I've been reading some about Bayesian statistics, and am a little confused. I never really covered Bayesian stuff as an undergrad, and am now a physicist, so I haven't had to learn it.(adsbygoogle = window.adsbygoogle || []).push({});

I have been trying to think of things in terms of more interesting examples than the trivial coin flipping / ball grabbing ones I've seen in books, but am apparently a little confused about how to interpret everything.

Let's suppose I have two events,

[tex]H[/tex] = I do something.

[tex]E[/tex] = someone tries to stop me from doing it.

Let my prior probability of doing something, [tex]P(H)=h[/tex].

Then, the probability I do something given that someone tries to stop me is,

[tex]P(H|E) = \frac{P(E|H)P(H)}{P(E)}[/tex]

where

[tex]P(E) = P(E|H)P(H) + P(E|!H)P(!H)[/tex] because the only options are me doing it or not ([tex]H[/tex] and [tex]!H[/tex]).

Let's call [tex]P(E|H) = x[/tex], [tex]P(E|!H)=y[/tex].

Then,

[tex]P(H|E) = \frac{P(E|H)P(H)}{P(E|H)P(H) + P(E|!H)P(!H)} = \frac{h x}{h x + y (1-h)}[/tex].

But, this means that if [tex]x\approx 1, y \approx 0[/tex], then [tex]P(H|E) > h[/tex].

[tex]P(E|H) = x = [/tex] the probability they try to stop me, given I do something,

[tex]P(E|!H) = y = [/tex] the probability they try to stop me, given I don't do something.

So, the probability of me doing somethingincreasesgiven that they try to stop me from doing it, provided that they would not try to stop me if I didn't do it?

This doesn't make any sense to me, because it clearly contradicts the purpose of event E.

Where has my interpretation gone wrong? How would I properly model something like this?

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# Bayesian statistics question

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