Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Usually in probability theory, for a random variable whose value is in $\mathbb{R}$, we talk about its cumulative distribution function $F(x)$ and then its density $f(x)$, in good enough cases $F'(x)=f(x)$.

That's the setup I'm familiar with, so I got annoyed when physicists talk about unnormalized "densities". E.g. if the probabilistic density of the position of a particle on $\mathbb{R}$ is equal to 1 everywhere, that means it is equally likely to appear anywhere. More generally you can imagine them talking about a non-negative function $f(x)$ being the density of something with the density $f(x)$ is not integrable on $\mathbb{R}$ but locally integrable, i.e. $\int_{[a,b]}f(x)dx$ make sense for $a\leq b, a, b\in\mathbb{R}$. As $\int_{\mathbb{R}}f(x)dx$ is undefined ($=\infty$), one cannot divide by it to normalize.

Is there a mathematical way to make sense of such statements? There is one I have in mind, namely, one can talk about the density of some random variable $X$ up to a scalar multiple, such that for any intervals $[a,b]\subset [c,d]$ we can express the conditional probability as a quotient of integrals:

$$P(X\in[a,b] \big| X\in[c,d])=\dfrac{\int_{[a,b]}f(x)dx}{\int_{[c,d]}f(x)dx}.$$

I only know very basic probability theory so I don't know if this makes sense. Am I allowed to interpret unnormalized probabilistic density functions this way? Is this what physicists mean? Or are there any other interpretations? Do I have to worry about something else when thinking about things in this way?

share|cite|improve this question
OK, how about this: The frequency $p$ with which a coin turns up "heads" is unknown. The probability that it is in any particular set $A\subseteq[0,1]$ is $\displaystyle\int_A\frac{c\,dp}{p(1-p)}$, where we assume $c$ is some infinitely small positive number such that that integral is $1$ when $A=[0,1]$. Toss the coin $10$ times and get $7$ heads. Conclude that the conditional probability distribution of $p$ given that outcome is $(\text{constant}\cdot p^6(1-p)^2\, dp)$ (a Beta distribution). I've seen a sober physicist argue that that's the right thing to do when...... – Michael Hardy Dec 3 '12 at 1:03
.....we don't initially know that either outcome is possible, and after we see one of the two possible outcomes, we don't yet know that the other one is possible. – Michael Hardy Dec 3 '12 at 1:04
@MichaelHardy What do you mean by this example? As an example for which what I tried doesn't work and physicists still like to think about? How do you "Conclude that the conditional probability distribution ..."? Did I miss something? – h__ Dec 3 '12 at 2:00
OK, I'll answer your last question first. If the prior probability measure is $f(p)\,dp$, and the likelihood function is $L(p) = \Pr(\text{observed data}\mid p)$, then the posterior probability measure, i.e. the conditional distribution of $p$ given the observed data, is $\text{constant}\cdot f(p)L(p)\,dp$. – Michael Hardy Dec 3 '12 at 14:45

The most coherent interpretation I know of what the physicists are doing in the real line example you mention is that they consider implicitely, for every positive $t$, a bona fide random variable $X_t$ with density $f_t$ where $f_t(x)=c_t^{-1}f(x)\,\mathbf 1_{[-t,t]}(x)$ and $c_t$ is the integral of $f$ on $[-t,t]$, and that they are convinced that the objects $X_t$ somehow converge when $t\to+\infty$ to... one does not know what kind of object exactly.

It happens that in most of the situations where (good) physicists are doing so, the properties of $X_t$ (of interest to them) somehow stabilize when $t\to+\infty$. Hence, although there is no random variable $X$ such that $X_t\to X$, nevertheless $X_t\approx X_s$ for every $s$ and $t$ large enough and this is all that is needed to proceed. So, in the end, there is (most often a pretty good amount of) reason in (them physicists') madness, although it may not be quite the brand of reason mathematicians are using.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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