3 ms·
> p(mud | rain) = 1.0 > p(rain | mud) = p(rain) I'm not sure these statements say what you think they do. Assuming P(rain) > 0, the first statement p(mu
by tr352 9y ago
> p(mud | rain) = 1.0
> p(rain | mud) = p(rain)
I'm not sure these statements say what you think they do. Assuming P(rain) > 0, the first statement
p(mud | rain) = 1.0
is equivalent to
p(mud & rain) = p(rain).
The second statement therefore implies
p(rain | mud) = p(mud & rain).
Thus
p(rain & mud) / p(mud) = p(mud & rain).
This, however, leads to the surprising conclusion
p(mud) = 1.0.
According to your theory, it is always muddy.
- ryanmonroe 9y agoI think what's meant is something like p(mud_{t+1} | rain_{t}) = 1 and p(rain_{t+1} | mud_{t}) = p(rain). Say during 10 units of time it rains at points 1, 3, 4, 6, and 10 and it's muddy IFF it rained in the previous period. Then the above probability statements are true and it's only muddy 40% of the time.
- YeGoblynQueenne 9y agoOK, but that's not a big problem. My theory only needs to say that mud doesn't cause rain. Judea Pearl says that only his addition to Bayesian probabilities can do that. I guess I could have thought of a better example, but I like how ryanmonroe's comment interprets it, below.