4
$\begingroup$

I have a doubt on how to relate uncountably infinite probability space and continuous random variables.

Take the following random experiment "choose a number from $[0,1]$".

I construct the associated probability space.

  • the sample space $\Omega\equiv[0,1]$

  • the $\sigma$-algebra equal to the Borel $\sigma$-algebra on $[0,1]$, $\mathcal{B}([0,1])$

  • the measure $\mathbb{P}:\mathcal{B}([0,1])\rightarrow [0,1]$ (this is not necessarily equal to the Lebesgue measure; there may be some numbers "more attractive" than others; also, it can't be the ratio of counting measures because $\Omega$ is uncountably infinite)

Consider now the random variable $$ X:\Omega\rightarrow \mathbb{R} $$

($\star$) I assume that this random variable is continuous

($\star \star$) (EDITED following a useful comment below) I assume that this random variable has cdf continuous on $\mathbb{R}$ and strictly monotone.


Clearly, ($\star \star$) $\rightarrow$ ($\star$).

I want to understand the different implications of these two assumptions on $(\Omega, \mathcal{F}, \mathbb{P})$. With this objective in mind, I have separated my question into 4 sub-questions.


1) Could you help me to formally understand the relation between $(\star)$ and $\Omega$ above? Is $\Omega$ uncountably infinite a necessary condition for $X$ being a continuous random variable?

2) Could you help me to formally understand the relation between $(\star)$ and $\mathbb{P}$ above?

3) Could you help me to formally understand the relation between $(\star \star)$ and $\Omega$ above? I think that all that matters here should be thorugh condition ($\star$), correct?

4) Could you help me to formally understand the relation between $(\star \star)$ and $\mathbb{P}$ above? Is this simply the relation among probability measure, pdf, cdf?

$\endgroup$
  • $\begingroup$ "Clearly, ($\star \star$) $\rightarrow$ ($\star$)" Well, no. For a counterexample, consider $X=BY$ where $B$ is Bernoulli $0$ or $1$, $Y$ is standard normal and $(B,Y)$ are independent. $\endgroup$ – Did Mar 11 '18 at 23:12
  • $\begingroup$ @Did Thanks. Could you help with the other points too? $\endgroup$ – STF Mar 11 '18 at 23:21
  • $\begingroup$ What do you mean "I assume this random variable is continuous"? Usually people talk about continuous random variables as having a CDF function $P[X\leq x]$ that is continuous in $x \in \mathbb{R}$. We usually don't care so much about continuity of the map $X:\Omega\rightarrow \mathbb{R}$. The CDF can be continuous even when the map is not. $\endgroup$ – Michael Mar 15 '18 at 14:57
  • $\begingroup$ In general, indeed if $P[X\leq x]$ is continuous over $x \in \mathbb{R}$, it means the sample space $\Omega$ must be uncountably infinite. If general if there is a $y \in \mathbb{R}$ such that $P[X=y]>0$ then, since $P[X\leq y] = P[X<y] + P[X=y]$, we see the CDF function cannot be continuous at $y$. $\endgroup$ – Michael Mar 15 '18 at 15:08
  • $\begingroup$ Your other questions are too loaded with undefined notation to be understood, such as "$\epsilon(\omega) = e$." $\endgroup$ – Michael Mar 15 '18 at 15:09
1
+50
$\begingroup$

Let $X:\Omega\rightarrow\mathbb{R}$ be a random variable. Recall that the CDF function $F_X(x)$ is defined $F_X(x)=P[X \leq x]$. The random variable $X$ is said to be continuous if the CDF function is continuous for all $x \in \mathbb{R}$.

Here are some exercises for you. Solving these will likely answer your residual questions:

1) Suppose there is a $y \in \mathbb{R}$ such that $P[X=y]>0$. Argue that $$|F_X(y) - F_X(w)|\geq P[X=y] \quad, \forall w<y$$and hence the CDF function for $X$ is not continuous at $y$.

2) Suppose $P[X\leq x]$ is continuous for all $x \in \mathbb{R}$. Assume the sigma algebra for $\Omega$ includes all single-point sets. Argue that $$P[\{\omega\}]=0 \quad, \forall \omega \in \Omega$$ Conclude that the sample space $\Omega$ is uncountably infinite.

3) Suppose the CDF of $X$ is discontinuous. Show there must be a $y \in \mathbb{R}$ such that $P[X=y]>0$.

4) Suppose $X=h(Y)$ for some random variable $Y$ and some (measurable) function $h:\mathbb{R}\rightarrow\mathbb{R}$. Show that if $Y$ has a discontinuous CDF, then $X$ must have a discontinuous CDF.

5) Give an example of random variables $X, Y$ such that $X=h(Y)$, $Y$ has a continuous CDF with flat parts (not strictly increasing), but $X$ has a continuous CDF that is strictly increasing.

6) Give an example of a random variable $X:\Omega\rightarrow\mathbb{R}$ such that $X(\omega)$ is not a continuous function of $\omega$, but the CDF of $X$ is continuous. This requires $\Omega$ to have some distance metric associated with it, so you can choose, for example, $\Omega = [0,1]$ with usual metrics of distance.

$\endgroup$
  • $\begingroup$ Note: For question 2, you can remove the assumption that $\Omega$ includes all single point sets to still conclude that if CDF is continouous then $\Omega$ is uncountably infinite: Show that $P[X=x]=0$ for all $x \in \mathbb{R}$, argue that there are uncountably many nonempty disjoint sets of the type $\{\omega \in \Omega: X(\omega) = x\}$. $\endgroup$ – Michael Mar 15 '18 at 18:13
1
$\begingroup$

The thing that you said is clearly true is not. That $X$ is a continuous function from $[0,1]$ into $\mathbb R$ does not follow from its c.d.f. being continuous and strictly monotone. For example, suppose the probability measure on $\Omega = [0,1]$ is Lebesgue measure and $$ X(\omega) = \begin{cases} \tan\dfrac \omega \pi & \text{at points where the value of that function is finite,} \\[8pt] 0 & \text{elsewhere.} \end{cases} $$ This is not a continuous function from $[0,1]$ into $\mathbb R,$ but its c.d.f. is continuous and strictly monotone, and in fact is absolutely continuous.

However, the statement that a random variable $X$ is "continuous"

  • is sometimes taken to mean precisely that its c.d.f. is continuous, and
  • is sometimes taken to mean its c.d.f. is absolutely continuous.

The topology of $\Omega,$ if any, is usually not a concern. In fact, usually $\Omega$ itself is not a concern; one works with c.d.f.s on the space into which $X$ maps. A function on $\Omega$ can have all sorts of discontinuities without any discontinuity of the c.d.f.

Absolute continuity in this context is equivalent to having a probability density function, i.e. a function $f$ such that for every Borel set $A,$ $$ \Pr(X\in A) = \int\limits_A f(u)\,du. $$ When such a density function exists, the c.d.f. is continuous and is equal to $$ F_X(x) = \Pr(X\le x) = \int_{-\infty}^x f(u)\,du. $$ In some cases a c.d.f. is continuous without having any absolutely continuous part, so there is no part that has a density function. The most well known instance of this is probably the Cantor distribution, defined as follows:

Toss a fair coin and put $X$ somewhere in the interval $[0,1/3]$ if you get tails and in the interval $[2/3,1]$ if you get heads.

Toss the coin again and put $X$ in the lower or upper third of that interval according as you get tails or heads.

Toss the coin again and put $X$ in the lower or upper third of that interval according as you get tails or heads.

and so on.

The c.d.f. of this distribution is continuous, as can be seen by observing that there are no point masses, i.e. there is no $x\in\mathbb R$ such that $\Pr(X=x)>0.$

However, the measure of the support of this distribution is less than $2/3,$ and less than $2/3$ of $2/3,$ and less than $2/3$ of $2/3$ of $2/3,$ and so on, so it is $0.$ Thus if you integrate any function $f$ over that support, you get $0.$ Hence this distribution has no density function, nor is it a weighted average of any distribution with a density function and any other distribution.

If a function $X$ from $[0,1]$ into $\mathbb R$ is continuous and the probability measure on that interval is Lebesgue measure, then the resulting c.d.f. is continuous. One place where I think this matters may be in pseudo-random number generators.

$\endgroup$

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.