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How does one go about proving that a function is the density of an absolutely continuous distribution. That is, what are the steps I need to go through to satisfy this definition?

For example, the standard cauchy distribution has $f(x):=\frac{1}{\pi}\frac{1}{1+x^2}$, $x\in\mathbb R$. What steps are required in this case?

Is it merely checking that it integrates to 1 over its potential x values?

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  • $\begingroup$ Each pdf or pmf by definition integrates to 1 no matter if it takes discrete or continuous variables. $\endgroup$
    – Tim
    Mar 18 '15 at 11:08
  • $\begingroup$ @Tim okay but is that the only condition that it needs to satisfy, or are there others i need to check for? Say you didnt know it was for the cauchy, but you just wanted to test it is a valid density for an absolutely continuous distribution, how would you go ahead $\endgroup$ Mar 18 '15 at 11:13
  • $\begingroup$ It has to output non-negative values (since there is no negative probability/density) that integrate to 1. $\endgroup$
    – Tim
    Mar 18 '15 at 11:17
  • $\begingroup$ @Tim so it suffices to show that it integrates to one and does not output negative values? Is this enough to prove that it is a probability distribution? $\endgroup$ Mar 18 '15 at 11:31
  • $\begingroup$ What is your definition of "absolutely continuous"? There are many, but one of them is that the distribution has a density (wrt Lebesgue measure), which is automatic since you are given the putative density function. This function must satisfy the axioms of probability, which include (1) no probability can be negative and (2) the total probability must be $1$. Although your question mentions (2), it does not refer to (1), which also must be checked. $\endgroup$
    – whuber
    Mar 18 '15 at 14:34
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The original question asks [taking some liberties] for the conditions under which one can prove that an absolutely continuous function is in fact the probability density function (PDF) of a probability distribution. While the definition of a probability distribution is useful, it is certainly not the whole answer. We also need to look at the relationship between PDFs and distributions.

A probability distribution is a function from a set of events to the unit interval, where as $f$ in this case is from $\mathbb{R} \rightarrow \mathbb{R}$. Points in $\mathbb{R}$ are not our events! Measurable sets in $\mathbb{R}$ are our events. So we attempt to define a distribution by defining the function

$$ P(A) = \int_A f d\mu = \int_A f(x)dx, $$ the last bit of notation being included to make some people more comfortable. Since $f$ is absolutely continuous (way more than measurable), $P$ is defined. The first two axioms from Tim's post, follow immediately if we assume:

$$ \int_{\mathbb{R}} f d\mu = 1, \\ f(x) \geq 0, \forall x \in \mathbb{R}. $$

Note that the second assumption there is stronger than we really need, but if $f$ is absolutely continuous, this is equivalent to the weaker statement $\int_A f d\mu \geq 0, \forall A$. The last axiom falls out of the countable additivity of the Lebesgue integral, namely:

$$ P(\cup_i A_i) = \int_{\cup_{A_i}}fd\mu = \sum_i \int_{A_i}fd\mu = \sum_i P(A_i). $$

This is one of those cases where, not so coincidentally, the structure of two mathematical objects (probability distributions and the Lebesgue integral) are in correspondence. These are import moments in math, and they can really provide a lot of insight.

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1.5 Definition. A function $\mathbb{P}$ that assigns a real number $\mathbb{P}(A)$ to each event $A$ is a probability distribution or a probability measure if it satisfies the following three axioms:

Axiom 1: $\mathbb{P}(A) \geq 0$ for every $A$
Axiom 2: $\mathbb{P}(\Omega) = 1$
Axiom 3: If $A_1,A_2,...$ are disjoint then $\mathbb{P}(\bigcup_{i=1}^\infty A_i) = \sum_{i=1}^\infty\mathbb{P}(A_i)$

Source: Wasserman, L. (2004). All of Statistics. Springer.

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  • $\begingroup$ How do we prove the third axiom for a continuous distribution $\endgroup$ Mar 18 '15 at 11:41

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