So there is something interesting I have come across today. Suppose we want to solve an inference problem and arrive at the following quantity:

$$p(C_k|\mathbf{x}) = \frac{p(\mathbf{x}|C_k)p(C_k)}{p(\mathbf{x})} \tag{1}$$

What you can do according to Bishop's PRML book is model the intersection distribution $p(\mathbf{x},C_k)$ and normalize it, to arrive at the posterior: $p(C_k|\mathbf{x})$. But I thought about the following:

$$p(\mathbf{x}|C_k) = \frac{p(C_k|\mathbf{x})p(\mathbf{x})}{p(C_k)} \tag{2}$$

In both cases: ($1$),($2$), the numerator is $p(\mathbf{x},C_k).$ So how do we make sure that by normalizing $p(\mathbf{x},C_k)$ we actually arrive at ($1$) and not ($2$)? Is it because, ($2$) does not normalize the intersection distribution. I would assume so, since there is only one constant that can do so. In which case, how can I show this mathematically?


1 Answer 1



$$\int p(C_k|\mathbf{x}) dC_k=\int \frac{p(\mathbf{x}|C_k)p(C_k)}{p(\mathbf{x})}dC_k = \frac{1}{p(\mathbf{x})} \int p(\mathbf{x},C_k)dC_k = \frac{p(\mathbf{x})}{p(\mathbf{x})}=1$$


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