Product/factor distribution.


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7 months ago by
Hi,

I'm referring to page 16 of this document: https://www.cl.cam.ac.uk/teaching/1718/DataSci/notes.pdf.

I am assuming that the formula for the product distribution there is \(\mathbb{P}(A=a,B=b,C=c,D=d) = \kappa\mathbb{P}(A=a|B=b,C=c,D=d)\mathbb{P}(B=b|C=c,D=d,A=a)\mathbb{P}(C=c|D=d,A=a,B=b)\mathbb{P}(D=d|A=a,B=b,C=c)\).

Is that the case, and if so, how would you prove it? I only see that \(\mathbb{P}(A=a,B=b,C=c,D=d) = \mathbb{P}(A=a|B=b,C=c,D=d)\mathbb{P}(B=b|C=c,D=d)\mathbb{P}(C=c|D=d)\mathbb{P}(D=d)\) but this doesn't help.

Thanks
Your second expression seems to be right to me, since it matches their definition of discriminatory model.

"A discriminative model is a type of descriptive model, written as the marginal distribution of some variable of interest conditional on features."

Is there a reason why you think it is the first expression?
written 7 months ago by Richie Yeung  

The reason I thought it was the first expression was that all four probabilities on the RHS were written in the notes as if they were conditional on something, which is not matched by the latter equation because of the last term.

Also, if you only got conditional distributions to begin with, how would you get things like P(D=d) out of that? If what you got were the terms P(A|B,C,D), P(B|C,D), P(C|D), P(D) or something that determines these terms, than the resulting joint distribution would be unique, wouldn't it? And the notes explicitly say the 'factor distribution' need not be the unique one consistent with the input.

written 7 months ago by Kuba Perlin  
I think its trying to say there could be multiple solutions to the factor distribution.

I think its only necessary to find one set of values that satisfy the marginals.

The "..." could be making the point that you could also write it as

\(\mathbb{P}(A=a,B=b,C=c,D=d) = \mathbb{P}(A=a|B=b,C=c,D=d)\mathbb{P}(B=b|C=c,D=d)\mathbb{P}(D=d|C=c)\mathbb{P}(C=c)\)

for example.
written 7 months ago by Richie Yeung  
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