Showing that a non-linear system has more than one solution

Question

Consider a $2\times 1$ random vector $$ V\equiv \begin{pmatrix} V_1\\ V_2 \end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix} 0\\ 0\end{pmatrix}, \begin{pmatrix} 1& \rho\\ \rho & 1 \end{pmatrix}\right) $$ with $\rho\neq 0$ and $\rho\in [-1,1]$ defined on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$. I am asking your help to solve the following exercise on identification in econometrics.

Consider the system of equations $$ \begin{cases} \mathbb{P}(V_1\geq V_2-b_1; \rho)=p_1\\ \mathbb{P}(V_1\geq V_2-b_2; \rho)=p_2\\ \mathbb{P}(V_1\geq V_2-b_3; \rho)=p_3\\ \end{cases} $$ where the unknowns are $(b_1,b_2, b_3, \rho)$ while $(p_1,p_2,p_3)\in [0,1]^3$ is known.

It is assumed that this system has at least one solution (i.e., the model is correctly specified).

I want to show that this system has at least two solutions denoted by $(b^*_1,b^*_2, b^*_3, \rho^*)$ and $(b^{**}_1,b^{**}_2, b^{**}_3, \rho^{**})$. Below I have reported what I have tried and my doubts.

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I started my analysis with deriving, for any $b\in \mathbb{R}$, the parametric expression of $$ \mathbb{P}(V_1\geq V_2-b). $$

From the assumption above we have

• $V_1 \mid V_2=v_2\sim \mathcal{N}(\rho v_2, (1-\rho^2))$ for any realisation $v_2$ of $V_2$

• $V_2\sim \mathcal{N}(0,1)$

Hence $$ \mathbb{P}(V_1\geq V_2-b)= E(\mathbb{P}(V_1\geq V_2-b \mid V_2))= \int_{\mathbb{R}} \Big[1-\Phi\Big(\frac{v-v\rho-b}{\sqrt{1-\rho^2}}\Big)\Big]\phi(v) \,\mathrm{d}v, $$ where $\Phi, \phi$ are the cdf and pdf of the standard normal.

Therefore the system of equations becomes $$ \begin{cases} \displaystyle\int_{\mathbb{R}} \Big[1-\Phi\Big(\frac{v-v\rho-b_1}{\sqrt{1-\rho^2}}\Big)\Big]\phi(v) \,\mathrm{d}v=p_1\\ \displaystyle\int_{\mathbb{R}} \Big[1-\Phi\Big(\frac{v-v\rho-b_2}{\sqrt{1-\rho^2}}\Big)\Big]\phi(v) \,\mathrm{d}v=p_2\\ \displaystyle\int_{\mathbb{R}} \Big[1-\Phi\Big(\frac{v-v\rho-b_3}{\sqrt{1-\rho^2}}\Big)\Big]\phi(v) \,\mathrm{d}v=p_3\\ \end{cases} $$ From here I do not know how to complete the exercise. Is non-uniqueness of solution obvious because I have three equations and four unknowns?

Answer 1

$\def\Φ{{\mit Φ}}$It suffices to prove that for any $-1 \leqslant ρ \leqslant 1$, there exists $b_1, b_2, b_3 \in \mathbb{R}$ satisfying$$ P(V_2 - V_1 \leqslant b_k; ρ) = p_k. \quad k = 1, 2, 3 $$

Because$$ \begin{pmatrix} V_1 \\ V_2 \end{pmatrix} \sim N\left( \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & ρ \\ ρ & 1 \end{pmatrix} \right), $$ then\begin{align*} V_2 - V_1 = \begin{pmatrix} -1 & 1 \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \end{pmatrix} &\sim N\left( \begin{pmatrix} -1 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & ρ \\ ρ & 1 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} \right)\\ &= N(0, 2(1 - ρ)). \end{align*} Since the distribution function of $N(0, 2(1 - ρ))$ is $F(x; ρ) = \Φ\left( \dfrac{x}{\sqrt{\smash[b]{2(1 - ρ)}}} \right)$, then$$ b_k = \sqrt{\smash[b]{2(1 - ρ)}} \Φ^{-1}(p_k). \quad k = 1, 2, 3 $$