4
$\begingroup$

Given a certain probability space $(\Omega,\mathcal{A},\mathbb{P})$ and a random variable $X:t\mapsto X(t)$ defined on it, which is the difference between the following statements: $$\color{blue}{\text{every }t}\text{ is }\color{red}{\text{almost surely}}\text{ a nondifferentiability point for }X(t)\tag{1}$$ $$\color{red}{\text{almost surely }} \color{blue}{\text{every }t}\text{ is a nondifferentiability point for }X(t)\tag{2}$$ ?


I would rewrite $(1)$ as: $$\tag{1.int}\forall t\text{, }\mathbb{P}(t\text{ is a nondifferentiability point for}X(t))=1$$ and $(2)$ as: $$\tag{2.int}\mathbb{P}(t\text{ is a nondifferentiability point for}X(t), \forall t)=1$$



First, I don't know whether $(1.\text{int})$ and $(2.\text{int})$ are correct "rewritings" of $(1)$ and $(2)$ (resp.).

In general, whichever the difference between $(1)$ and $(2)$, which is the gist of such a difference from a mathematical standpoint? I cannot grasp it.




Could you please give an example of a random variable for which $(1)$ holds true, but $(2)$ does NOT hold true? (or viceversa)

$\endgroup$
3
  • $\begingroup$ You already know the correct interpretation! $\endgroup$ Oct 8, 2020 at 11:24
  • $\begingroup$ Good, but I cannot understand what it does mean "practically": could you give me an example of a random variable for which $(1)$ holds true, but $(2)$ does NOT hold true? @KaviRamaMurthy $\endgroup$ Oct 8, 2020 at 11:29
  • 1
    $\begingroup$ For any fixed $t_0$, note that the set in (2.int) is a subset of the set in (1.int) at $t=t_0$. So (2.int) implies (1.int) by subadditivity. $\endgroup$ Oct 8, 2020 at 11:33

1 Answer 1

5
$\begingroup$

On $(0,1)$ with Lebesgue measure let $X_t(\omega)=|t-\omega|$. Then $P(X_t \, \text {is differentiable at } \, t)=1$ for each $t$ and $P(X_t \, \text {is differentiable at every point} \, t)=0$.

(2) implies (1) always.

$\endgroup$
2
  • $\begingroup$ Why, for $X_t(\omega)=|t-\omega|$, does it hold true that $P(X_t \text{ is differentiable at }t)=1$ for each $t$? $\endgroup$ Oct 8, 2020 at 17:15
  • 1
    $\begingroup$ @Strictly_increasing Because $P$ here is the Lebesgue measure and Lebesgue measure of $(0,1) \setminus \{t\}$ is $1$. $\endgroup$ Oct 8, 2020 at 23:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .