Write $\log(1+(x-y)^2)$ in finite sum of $p_i(x)q_i(y)$

Question

Main question

Is it possible to write $\log(1+(x-y)^2)$ in finite sum of $p_i(x)q_i(y)$, where $p_i,q_i:\mathbb{R}\to\mathbb{R}$ are arbitrary functions?

Background

One of my friends have encountered a problem. In virtually any textbook available, it is assumed that the collection of Cauchy distributions, \begin{equation*} \mathcal{P}=\left\{f(x\mid\theta)=\frac{1}{\pi(1+(x-\theta)^2)}:\theta\in\mathbb{R}\right\} \end{equation*} is not a $k$-parameter exponential family for any $k$.

Similar questions on MSE

I have read Why Cauchy Distribution isn't Exponental Family? and How to prove Cauchy Distribution doesn't belong to Exponential Family?, but the answers are not satisfying. What I want is a completely rigorous proof of the statement.

Case when $k=1$

I have solved the case for $k=1$. If it was a $1$-parameter exponential family, we shall have $$1+(x-y)^2=h(x)l(y)e^{p(x)q(y)}\text{ }\forall x,y\in\mathbb{R}.$$ Replacing $h$ by $\vert h\vert$ and $l$ by $\vert l\vert$ when necessary and taking logarithm, $$\log(1+(x-y)^2)=f(x)+g(y)+p(x)q(y)\text{ }\forall x,y\in\mathbb{R}.$$ Now, suppose $x_1\neq x_2$. $$\log(1+(x_1-y)^2)-\log(1+(x_2-y)^2)=f(x_1)-f(x_2)+(p(x_1)-p(x_2))q(y)\text{ }\forall y\in\mathbb{R}.$$ Suppose $p$ is not injective, then $\log(1+(x_1-y)^2)-\log(1+(x_2-y)^2)$ would be a constant, which is not true. Similarly, $q$ is injective. But this means that \begin{equation*} \frac{\log(1+(x_1-y)^2)-\log(1+(x_2-y)^2)-(f(x_1)-f(x_2))}{p(x_1)-p(x_2)} \end{equation*} is an injective function of $y$ for any $x_1\neq x_2$. This is not true, and we arrive at a contradiction.

General case

The general case is equivalent to have functions $p_i,q_i:\mathbb{R}\to\mathbb{R}$ for $i=1,\dots,k,$ so that \begin{equation*} 1+(x-y)^2=h(x)l(y)\exp\left(\sum_{i=1}^kp_i(x)q_i(y)\right)\text{ }\forall x,y\in\mathbb{R}. \end{equation*} By taking $p_{k+1}(x)=h(x),q_{k+1}(y)=1$ and $p_{k+2}(x)=1,q_{k+2}(y)=l(y)$, we may ignore the things outside $\exp$ and write $$\log(1+(x-y)^2)=\sum_{i=1}^{k+2}p_i(x)q_i(y).$$ And for this I barely have an idea of how to do.

Any help would be appreciated!

Answer 1

The answer is No.

Lemma. Let $f(x,y)=\mathbb{log}(1+(x-y)^2)$. For all $n\ge1$, there exists distinct real numbers $x_i,y_i(1\le i \le n)$ that $\mathbb{det}(A)\ne 0$ for $A=(a_{ij})=(f(x_i,y_j))$.

Proof by induction. $n=1$ case is trivial. Now assume $n-1$ case proved, that is the $n-1\times n-1$ matrix $A_0=(f(x_i,y_j)),i,j\le n-1$ being nonsingular. Choose an arbitrary $y_n$. Consider $x_1,\cdots,x_{n-1},y_1,\cdots,y_n$ as fixed, $x_n$ as a variable and denote the $n\times n$ matrix $A(x_n)=(f(x_i,y_j))$. By the laplace expansion on the last row of $A(x_n)$, we get $$\mathbb{det}(A(x_n))=\sum_{i=1}^n {(-1)^i A_i f(x_i,y_{n+1-i})}$$ where $A_i$ is the determinant of $n-1\times n-1$ submatrix of $A(x_n)$ in the first $n-1$ rows. (Obviously there are $n$ submatrices). By induction we have $A_1=\mathbb{det}(A_0)$ nonzero. Hence $\mathbb{exp}(\mathbb{det}(A(x_n)))$ is of the form $\prod{(1+(x_n-y_i)^2)^{\pm A_i}}$. Since all $y_i$ are distinct and $A_1 \ne 0$, there should exist $x_n$ that make $\mathbb{exp}(\mathbb{det}(A(x_n)))\ne 1$ and all $x_i,y_i$ distinct (the detailed proof should be very complicated). Hence $\mathbb{det}(A(x_n))\ne 0$, as claimed.

Now we solve the original problem. If we have $f(x,y)=\sum_{i=1}^n {p_i(x)q_i(y)}$, then for any $2n+2$ real numbers $x_1,\cdots,x_{n+1},y_1,\cdots,y_{n+1}$, consider the matrix $A=(f(x_i,y_j))$. It has determinant

$$\mathbb{det}A=\sum_{1\le k_i \le n,i=1,\dots,n+1} {\mathbb{det}((p_{k_i}(x_i)q_{k_i}(y_j)))}$$

By pigeonhole principle, each of the matrices on the RHS has two rows (corresponding to $i_1$ and $i_2$) that $k_{i_1}=k_{i_2}$, hence proportional, hence has determinant $0$. So $A$ also has determinant $0$, which contradicts the lemma.