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

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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

share|cite|improve this question

asked 6 hours ago

eric_kernfeld

2,414421

add a comment | 

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

share|cite|improve this question

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

share|cite|improve this question

asked 6 hours ago

eric_kernfeld

2,414421

share|cite|improve this question

asked 6 hours ago

eric_kernfeld

2,414421

share|cite|improve this question

share|cite|improve this question

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

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Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracP(XP(X) P(C) + P(H) [1 - P(C)],
$$
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.)

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answered 6 hours ago

tddevlin

1,124416

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  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 | 

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1 Answer
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active

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1 Answer
1

active

oldest

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active

oldest

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active

oldest

votes

up vote
3
down vote

Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracP(XP(X) P(C) + P(H) [1 - P(C)],
$$
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.)

share|cite|improve this answer

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 | 

up vote
3
down vote

Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracP(XP(X) P(C) + P(H) [1 - P(C)],
$$
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.)

share|cite|improve this answer

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 | 

up vote
3
down vote

up vote
3
down vote

Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracP(XP(X) P(C) + P(H) [1 - P(C)],
$$
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.)

share|cite|improve this answer

answered 6 hours ago

tddevlin

1,124416

Well by distributing the $P(H)$ term, we obtain
$$
P(H|X) = fracP(XP(X) P(C) + P(H) [1 - P(C)],
$$
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.)

share|cite|improve this answer

answered 6 hours ago

tddevlin

1,124416

share|cite|improve this answer

share|cite|improve this answer

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 | 

• 
  
  
  

  
  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 | 

 

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