# When can we differentiate the distribution function to get the density?

Good morning Math Stack Exchange

In my journey to learn Measure Theory, I wish to understand with clarity the link between a random variable's distribution function (df) and it's distribution. I believe some would call the distribution a density but in Resnick's "A Probability Path", the book I'm using, he calls it distribution.

I've been reading the chapter on Integration and Expectation and oddly enough, there isn't a theorem on how we can get the distribution from its df. Undergraduate classes would be quick to say "differentiate the df to get the distribution". But I'm sure there are numerous df which are not continuous hence not differentiable, i.e., $F(x)=\sum\frac{1}{2^n}1_{x\le n}$. For such df's, is it possible, and under what conditions, for us to get the df?

To me, I feel that we can easily get the df from the measure on defined on $(\Omega,\mathcal{F})$ namely $F(x)=P[X<x]$. The problem comes in getting the distribution. Most of the time, I need it so I can integrate in $\mathbb{R}$ to prove that, say, something is in $L_1$. Can we just differentiate $F(x)$?

Thank you!

-
What do you mean when you say "getting the distribution"? The density? Are you aware of the fact that many distributions do not have a density function? – Rasmus Oct 23 '11 at 8:41
Hello Rasmus. I do mean the density and am also aware the fact that many distributions do not have a density function. My question then is that 1. If the distribution function is differentiable, is it the density and 2. If a random variable does not have a density, does that mean I can't integrate in R to get expectation? – Donny Lee Oct 23 '11 at 16:56

There is indeed a nice characterization for probability measures on $\mathbb{R}$ to have a density function:

For probability measures $\mu$ on $\mathbb{R}$ that come from a density function in $L_{\rm loc}^1(\mathbb{R})$ (which is the largest space of measures that can reasonably be called functions), the distribution function $x\mapsto \mu((-\infty, x))$ is absolutely continuous.

On the other hand, each absolutely continuous distribution function is differentiable almost everywhere, the derivative is in $L_{\rm loc}^1$ and it is the density function.

Note that if the distribution function is merely differentiable almost everywhere, the derivative will not be the density function (for example, for distribution functions with jumps, the measure has atoms in the jump points).

-

Good morning Donny Lee and welcome to Math Stack Exchange!

There is a rigorous foundation for "differentiate the distribution function to get the density" but one has to be careful even if the distribution function is continuous and almost everywhere differentiable. From wikipedia:

... if a real function $F$ on $[a, b]$ admits a derivative $f(x)$ at every  point $x$ of $[a, b]$ and if this derivative $f$ is Lebesgue integrable on $[a, b]$, then $$F(b) - F(a) = \int_a^b f(t) \, \mathrm{d}t. \qquad\mathrm{Rudin (1987, th. 7.21)}$$ This result may fail for continuous functions $F$ that admit a derivative $f(x)$ at almost every point $x$, as the example of the Cantor function shows. But the result remains true if $F$ is absolutely continuous: in that case, $F$ admits a derivative $f(x)$ at almost every point $x$ and, as in the formula above, $F(b) − F(a)$ is equal to the integral of $f$ on $[a, b]$.

-