We all know that given a probability space $(\Omega,P,\mathcal F)$ and a random variable $X$, where $\mathcal F$ can be defined as $\sigma(X)$, we can determine a distribution function $F$ of $X$.

We also know that given a distribution function $F$, we can determine a probability space $(\Omega,P,\mathcal F)$ and a random variable $X$ on this space by $P(X\leq x)=F(x)$.

Well, I am wondering that if given a random variable $X(\omega):\Omega\mapsto R^k$, and an arbitrary distribution function $F$, can we find a probability measure $\tilde P$, or say a probability space $(\Omega,\tilde P,\tilde{\mathcal F})$ such that $X$ admits the distribution function $F$?

I have find Theorem 2.2.4 in A Course in Probability Theory by Kai Lai Chung, Third edition, which is described as follows:

Given the probability measure $\mu$ on $\mathcal F$, there is a unique distribution function $F$ satisfying $$\forall x \in R:\mu((-\infty,x])=F(x).$$ Conversely, given the distribution function $F$, there is a unique probability measure $\mu$ satisfying the previous identity.

I am not sure if this Theorem 2.2.4 can solve my question. And this theorem only show the case of real valued random variables, if it holds true for $R^k$ valued random variables?

I would appreciate it if someone can give me some tips.

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    $\begingroup$ I don't quite understand the question. You say "its distribution function" - that suggests that $F$ is already the distribution function of $X$ with respect to $P$, so then your problem is trivial by taking $\tilde{P}=P$. Did you mean for $F$ to be an arbitrary distribution function? It won't in general be possible, given arbitrary $X$ and $F$, to find a measure $\tilde{P}$ such that $F(x) = \tilde{P}(X \le x)$. For instance, suppose that $X$ were constant. $\endgroup$ – Nate Eldredge Feb 24 '16 at 15:31
  • $\begingroup$ @NateEldredge Right, I mean an aebitrary distribution function. Here is my another quesition, if a $\mathcal F_t$ measurable continuous process $B_t$ satisfies $B_0=0$, $E[B_t]=0$ and the increment $B_t-B_s$ is independent of $\mathcal F_s$ but $E[B^2_t]\neq t$, can I find a probability measure to make $B$ a Brownian motion? $\endgroup$ – XIAO Lishun Feb 24 '16 at 16:02
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    $\begingroup$ Ask this as a new question, please. But again, in general no: $B_t$ might be identically 0, in which case it will be identically 0 no matter what measure you use. $\endgroup$ – Nate Eldredge Feb 24 '16 at 16:31

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