0
$\begingroup$

Suppose $X$ is an absolutely continuous real random variable, (that is, there exist a non-negative integrable function $f$, such that $\int_\mathbb{R} f=1$ and for every interval $I\subseteq \mathbb{R}$, $\int_I f = P(X \in I)$, such $f$ is called a density function of $X$) and let $F$ be the cumulative distribution function.

I was told that the density function of $X$ is never completely determined, because if $f$ is a density function of $X$, changing its values on a finite set you still get a density function of $X$. But I know that where $F$ is differentiable, $F'(x)=f(x)$ for every density function $f$. I was thinking that if $F$ is differentiable everywhere, then there should be an only density function. What's wrong?

$\endgroup$
0
$\begingroup$

The probability density function is defined for a specific measure $\mu$ on $\mathbb{R}$ (usually, the Lebesgue measure), and the integrals $\int_\mathbb{R} f$ and $\int_{I}f$ should be written as $\int_\mathbb{R} f d \mu$ and $\int_I f d\mu$. The probability density function is defined $\mu$-almost everywhere uniquely (see https://en.wikipedia.org/wiki/Probability_density_function#Discussion), as it is, strictly speaking, the Radon–Nikodym derivative $\frac{d P}{d \mu}$ of the probability measure $P$ w.r.t. $\mu$. However, in case of Lebesgue measure $\mu$ $F'(x)$ is one of those $\mu$-a.e. probability density functions.

Edit: essentially, a probability density function on a real line is not unique. The wrong equation is $F'(x) = f(x)$, the correct version of which is $\int_{-\infty}^x f(t) dt = F(x)$ for any real $x$ (and where Lebesgue, not Riemann integration is used).

$\endgroup$
  • $\begingroup$ I'm sorry, I'm a first year math student, and our probability course doesn't use measure. Is it possible to give a more elementary explanation? $\endgroup$ – Angelo Russo Jul 2 '15 at 12:42
  • $\begingroup$ @AngeloRusso, I added a simplified explanation. $\endgroup$ – Budenn Jul 2 '15 at 13:12
  • $\begingroup$ I think you mean $\int_{-\infty} ^x f(t)dt = F(x)$ for all real $x$. So we can't differentiate both sides (we don't know that $f$ is continuous, I got confused at this point). Just a last question: why do you say that lebesgue integral is used? Isn't it the usual improper integral? Thanks! $\endgroup$ – Angelo Russo Jul 2 '15 at 13:32
  • $\begingroup$ @AngeloRusso, yes, you are right about the formula. The result of differentiation is not unique, because $f$ can be changed in a countable amount of points and the integral's value will not change. (However, for real line Lusin's theorem states that the result of differentiation will be "nearly continuous".) Lebesgue integral is used because it allows a wider class of distributions and sample spaces, but for real line if improper Riemann integral converges absolutely, then the corresponding Lebesgue integral has the same value. $\endgroup$ – Budenn Jul 2 '15 at 14:18
0
$\begingroup$

An elementary example can be found by using a uniformly-distributed random variable with positive probability on [0,1] versus the uniform distribution with support (0,1). Both RVs have the same CDF (theyre equal at every point), but different densities (as a result of the behaviour at the endpoints).

$\endgroup$
  • 1
    $\begingroup$ This might be nitpicking, but support is often defined to be closed, so in your example both variables can have the same support, depending on exact definition. $\endgroup$ – Budenn Jul 2 '15 at 13:16
  • $\begingroup$ In it is arguable the distribution or a real-valued random variable is fully expressed by the cumulative distribution function, so these two examples have the same distribution and so the same support. $\endgroup$ – Henry Jul 2 '15 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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