# Show that a linear system with integral equations has at least one solution

I would like your help to show that the system below has/does not have at least one solution.

Let $$\mathcal{Y}\equiv \{0,1\}$$. Let $$\mathcal{V}\subseteq \mathbb{R}$$, not finite. Let $$w: \mathcal{V}\rightarrow \mathbb{R}$$ denote a probability density function on $$\mathcal{V}$$. Let $$q: \mathcal{Y}\rightarrow [0,1]$$ denote a probability mass function on $$\mathcal{Y}$$.

Consider the system of equations below. The unknown is the mixed joint density $$f: \mathcal{Y}\times \mathcal{V}\rightarrow \mathbb{R}$$. The functions $$w$$ and $$q$$, and the sets $$\mathcal{V},\mathcal{Y}$$ are known. $$(*) \quad \begin{cases} &(1) \quad \sum_{y\in \mathcal{Y}}f(y,v) =w(v) \quad \forall v \in \mathcal{V},\\ &(2) \quad \int_{v\in \mathcal{V}} f(y,v) dv=q(y)\quad \forall y\in \mathcal{Y},\\ &(3) \quad\int_{v\in \mathcal{V}} f(1,v) *v \text{ }dv \geq 0,\\ &(4) \quad\int_{v\in \mathcal{V}} f(0,v) *v\text{ }dv \leq 0.\\ \end{cases}$$

Question: Show that $$(*)$$ has/does not have at least one solution. Does the answer change if $$\mathcal{V}\equiv \mathbb{R}$$ (for instance, $$w$$ is the normal density function)?

Note: The case with $$\mathcal{V}$$ finite (hence, $$w$$ and $$f$$ are probability mass functions) has been studied here. The answer shows that the system may not have a solution. I want to generalise the claim (or prove the opposite) when $$\mathcal{V}$$ is not finite.

There might be no solution. In fact, it is possible to write a necessary and sufficient condition for its existence.

First of all, we need to assume that $$\int_{\mathbb R} |v| \, w(v) dv<\infty$$.

When $$q(0)=0$$, then it is obvious that $$(*)$$ has a solution iff $$\int_{\mathbb R} v\, w(v) dv \ge 0$$ (since we must have $$f(1,v) = w(v)$$ a.e.). Similarly, if $$q(1) = 0$$, then $$(*)$$ has a solution iff $$\int_{\mathbb R} v\, w(v) dv \le 0$$.

Let us study the case $$q(0)\cdot q(1) >0$$. Let $$F^{-1}(z)$$ be the quantile function corresponding to $$w$$, i.e. $$F^{-1}(z) = \inf\left\{x\in \mathbb{R}: \int_{-\infty}^x w(v) dv\ge z\right\},\quad z\in(0,1).$$

Claim If $$q(0)\cdot q(1) >0$$, then $$(*)$$ has a solution iff $$\int_{-\infty}^{A}v\, w(v) dv\le 0 \tag{\le}$$ and $$\int_{A}^\infty v\, w(v) dv\ge 0 \tag{\ge},$$ where $$A= F^{-1} (q(0))$$.

Proof. Let $$(\le)$$ fail, i.e. $$\int_{-\infty}^{A}v\, w(v) dv >0.$$ Then, clearly, $$A>0$$ and $$\int_{-\infty}^{\infty} v\, f(0,v) dv \ge \int_{-\infty}^A v\,f(0,v) dv + A \int_A^\infty f(0,v) dv \\ \overset{(2)}{=} \int_{-\infty}^{A} v f(0,v) dv + Aq(0)- A \int_{-\infty}^A f(0,v) dv = Aq(0) + \int_{-\infty}^{A} (v-A) f(0,v) dv \\ = Aq(0) - \int_{-\infty}^{A} \int_v^A du\, f(0,v) dv = A\int_{-\infty}^A w(v) dv - \int_{-\infty}^A \int_{-\infty}^u f(0,v) dv\,du \\ \ge A\int_{-\infty}^A w(v) dv - \int_{-\infty}^A \int_{-\infty}^u w(v) dv\,du = A\int_{-\infty}^A w(v) dv + \int_{-\infty}^A (v-A) w(v) dv \\ = \int_{-\infty}^A v\,w(v) dv>0,$$ so $$(4)$$ fails.

Similarly, if $$(\ge)$$ fails, then so does $$(3)$$.

On the other hand, if both $$(\le)$$ and $$(\ge)$$ hold, then $$f(0,v) = w(v) \mathbf{1}_{(-\infty, A)}(v), \quad f(1,v) = w(v) \mathbf{1}_{[A, +\infty)}(v) \tag{f}$$ are easily seen to satisfy $$(*)$$.

Remarks

1. There's no need to consider the case $$q(0)\cdot q(1) = 0$$ individually: in this case $$A$$ is either $$-\infty$$ or $$+\infty$$, so one of $$(\le)$$ and $$(\ge)$$ holds trivially, and we need just another one, as it is written above.
2. Where do $$(\le)$$ and $$(\ge)$$ come from? In fact, $$(f)$$ gives the decomposition $$w(v) = f(0,v) + f(1,v)$$ satisfying (2) and such that $$\int_{\mathbb R} v f(0,v) dv$$ is minimal (correspondingly, $$\int_{\mathbb R} v f(1,v) dv$$ is maximal).
3. It is easy to see that the solution is unique iff at least one of $$(\le)$$ or $$(\ge)$$ is equality.
4. Similarly, one can write a necessary and sufficient condition in the case where the corresponding distribution does not have a density (naturally, $$(*)$$ needs to be adapted correspondingly). Namely, denoting the corresponding (right-continuous) cdf by $$F$$, $$(\le)$$ and $$(\ge)$$ transform to $$\int_{(-\infty, A)} v\, dF(v) + A \cdot\big(q(0) - F(A-)\big)\le 0$$ and $$\int_{(A, \infty)} v\, dF(v) + A\cdot \big(F(A) - q(0)\big)\ge 0.$$