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This may be wrong, but I have often heard some saying " we mainly care about CDFs". Similarly, in textbooks, one sees $X \sim N(0,1)$, without any reference to sample space. But why - and how do we justify this?

My thoughts: with my limited knowledge in probability, two results:

  • Any distribution function $F:\mathbb{R} \rightarrow [0,1]$, yields a Lebesgue-Stiltjes measure $\mu_F$. Considering the space $(\mathbb{R}, B_{\mathbb{R}}, \mu_F)$ with randon variable $X$ being identity, we obtain, $$P(X \le t ) = \mu((-\infty, t]) = F(t)$$

This implies it is sufficient in specifying cdf $F$, and say there exists a RV, $X$ with cdf $F$.

  • (Skorokhod's construction, in Williams) Let $F: \mathbb{R} \rightarrow [0,1]$ be a cdf. $U\sim U[0,1]$. $$ X^-:= \sup \{ y \in \mathbb{R} \,: \, F(y) < U \} $$ is a RV on $[0,1]$ with same distribution as $F$.

I believe there is also a generalization to joint variables.

But these results are not satisfying, I don't see how they are canonical.

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    $\begingroup$ When you have a measurable function $X:\Omega\to E$ and a measure $P$ on $\Omega$, you can define the push-forward measure $X_*(P)$ on $E$ defined as $X_*(P)(A)=P(X^{-1}(A))$, for each measurable set of $E$. In the case of $E=\mathbb{R}$, and $P$ a probability, it is enough to know $X_*(P)$ on sets of the form $(-\infty,a]$. If you are going to ask only questions about the sizes of sets of the form $X^{-1}(A)$, then all you need to know is $X_*(P)$ on $E$. $\endgroup$
    – user551819
    Apr 18 '18 at 18:25
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    $\begingroup$ Given $E=\mathbb{R}$ and a CDF, $F$, on it, you are right that there are many $X:\Omega\to E$, and measures $P$ on $\Omega$, such that $F(a)=X_*(P)((-\infty,a])$. For example, having one such $X$ you can always pick an element $\omega\in\Omega$ and add one (or many) element $\omega_1$ to $\Omega$ and define $X(\omega_1)=X(\omega)$. The push-forward measure will be the same. If you are going to be studying the elements of $\Omega$ then you cannot forget about it. The claim holds as long as all you will ask is questions about $P(X^{-1}(A))$. $\endgroup$
    – user551819
    Apr 18 '18 at 18:41
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    $\begingroup$ CWL: Please explain what is not canonical here. $\endgroup$
    – Did
    Apr 18 '18 at 22:08
  • $\begingroup$ Regarding the two results I stated: Given a cdf $F$, the existence of a random variable $X$ that has such cdf has domain a subset of $\mathbb{R}$. (at least in the first case, how does one go about constructing $X$ for a given sample space $\Omega$?) $\endgroup$
    – Bryan Shih
    Apr 19 '18 at 7:24
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Because of the Kolmogorov existence theorem. It tells us that given any distributions (not defined in a dumb way), there exists a stochastic process (in particular, a random variable) with those distributions.

Often we do not care about the actual probability space. The entire point of a random variable is to assign numbers to outcomes. Real numbers are easier to deal with than generic probability spaces. Although there are times where the space itself is important.

Taken from Wikipedia:

Since the two conditions are trivially satisfied for any stochastic process, the power of the theorem is that no other conditions are required: For any reasonable (i.e., consistent) family of finite-dimensional distributions, there exists a stochastic process with these distributions.

The measure-theoretic approach to stochastic processes starts with a probability space and defines a stochastic process as a family of functions on this probability space. However, in many applications the starting point is really the finite-dimensional distributions of the stochastic process. The theorem says that provided the finite-dimensional distributions satisfy the obvious consistency requirements, one can always identify a probability space to match the purpose. In many situations, this means that one does not have to be explicit about what the probability space is. Many texts on stochastic processes do, indeed, assume a probability space but never state explicitly what it is.

The theorem is used in one of the standard proofs of existence of a Brownian motion, by specifying the finite dimensional distributions to be Gaussian random variables, satisfying the consistency conditions above. As in most of the definitions of Brownian motion it is required that the sample paths are continuous almost surely, and one then uses the Kolmogorov continuity theorem to construct a continuous modification of the process constructed by the Kolmogorov extension theorem.

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  • $\begingroup$ Quite unrelated to the question asked. Kolmogorov's existence is about assembling many random variables into a process. They are asking about forgetting about the random variable. $\endgroup$
    – user551819
    Apr 18 '18 at 18:45
  • $\begingroup$ @totoro Kolmogorov is about the existence of such random variables. Read: The theorem says that provided the finite-dimensional distributions satisfy the obvious consistency requirements, one can always identify a probability space to match the purpose. In many situations, this means that one does not have to be explicit about what the probability space is $\endgroup$
    – user223391
    Apr 18 '18 at 18:47
  • $\begingroup$ @totoro The theorem says, given any distributions, there exists a probability space and stochastic process such that the pushfowards are those distributions. $\endgroup$
    – user223391
    Apr 18 '18 at 18:49
  • $\begingroup$ Yes, the question already quotes a particular case of it. But they are asking about whether 'they are canonical'. If the construction were unique that would totally explain why dropping the sample space and the random variable is OK. Since it is not, it doesn't say anything about why the random variable is dropped. $\endgroup$
    – user551819
    Apr 18 '18 at 18:57

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