Can any continuous function of two variables be described as a joint of two continuous functions?

Question

(1) Would this be a valid definition of a joint function $f$ of two functions $g$ and $h$? $$g: \mathbb{X} \rightarrow \mathbb{R};\,\,\, h: \mathbb{Y} \rightarrow \mathbb{R}$$ $$f: \mathbb{X} , \mathbb{Y} \rightarrow \mathbb{R};\,\,\,f(x,y) = g(x) \cdot h(y) $$

This is how, e.g., some define (implement) multivariate probability mass functions.

• This thing done to two binomial distributions
• It must be possible to describe bivariate normal distribution as this "product" of two normal distributions

(2) Whatever the proper name of this operation on two functions, question: would it be a correct definition of the bivariate normal distribution to be such $f$ of two normal distribution functions $g$ and $h$?

(3) The main question. Can any continuous function of two variables $f$ be described as such "product" of two continuous functions $g$ and $h$? What special properties would those that cannot carry?

Answer 1

Clearly, not any two-variable map can be written as a product of two one-variable maps.

For example, the map $f(x,y) = x^2 + y^2$ defined on $\mathbb R^2$, vanishes only at $(0,0)$. If $f(x,y) = g(x)h(y)$, then either $g(0)=$ or $h(0)=0$. In the first case you'll have $f(0,y)=0$ for any $y \in \mathbb R$. Which isn't the case. Similar argument if $h(0)=0$. This proves that $f$ can't be written as the product $g(x)h(y)$.

More generally, a map $(x,y) \mapsto g(x)h(y)$ vanishes on lines. This is not the case for general two-variable maps.

Answer 2

Almost no functions from $\Bbb R^2$ to $\Bbb R$ are the product of two functions in the way that you describe.

Consider this: suppose $f$ is non-zero on the $x-$ and $y-$axes, with $f(x,y)=g(x)h(y)$ for all $x,y$. Then $f$ is completely determined by its values on the $x-$ and $y-$axes, because $g(x)=f(x,0)/h(0)$ and $h(y)=f(0,y)/g(0)$. So any other function that agrees with $f$ on the $x-$ and $y-$axes must fail to be a 'joint' function.

As an example, suppose that $f(x,y)$ is a 'joint' function that equals $1$ on the $x-$ and $y$-axes. Then $f(x,y)$ must equal $1$ for all $x,y$. In particular, the function $e^{xy}$ (from a comment by @TheSilverDoe) is equal to $1$ on the $x-$ and $y-$ axes; so it can't be a 'joint' function, because it is not equal to $1$ everywhere.