Paradox of Probability Distribution Functions (Adjusted for comments) This is a problem I have tried to work out for a while:

*

*We can not take the integral of a probability density function at a single point - as a matter of fact, we can not take the integral of any function at a single point. This means that if we have a Normal Probability Density Function, we can not find out the probability of observing a specific point $x$ as it will be zero - we can only find out the probability of observing a value in a range of points between $x_1$ and $x_2$.


*But at the same time, we usually say that the mode of a probability distribution (e.g. a Normal Distribution where the mode is equal to the mean) is the most frequent point. For example, if humans have their height distributed according to a Normal Distribution with 160 cm (and some standard deviation "sigma") - we naturally expect that the most common height of a human will be 160 cm. But using the first point listed above, it is impossible to calculate the probability of observing a human having a height exactly of 160 cm. We can either find out the probability of observing a human having a height between 159-161 cm - or if we really wanted, we could find out the probability of observing a human having a height between 159.999 cm and 160.0001 cm. But going ahead with the latter option will most likely result in a very small probability (e.g. 0.001), even though that range passes through the most frequent height according to our distribution.


*On the other hand, we could calculate the density or likelihood of observing a human with 160 cm - but unlike the probability distribution function, the likelihood function is not between 0 and 1, and is also more difficult to explain (e.g. a likelihood of 453 vs a likelihood of 8731).
This brings me to my questions: If the mode  height is the most frequent height in a (Normal) Probability Distribution Function - why can't we find the probability of observing a human with that height?
Or how can the mode of a normal distribution be the most frequent point despite having a probability of 0?
 A: First, you're correct: For any continuous random variable $X$, and any point $a$, it is always necessarily true that $\mathbb P(X = a) = 0$.
The mode of a continuous random variable is commonly defined to be the location of the maximum of the density function. This is also commonly understood to be the "most frequent value," which is slightly absurd because as noted above, no particular value is realized with positive probability. But what is true, under mild regularity assumptions, is that if you consider small intervals of fixed width, those that contain $m$ will have a higher probability of containing $X$ than those that do not. (Note also that these probabilities will be positive.) It is in this sense that the "most frequent value" understanding is justified. It's not that the mode is a possible value; it's just that you're more likely to find $X$ near the mode than somewhere farther away from it.
(*) To make this rigorous: if $X$ has a continuous density function $f(x)$, and $m$ is the mode of $X$, then there exists some $\epsilon_0$ such that:

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*if $I$ and $J$ are intervals of equal width $\epsilon \leq \epsilon_0$,

*and $m \in I$,

*and $m \not \in J$,

then $\mathbb P(X \in I) > \mathbb P(X \in J)$.
