Let $\Phi: \mathbb{R} \to [0,1]$ define the standard normal cdf function. I am trying to find some functions $f$ and $g$ -- if they exist -- such that $$\Phi(x)f(y)=g(x+y)$$ for all $(x,y)\in \mathbb{R}^2$. I don't care about an explicit form for $g$, but would like to find some $f$ so that the above holds.
Any hint regarding the existence of such $f$ and $g$ or the form of $f$ would be much appreciated!
On
Take derivatives with respect to $y$ and with respect to $x$. Subtract the resulting equations. You obtain $$ \phi'(x)f(y)-\phi(x)f'(y)=0\\ \Leftrightarrow \frac{d}{dx}\ln(\phi(x))=\frac{d}{dy}\ln(f(y)) $$ Conclude that both sides must be constant. But the left-hand side isn't
On
If that functional equation is to be true for all $(x,y)$, then in particular it should hold identically along the $x$- and $y$-axes. If $y=0$, then $\Phi(x)f(0)=g(x)$ for all $x$; if $x=0$, then $\Phi(0)f(y)=g(y)$ for all $y$. Between these, we conclude $f(x)=\dfrac{f(0)}{\Phi(0)}\Phi(x)$ for all $x$ i.e. $f(x)\propto \Phi(x)$. But then $\Phi(x)f(y)\propto \Phi(x)\Phi(y)$, which is not a function of $x+y$.
This isn't possible. If the condition $$\Phi(x)f(y)=g(x+y)\tag1$$ holds for all $x,y$, then taking $x=-y$ yields $$f(y)={g(0)\over \Phi(-y)}={c\over 1-\Phi(y)}$$ as the explicit form for $f$ (with $c$ a constant). Plugging this into (1), we find that $${c\Phi(x) \over 1-\Phi(y)}=g(x+y)$$ cannot hold for any $g$: The value of the LHS when $(x,y)=(1,0)$ is different from the value when $(x,y)=(0,1)$, but the RHS is $g(1)$ in both cases.