Suppose $X_1, X_2, Y_1, Y_2$ are independent random variables on the same probability space with densities $f_1,f_2,g_1,g_2$ respectively.
If $$ \int_{x} f_1(x)f_2(z-x) \,dx = \int_{y} g_1(y)g_2(z-y) \,dy$$ for all feasible $z$
then is it true that $f_1 = g_1$ or $g_2$ and $f_2 = g_2$ or $g_1$ correspondingly?
This question raised when I was pondering over providing an example for two different sets of two densities which result in the same when convolution operation is done as formulated above.
Thanks in advance for any kind of help and ideas.
No.
As counterexample take all rv's independent and normal with parameters $\langle\mu_i,\sigma_i^2\rangle$ for the $X_i$ and $\langle\nu_i,\rho_i^2\rangle$ for the $Y_i$.
To achieve that $X_1+X_2$ and $Y_1+Y_2$ (having the convolutions as PDF) have the same distribution it is sufficient if $$\mu_1+\mu_2=\nu_1+\nu_2\text{ and }\sigma^2_1+\sigma^2_2=\rho^2_1+\rho^2_2$$
It is not necessary that $\{\langle\mu_1,\sigma_1^2\rangle,\langle\mu_2,\sigma_2^2\rangle\}=\{\langle\nu_1,\rho_1^2\rangle,\langle\nu_2,\rho_2^2\rangle\}$.