Can one decompose a random variable $X$ into a mixture of $Y$ and $Z$ with prespecified mixing probabilty $p$ subject to the constraint $E[Y] = \mu$?

Question

Let $X$ be a random variable with known distribution function $F$ and let $Q \sim Bernoulli(p)$. Conditional on $Q=1$, $X = Y$ where $Y \sim G$. Conditional on $Q = 0$, $X = Z$ where $Z \sim H$.

Now suppose we fix a value for $p$, strictly between zero and one, along with a mean $\mu$ for the random variable $Y$. Given $F$, $p$, and $\mu$, does there always exist a corresponding pair $(G,H)$ that satisfies the mixture model? If not, are there conditions on $F$ that would ensure the existence of such a pair?

Update: In the first answer below, kimchee lover gives two examples of a discrete $X$ in which the decomposition fails. This is a helpful start, but I'm particularly interested in settings where the support set of $X$ is fairly rich. I'd be very happy with a characterization that applies to continuously distributed $X$ even if one cannot say something in general for discrete RVs.

Answer 1

Not always. Suppose $P(X=\nu)=1$, where $\nu$ is a constant, different from $\mu;$ in other words $F=\delta_\nu.$ Since $F$ is an extreme point in the space of probability measures, both $G$ and $H$ must equal $F$. But the expectation of $G$ is then wrong.

More elaborately, let $F= (\delta_0 + \delta_1)/2$ put half its mass at $0$ and half at $1$. Then $G$ and $H$ must also be supported on $\{0,1\}$, and hence $P(Y=1) = \mu$ and thus $P(Y=0)=1-\mu$. Already we know that $\mu$ must satisfy $0\le\mu\le1$. Similarly, $P(Z=1)$ must then solve $1/2 = p\mu + P(Z=1)$ which implies $$0 \le \frac{1/2-\mu p}{1-p} \le 1.$$

If I can think of a simple condition involving $F$, $p$, and $\mu$ alone, allowing a decomposition, I'll post it.

ADDED 2 hours later.

Suppose $F$ is continuous, with support $\mathbb R$. Since $F$ is a non trivial mixture of $G$ and $H$, we have $G\ll F$ and $H\ll F$, and thus $G$ and $H$ have Radon-Nikodym derivatives with respect to $F:$ call them $\gamma$ and $\eta$. The desired decomposition thus requires $p\gamma(x) +(1-p)\eta(x)=1$ for almost all (w.r.t. $F$) values of $x$, and $\int_{\mathbb R} \gamma(x) F(dx)=\int_{\mathbb R} \eta(x) F(dx) = 1$ and $\int_{\mathbb R} x \gamma(x) F(dx)=\mu$. And, of course, $\gamma(x)\ge 0$ and $\eta(x)\ge 0$, for almost all $x$.This is, in a sense, a linear programming problem (or feasibility problem). The unknown function $\eta$ can be eliminated from this setup by noting that given $\gamma$, we find $\eta=(1-p\gamma)/(1-p)$, and the non-negativity constraint on $\eta$ can be transferred to $\gamma$ by requiring $0\le \gamma(x) \le 1/p$.

I think the following sketch shows when such a $\gamma$ exists. For each $t\in [0,1-p]$ let $A(t) = F^{-1}(1-p-t)$ and $B(t) = F^{-1}(1-t)$. Thus, $p$ of $X$'s mass lies between $A(t)$ and $B(t)$, and $t$ of the mass lies above $B(t)$. Now consider the decreasing function $m(t) = \int_{A(t)}^{B(t)} x F(dx)/p$. If $\mu = m(t)$ for some $t$, we are done: let $\gamma=I_{[A(t),B(t)]}/p$. If there is no such $t$, I think the problem is infeasible. Which could happen if $\mu$ is too big.