Help with Bayes Theorem sought...
Jul. 5th, 2001 03:23 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
D'oh! I worked it out. Sorry for wasting the time of anyone who thought about this.
Obsolete plea for help with my cack-handed probability theory follows. The correct formula is here:
http://www.livejournal.com/talkread.bml?itemid=6362421
gothslut,
kitty_goth,
ajva, anyone?
I'm trying to do some sums using Bayes Theorem and I'm not getting the results I expect. I thought I knew this stuff, but it's all going wrong - help!
Consider two events, L and V. Bayes theorem states
P(V|L)P(L) = P(V & L) = P(L|V)P(V)
therefore P(L)/P(L|V) = P(V)/P(V|L)
= (P(V|L) + P(V|~L))/P(V|L) <- here's the error
= 1 + P(V|~L)/P(V|L)
therefore P(V|~L)/P(V|L) = P(L)/P(L|V) -1, yes?
But this doesn't ring true. Suppose V is an illness. 6 people on earth do something that guarantees that you get the illness. The other 6 billion of us take a one-in-a-billion risk of it, so there are 6 more sufferers who took that risk and lost. Let L be the probability we're in the low risk group (about 1-1E-9) and V be the probability we have the illness. I've stated that P(V|~L) = 1 (all those outside the low risk group get it) and P(V|L) = 1E-9 (only one in a billion get it in the low risk group). So the left hand side should be a billion, indicating that those outside the low risk group run a billion times greater risk.
However, half the sufferers come from the low risk group, ie P(L|V) = 0.5. P(L) is almost 1, so the right hand side comes out as 1/0.5 -1 = 1.
What am I doing wrong? I keep checking my algebra but to no avail.
I'm sure you can guess why I'm doing these sums, BTW...
Obsolete plea for help with my cack-handed probability theory follows. The correct formula is here:
http://www.livejournal.com/talkread.bml?itemid=6362421
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I'm trying to do some sums using Bayes Theorem and I'm not getting the results I expect. I thought I knew this stuff, but it's all going wrong - help!
Consider two events, L and V. Bayes theorem states
P(V|L)P(L) = P(V & L) = P(L|V)P(V)
therefore P(L)/P(L|V) = P(V)/P(V|L)
= (P(V|L) + P(V|~L))/P(V|L) <- here's the error
= 1 + P(V|~L)/P(V|L)
therefore P(V|~L)/P(V|L) = P(L)/P(L|V) -1, yes?
But this doesn't ring true. Suppose V is an illness. 6 people on earth do something that guarantees that you get the illness. The other 6 billion of us take a one-in-a-billion risk of it, so there are 6 more sufferers who took that risk and lost. Let L be the probability we're in the low risk group (about 1-1E-9) and V be the probability we have the illness. I've stated that P(V|~L) = 1 (all those outside the low risk group get it) and P(V|L) = 1E-9 (only one in a billion get it in the low risk group). So the left hand side should be a billion, indicating that those outside the low risk group run a billion times greater risk.
However, half the sufferers come from the low risk group, ie P(L|V) = 0.5. P(L) is almost 1, so the right hand side comes out as 1/0.5 -1 = 1.
What am I doing wrong? I keep checking my algebra but to no avail.
I'm sure you can guess why I'm doing these sums, BTW...
So..
Date: 2001-07-05 08:16 am (UTC)M xx
Re: So..
Date: 2001-07-05 08:34 am (UTC)http://www.livejournal.com/talkread.bml?itemid=6362421
I'm pretty sure the second formula is correct; at least it rings true...