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.


1 Answer 1


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 $(*)$.


  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 $$ \int_{(A, \infty)} v\, dF(v) + A\cdot \big(F(A) - q(0)\big)\ge 0. $$

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