I usually want to read your posts on Probability and Bayesianism, but stop in the middle, not because it’s not clear, but because I feel like I need to be working on something else. This is pretty much the first post of this sequence that I finished, and I’m glad that I did, because it makes me feel that my Bayesian understanding just improved!
What we’re basically doing here is making a type distinction between probability and likelihoods, to help us remember that likelihoods can’t be converted to probabilities without combining them with a prior.
It probably depends on the background of the reader, but for someone with a theoretical computer science one like me, that’s an amazingly helpful explanation.
Trying to understand your second picture, when comparing P(A|B) and L(B|A), there should be at least one value at which they agree, right? In your picture it’s the 10%. As far as I understand, it’s because P(A|B) assumes some fixed B (let’s say bi), and L(B|A) assumes some fixed A (let’s say aj). By your definition, L(bi|A)=L(bi|aj)=P(aj|bi)=P(aj|B), so there should be at least one value of L(B|A)equal to a value of P(A|B). For your drawing, we have bi=b2, and aj is either a1 or a4.
Is that correct, or am I completely of the mark?
I have a pet theory that some biases can be explained as a mix-up between probability and likelihood. (I don’t know if this is a good explanation.)
This looks like a very interesting theory, and the two examples given are pretty convincing. So I’m curious whether you know of cases where it fails to explain the bias you mentioned?
Trying to understand your second picture, when comparing P(A|B) and L(B|A), there should be at least one value at which they agree, right?
Ahh, I knew there was a bit of a risk there… I didn’t really start with a conditional distribution and pull out rows/columns, I just made up numbers to illustrate the summing to 1 / not summing to 1 distinction. Fortunately, as you mentioned, I happened to place one matching number in the two, allowing us to deduce that I’m illustrating the probabilities of different A given B=b2, and then the likelihoods of different B given A=a1, as you demonstrated.
This looks like a very interesting theory, and the two examples given are pretty convincing. So I’m curious whether you know of cases where it fails to explain the bias you mentioned?
No, but I must admit that I haven’t read deeply into the literature.
I usually want to read your posts on Probability and Bayesianism, but stop in the middle, not because it’s not clear, but because I feel like I need to be working on something else. This is pretty much the first post of this sequence that I finished, and I’m glad that I did, because it makes me feel that my Bayesian understanding just improved!
It probably depends on the background of the reader, but for someone with a theoretical computer science one like me, that’s an amazingly helpful explanation.
Trying to understand your second picture, when comparing P(A|B) and L(B|A), there should be at least one value at which they agree, right? In your picture it’s the 10%. As far as I understand, it’s because P(A|B) assumes some fixed B (let’s say bi), and L(B|A) assumes some fixed A (let’s say aj). By your definition, L(bi|A)=L(bi|aj)=P(aj|bi)=P(aj|B), so there should be at least one value of L(B|A) equal to a value of P(A|B). For your drawing, we have bi=b2, and aj is either a1 or a4.
Is that correct, or am I completely of the mark?
This looks like a very interesting theory, and the two examples given are pretty convincing. So I’m curious whether you know of cases where it fails to explain the bias you mentioned?
Typo
Ahh, I knew there was a bit of a risk there… I didn’t really start with a conditional distribution and pull out rows/columns, I just made up numbers to illustrate the summing to 1 / not summing to 1 distinction. Fortunately, as you mentioned, I happened to place one matching number in the two, allowing us to deduce that I’m illustrating the probabilities of different A given B=b2, and then the likelihoods of different B given A=a1, as you demonstrated.
No, but I must admit that I haven’t read deeply into the literature.
Thanks!