Let $F:\mathbb R\to [0,1]$ a function s.t. $F$ is right continuous, $\lim_{x\to -\infty }F(x)=0$, $\lim_{x\to \infty }F(x)=1$ and non decreasing.

I have a theorem in my lecture that says :

A function that the above properties is the distribution of some random variable.

I'm a bit confused by this theorem. Does it mean that

A) There exist a probability space $(\Omega ,\mathcal F,\mathbb P)$ and a random variable $X:\Omega \to \mathbb R$ s.t. $$F(x)=\mathbb P(X\leq x),$$

B) Or, in the previous theorem a probability space is already fixed ? I.e the theorem is :

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space. If a function $F$ has the properties above, then there is a r.v. $X:\Omega \to \mathbb R$ s.t. $F(x)=\mathbb P(X\leq x)$.

Why such a question ? My previous question could look a bit weird, but when in an exercise we say : Let $(\Omega ,\mathcal P,\mathbb P)$ a probability space and let $X,Y$ two random variable s.t. $X$ is normally distributed and $Y$ is exponentially distributed.

The question arise is : why two such r.v. exist on the same space ? If B) is not true, even if $X$ is normally distributed on $(\Omega ,\mathcal F,\mathbb P)$ (take $F(x)=\int_{-\infty }^x e^{-x^2/2}/\sqrt{2\pi}dx$) and $Y$ is normally distributed on $(\tilde \Omega ,\tilde {\mathcal F},\tilde {\mathcal P})$, ((take $F(x)=\int0^\infty \lambda e^{-\lambda x}dx$) there is no reason that such two r.v. exist on the same space.


1 Answer 1


Given $F$ there exists some probability space on which there is a random variable with distribution $F$. You cannot always do this on an arbitrary space. For example if the probability space consists of a finite set we cannot construct $X$ with normal distribution on it. [However, it can be shown that we can always find $X$ on $(0,1)$]. To construct $X$ and $Y$ with given distributions on the same space we can use product spaces. Kolomogorov's Existence Theorem (also known as Kolomogorov's Consistency Theorem) guarantees existence any number of random variables with given distributions on a single space.

  • $\begingroup$ Thank you for your answer :) $\endgroup$
    – user657607
    Commented Mar 27, 2019 at 6:21
  • $\begingroup$ At the end, if on $((0,1),\mathcal B(0,1),\mathbb P)$ where $\mathbb P$ the Lebesgue measure, we can construct for all $F$ a random variable, this guarantee the existence of any number of random variable with any distribution, no need Kolmogorov, do we ? $\endgroup$
    – user657607
    Commented Mar 27, 2019 at 6:25
  • $\begingroup$ Kolmogorov's Theorem achieves much more than just constructing random variables with given distributions. We often want additional properties like independence, for example. Kolmogorov's Theorem is a basic result on which the entire theory of stochastic processes rests. $\endgroup$ Commented Mar 27, 2019 at 6:30
  • $\begingroup$ @KaviRamaMurthy: Why $((0,1),\mathcal B(0,1),\mathbb P)$ and not $([0,1], \mathcal B([0,1],\mathbb P)$ whith Lebesgue measure ? Such a probability space doesn't work in the construction ? $\endgroup$
    – user657324
    Commented Mar 27, 2019 at 14:51
  • $\begingroup$ @user657324 If you are asking why I am taking open interval and not closed interval the answer is you can take either. Since single points have Lebesgue measure $0$ it makes no difference. $\endgroup$ Commented Mar 27, 2019 at 23:04

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