What's the name of the theorem that tells us that the total area under any probability density function, discrete or continuous, equals 1?

My stats book actually defines a PDF by requiring that $$\sum_{x}f(x)=1\quad\text{or}\quad\int_{-\infty}^{\infty}f(x)=1$$

In other words, $f(x)$ is a PDF only if the above is true (along with a few other requirements).

Rather than using this as a definition, is it possible to prove that it's true for any PDF? If so, has it been done before? I'd like the name of the theorem/proof so I can reference it.

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    $\begingroup$ This is one of the Kolmogorov axioms of probability: see "Second Axiom" at en.wikipedia.org/wiki/Probability_axioms. If you want a "proof," please stipulate what alternative axiomatization you are using. $\endgroup$
    – whuber
    Dec 16, 2012 at 21:46
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    $\begingroup$ Thanks. However, even if that comment might have cleared things up for you--and I am glad if it did--it doesn't appear to answer the question you have asked: it cites no theorem or proof. Unless your question is clarified to indicate what axioms you wish to use for the proof--or perhaps changed to indicate that a set of axioms would be fine, too--it appears to be unanswerable. $\endgroup$
    – whuber
    Dec 16, 2012 at 22:01
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    $\begingroup$ The answer to your question (and implicit in @whuber 's comment, is "No". Just like you can't prove that a rectangle has four sides. $\endgroup$
    – Peter Flom
    Dec 16, 2012 at 22:10
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    $\begingroup$ These are not unprovable, they are trivially provable. $\endgroup$ Dec 17, 2012 at 0:06
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    $\begingroup$ I think the point @whuber makes is a good one which is that to answer the question we need to know which axioms are acceptable as the basis for the proof, if the statement is not to be accepted as a defining axiom itself. Equivalent of the point made about defining a rectangle as an enclosed figure with four right angles (then you might be able to prove it has four sides). $\endgroup$ Dec 17, 2012 at 10:12

2 Answers 2


Already well answered in comments. The requirement that the integral $\int_{-\infty}^\infty f(x)\; dx=1$ (together with $f(x)\ge 0\;\;\text{for all $x$}$) is part of the definition of a probability density function. And it must be so since that integral represents the total probability, and part of Kolmogorov's axioms for probability is that the total probability is always 1.

So a proof from those axioms is very short, it just consists in pointing out that the statement is part of the axioms (as above). Somebody will say that that is not really a proof, I guess that will depend on your exact definition of proof.


What's the name of the theorem ...

This requirement is imposed by the norming axiom of probability theory, which holds that any probability measure has unit value ovver the sample space ---i.e., $\mathbb{P}(\Omega) = 1$. The restriction on the density is then obtained by recognising that, for a random variable $X: \Omega \rightarrow \mathbb{R}$, we have:

$$\mathbb{P}(X^{-1}(\mathcal{A})) = \text{Pr} (\mathcal{A}) = \int \limits_\mathcal{A} f_X(x) dx.$$

It is worth noting that the "axioms" of probability are usually taken as the starting place for work in this field, and within this limited context, they are not derived from earlier results. Within the usual probability framework these are not treated as results that you prove via theorems; you merely assert them as axioms. If you would like to treat them this way then you will need to go deeper into the literature on measurement of uncertainty, and look at how the standard probability measure is derived from more primitive conditions/desiderata. For example, Jaynes (2003) derives the probability "axioms" as consequences of consistency requirements for measuring uncertainty.


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