Jtdylktuy

I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?

I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.

But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.

Definition (Gut, 2005): A distribution function F is

• discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.


• continuous iff it is continuous for all x.


• absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.
  


• singular iff F $neq$ 0, F' exists and equals 0 a.e.
  

By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.

However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.