1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.