XKCD's modified Bayes theorem: actually kinda reasonable?
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I know this is from a comic famous for taking advantage of certain analytical tendencies, but it actually looks kind of reasonable after a few minutes of staring. Can anyone outline for me what this "modified Bayes theorem" is doing?
https://xkcd.com/2059/
bayesian hierarchical-bayesian
asked 6 hours ago
eric_kernfeld
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up vote
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I know this is from a comic famous for taking advantage of certain analytical tendencies, but it actually looks kind of reasonable after a few minutes of staring. Can anyone outline for me what this "modified Bayes theorem" is doing?
https://xkcd.com/2059/
bayesian hierarchical-bayesian
asked 6 hours ago
eric_kernfeld
2,414421
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I know this is from a comic famous for taking advantage of certain analytical tendencies, but it actually looks kind of reasonable after a few minutes of staring. Can anyone outline for me what this "modified Bayes theorem" is doing?
https://xkcd.com/2059/
bayesian hierarchical-bayesian
asked 6 hours ago
eric_kernfeld
2,414421
I know this is from a comic famous for taking advantage of certain analytical tendencies, but it actually looks kind of reasonable after a few minutes of staring. Can anyone outline for me what this "modified Bayes theorem" is doing?
https://xkcd.com/2059/
bayesian hierarchical-bayesian
bayesian hierarchical-bayesian
asked 6 hours ago
eric_kernfeld
2,414421
asked 6 hours ago
eric_kernfeld
2,414421
asked 6 hours ago
eric_kernfeld
2,414421
asked 6 hours ago
eric_kernfeld
2,414421
asked 6 hours ago
eric_kernfeld
2,414421
2,414421
add a comment |Â
add a comment |Â
1 Answer
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Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracH)P(H)P(X) P(C) + P(H) [1 - P(C)],
$$
which we can interpret as the Law of Total probability applied to the event $C =$ "you are using Bayesian statistics correctly." So if you are using Bayesian statistics correctly, then you recover Bayes' law (the left fraction above) and if you aren't, then you ignore the data and just use your prior on $H$.
I suppose this is a rejoinder against the criticism that in principle Bayesians can adjust the prior to support whatever conclusion they want, whereas Bayesians would argue that this is not how Bayesian statistics actually works.
(And yes, you did successfully nerd-snipe me. I'm neither a mathematician nor a physicist though, so I'm not sure how many points I'm worth.)
answered 6 hours ago
tddevlin
1,124416
3
A clever joke that's embedded in the formula above is that if you're not using Bayesian statistics correctly, your inference is completely independent of the truth.
– Cliff AB
4 hours ago
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracH)P(H)P(X) P(C) + P(H) [1 - P(C)],
$$
which we can interpret as the Law of Total probability applied to the event $C =$ "you are using Bayesian statistics correctly." So if you are using Bayesian statistics correctly, then you recover Bayes' law (the left fraction above) and if you aren't, then you ignore the data and just use your prior on $H$.
I suppose this is a rejoinder against the criticism that in principle Bayesians can adjust the prior to support whatever conclusion they want, whereas Bayesians would argue that this is not how Bayesian statistics actually works.
(And yes, you did successfully nerd-snipe me. I'm neither a mathematician nor a physicist though, so I'm not sure how many points I'm worth.)
answered 6 hours ago
tddevlin
1,124416
3
A clever joke that's embedded in the formula above is that if you're not using Bayesian statistics correctly, your inference is completely independent of the truth.
– Cliff AB
4 hours ago
add a comment |Â
up vote
3
down vote
Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracH)P(H)P(X) P(C) + P(H) [1 - P(C)],
$$
which we can interpret as the Law of Total probability applied to the event $C =$ "you are using Bayesian statistics correctly." So if you are using Bayesian statistics correctly, then you recover Bayes' law (the left fraction above) and if you aren't, then you ignore the data and just use your prior on $H$.
I suppose this is a rejoinder against the criticism that in principle Bayesians can adjust the prior to support whatever conclusion they want, whereas Bayesians would argue that this is not how Bayesian statistics actually works.
(And yes, you did successfully nerd-snipe me. I'm neither a mathematician nor a physicist though, so I'm not sure how many points I'm worth.)
answered 6 hours ago
tddevlin
1,124416
3
A clever joke that's embedded in the formula above is that if you're not using Bayesian statistics correctly, your inference is completely independent of the truth.
– Cliff AB
4 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracH)P(H)P(X) P(C) + P(H) [1 - P(C)],
$$
which we can interpret as the Law of Total probability applied to the event $C =$ "you are using Bayesian statistics correctly." So if you are using Bayesian statistics correctly, then you recover Bayes' law (the left fraction above) and if you aren't, then you ignore the data and just use your prior on $H$.
I suppose this is a rejoinder against the criticism that in principle Bayesians can adjust the prior to support whatever conclusion they want, whereas Bayesians would argue that this is not how Bayesian statistics actually works.
(And yes, you did successfully nerd-snipe me. I'm neither a mathematician nor a physicist though, so I'm not sure how many points I'm worth.)
answered 6 hours ago
tddevlin
1,124416
Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracH)P(H)P(X) P(C) + P(H) [1 - P(C)],
$$
which we can interpret as the Law of Total probability applied to the event $C =$ "you are using Bayesian statistics correctly." So if you are using Bayesian statistics correctly, then you recover Bayes' law (the left fraction above) and if you aren't, then you ignore the data and just use your prior on $H$.
I suppose this is a rejoinder against the criticism that in principle Bayesians can adjust the prior to support whatever conclusion they want, whereas Bayesians would argue that this is not how Bayesian statistics actually works.
(And yes, you did successfully nerd-snipe me. I'm neither a mathematician nor a physicist though, so I'm not sure how many points I'm worth.)
answered 6 hours ago
tddevlin
1,124416
answered 6 hours ago
tddevlin
1,124416
answered 6 hours ago
tddevlin
1,124416
answered 6 hours ago
tddevlin
1,124416
1,124416
3
A clever joke that's embedded in the formula above is that if you're not using Bayesian statistics correctly, your inference is completely independent of the truth.
– Cliff AB
4 hours ago
add a comment |Â
3
A clever joke that's embedded in the formula above is that if you're not using Bayesian statistics correctly, your inference is completely independent of the truth.
– Cliff AB
4 hours ago
3
3
A clever joke that's embedded in the formula above is that if you're not using Bayesian statistics correctly, your inference is completely independent of the truth.
– Cliff AB
4 hours ago
A clever joke that's embedded in the formula above is that if you're not using Bayesian statistics correctly, your inference is completely independent of the truth.
– Cliff AB
4 hours ago
add a comment |Â
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