong in your calculations and treated a conditional probability as a kind of event: note how you first calculate $P(H|Aa)$, and then plug in that value for $P(H)$ later ... as if $H = H|Aa$ ...
$endgroup$
– Bram28
Jul 25 '17 at 20:44
$begingroup$
Why can't I treat it as a kind of event? Isn't it adding new information? Whether I start with calculating with arguments against or I start with calculating for arguments for, I end up with the same ultimate result. (Well, I got 0.0067% instead of 0.0068%, but I will chalk the difference up to my rounding for intermediate steps.)
$endgroup$
– Aaliyah
Jul 25 '17 at 21:16
$begingroup$
Why can't I treat it as a kind of event? Isn't it adding new information? Whether I start with calculating with arguments against or I start with calculating for arguments for, I end up with the same ultimate result. (Well, I got 0.0067% instead of 0.0068%, but I will chalk the difference up to my rounding for intermediate steps.)
$endgroup$
– Aaliyah
Jul 25 '17 at 21:16
$begingroup$
I think I may have found the simplification: P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)] P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999] P(H|Aa, Af) = 0.000067 Which is the same result. Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
$endgroup$
– Aaliyah
Jul 25 '17 at 21:39
$begingroup$
I think I may have found the simplification: P(H|Aa, Af) = P(Aa|H) * P(Af|H) * P(H) / [P(Aa|H) * P(Af|H) * P(H) + P(Aa|H’) * P(Af|H’) * P(H’)] P(H|Aa, Af) = 0.01 * 0.8 * 0.001 / [0.01 * 0.8 * 0.001 + 0.6 * 0.2 * 0.999] P(H|Aa, Af) = 0.000067 Which is the same result. Is there any reason I shouldn't be doing this? I am looking at evidence for and against as "if H is true, how likely would I expect to see Aa? Not likely. But Aa is present, so that's new information."
$endgroup$
– Aaliyah
Jul 25 '17 at 21:39
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2371536%2fhelp-simplifying-bayes-theorem-for-multiple-conditions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
