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I know that if you have a set of data, say $data=(x_1,x_2,x_3,...,x_n)$ that you think came from a discrete distribution, you can calculate the probability of that data occurring by just multiplying the PDF. e.g.:

if:$$data \sim binomial(n,p) = p(x)$$

then the probability of some $data=(x_1,x_2,x_3,...,x_n)$, would be: $$prob(data|binomial(n,p)) = p(x_1)p(x_2)...p(x_n)=\prod_i p(x_i)$$

However, how do you do the same thing for a continuous distribution? If the probability at any given point is $0$, then the probability that data existed at exactly a single point is $0$ and thus $prob(data|distribution)=0$. The best you could do is say the probability is some value $p(x_i)dx$, but this is still infinitesimally small.

It is my understanding that in order for a continuous distribution to "behave" well as a measure on some sigma algebra, it cannot have a probability associated with a single point.

Could someone help me understand where my logic is flawed?

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Don't see any flaw in your logic. The probability of getting the observed data is zero, but that's okay.

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