Existence of a random variable satisfying a condition on its distribution

Question

Let $X, Y : [0,1] \to \mathcal{X}$ be two random variables. Here, $[0,1]$ is the interval with the Lebesgue $\sigma$-algebra and $\mathcal{X}$ is a topological space with the Borel $\sigma$-algebra.

The distribution of $X$ is by definition a measure $\mu_X$ on $\mathcal{X}$ defined as $\mu_X(A) = \mathbf{P}(X^{-1}(A))$, where $\mathbf{P}$ is the Lebesgue measure on $[0,1]$.

Are there results on the existence of a random variable $Z : [0,1] \to \mathcal{X}$, whose distribution $\mu_Z$ satisfies $$\frac{1}{2} \mu_X + \frac{1}{2}\mu_Y = \mu_Z.$$

Or more generally. Given probability distributions of two random variables, is every convex combination of these distribution also a distribution of a random variable (with the same domain)?

Answer 1

Let's try the following construction. We divide the unit interval in pieces $[0,1/2]$ and $(1/2,1]$. Then define $Z(x)$ by $Z(x) = X(2x)$ if $x \in [0,1/2]$ and $Z(x) = Y(2(x-1/2))$ if $x \in (1/2,1]$. Now $$\begin{eqnarray*}P(Z(x) \in A) & = & P(\{x \in [0,1/2]\} \cap \{X(2x) \in A\}) + P(\{x \in (1/2,1]\} \cap \{Y(2(x - \frac{1}{2}) \in A\}) \\ & = & \frac{1}{2} P(X(x) \in A) + \frac{1}{2} P(Y(x) \in A).\end{eqnarray*}$$ A similar construction should work for other convex combinations.

Answer 2

If $(S,\mathcal{B},\mu)$ is a measurable space with $\mu(S) = 1$, then there always exists a probability space $\Omega$ and a random variable $X : \Omega \rightarrow S$ with distribution $\mu$. Just take $\Omega = S, \mathcal{F} = \mathcal{B}$, $P = \mu$ and define $X(s) = s$ for all $s \in \Omega$. Then for any $A \in \mathcal{B}$, \begin{gather*} P\circ X^{-1}(A) = P(X \in A) = P(\{s : X(s) \in A\}) = \mu(\{s : s \in A\}) = \mu(A). \end{gather*} It's important that the question just asks for existence of a random variable.