# Existence of a random variable satisfying a condition on its distribution

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)?

• Isn't it so that for any probabilitymeasure a random variable can be defined such that the measure induced by it coincides with this probabilitymeasure? If so then it is enough to prove that $\frac{1}{2}\mu_X+\frac{1}{2}\mu_Y$ is indeed a probabilitymeasure. Aug 4, 2014 at 8:14
• Do you have a reference for this? Any convex combination of probability measures is a probability measure, so $\frac{1}{2}\mu_X + \frac{1}{2}\mu_Y$ is a probability measure. Aug 4, 2014 at 8:16
• If $F$ is a distributionfunction and $X$ is defined on $[0,1]$ by $X(\omega)=\sup\{x\mid F(x)<\omega\}$ then a rv is created with $F_X=F$. Of course this is a special case and maybe you allready are familiar with that. I just thought: can this be generalized? If so then you are ready. I don't know yet. Aug 4, 2014 at 8:22
• I suspect this might fail if the space $\mathcal{X}$ is complicated compared to $[0,1]$. Something analogous to the following: Let the domain be two points distributed uniformly. As the range we take any space with at least $3$ points. For $X$ choose a mapping that maps the first point to $0$ and the second one to $1$. For $Y$ choose a mapping that maps the first point to $0$ and the second one to $\frac{1}{2}$. The convex combination of the distributions now supports $3$ points, so there does not exist a random variable from the two point set to the three point set with this distribution. Aug 4, 2014 at 12:08
• On the other hand taking convex combinations is a finite operation, so I am really not sure. Aug 4, 2014 at 12:11

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.
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.
• I am sorry that my question was unclear, but what I want is actually a random variable with domain $[0,1]$, the same as the domains of $X$ and $Y$. I will update the question. Aug 4, 2014 at 14:26