8

Question:

Suppose that $X: \Omega \to \mathbb{R}$ is a random variable and it's distribution function $F(x) = \mathbf{P}(\xi \le x)$ is differentiable for all $x$. Is it true that $F'(x)$ is density?

What do I know:

Remark 1.

If $F'$ is continuous then $F'$ is density - see, e.g., A random variable $X$ with differentiable distribution function has a density

Hence if there's counterexample $F$ then $F'$ is not continuous everywhere.

Remark 2.

Absolutely continuous measures on $\mathbb{R}$ are precisely those that have densities. And if $g$ is differentiable everywhere then it doesn't follow that $g$ is absolutely continuous.

Does the everywhere differentiability of $f$ imply it is absolutely continuous on a compact interval?

Unfortunately, the example from the link doesn't help.

Remark 3 (close to remark 2).

In order to make a counterexample it's sufficient to find $F$ such that $F'$ is not intergrable. It's easy to find differentiable function with nonintergable derivative. Consider $g(x) = x^2 \sin(\frac{1}{x^2})$ is differentiable for all $x$ and $\int_{0}^1 g(x) dx$ doesn't exist. Unfortunatelly, $g(x)$ is not a counterexample in our problem, because it's not monotone and hence it's not a distribution function. A function $(g(x) + 1000 x) \cdot const$ also doesn't "work".

Addition: I found the similar question here https://math.stackexchange.com/questions/3387905/derivation-of-distribution-function-is-density-function/3387913 but there're no proofs there, so in fact there's still no answer.

Addition2: If $F'$ exists for all $x$ then it doesn't follow that $F'$ is continious. Counterexample: we may consider $\tilde{F}(x) = \frac{g(x) - g(-5)}{g(5)}I_{|x|\le 5} + I_{x > 5}$ where $g(x) = x^2 \sin (\frac{1}x) + 5x + 10$ and make $F(x)$ which is smooth, nondecreasing and which coinsides with $F$ for all $x \in \mathbb{R} \backslash (U_{\frac1{100}}(-5) \cup U_{\frac1{100}}(5))$.

Sha Vuklia
  • 4,356
Botnakov N.
  • 5,722
  • 2
    Absolute continuity is equivalent to the weak derivative being an integrable function. The monotonicity deals with the integrability issue, because the absolute value of the derivative is just the same as the derivative.

    The issue is that you need the weak derivative to be a function at all, which it isn't in the case of e.g. the Cantor function, where the weak derivative is the formal density of the Cantor distribution. (This example shows that a.e. differentiability is not sufficient for your question.) Ultimately what you need is that the integral of the pointwise derivative is $1$.

     – Ian Nov 01 '21 at 17:24
  • @MikeEarnest I misread something and missed the condition that made it true.

    OP: For your example in Remark 3, just look for Cantor's devil staircase. It is quite hard to come up with function that are increasing, continuous and differentiable almost everywhere and yet, their derivative be zero a.e. (en.wikipedia.org/wiki/Cantor_function)

     – William M. Nov 01 '21 at 18:30

1 Answers1

3

Use $F(x)$ to define a finite measure $\mu$ on the real line via the usual assignment of $\mu((a, b)) = F(b) - F(a)$. By Lebesgue-Radon-Nikodym, we can decompose $\text d\mu = h \text dm + \text d\lambda$ s.t. $h \in L^1(m)$ and $\lambda \perp m$. Then, $m$-almost everywhere (see Rudin's Real and Complex Analysis Chapter 7, theorem 7.14) we have $F' = D\mu = h$, whence $F' \in L^1(m)$. Theorem 7.21 from the same book shows that $F(x) = \int_{-\infty}^x F'(s) \text ds$.

GuPe
  • 7,518
  • Where are you using the (classical) differentiability of $F$? Which bit of your argument breaks for the Heaviside function or the Devil's staircase? – Bananach Nov 01 '21 at 23:57
  • 1
    @Bananach one of the hypothesis for theorem 7.21 is that $F'$ exist everywhere. – GuPe Nov 02 '21 at 00:03
  • 2
    So, to prove the seemingly innocuous fact that a differentiable CDF is the integral of its derivative, you need to develop measure theory, then combine the Lebesgue-Radon-Nikodym theorem, Vitali covering lemma, Hardy-Littlewood maximal inequality, Lebesgue differentiation theorem, and Rudin's 7.14 and 7.21. Pretty surprising since the original statement can be understood by a high school calculus student. – Mike Earnest Nov 02 '21 at 02:44
  • Thank you very much! – Botnakov N. Nov 02 '21 at 11:14
  • 1
    @MikeEarnest There are some corners you can cut out of that progression but yeah. Type II discontinuities are a pain, aren't they? As soon as you assume there's none of those floating around, you're back to the $C^1$ FTC. (Also, I think I recall this being on an analysis qualifying exam I took once.) – Ian Nov 02 '21 at 21:42