Joint characteristic function only about $\omega_1+\omega_2$

Question

Consider joint characteristic function of 2 random variables $X_1,X_2$ to be $\Phi(\omega_1,\omega_2)$, how to prove that $X_1=X_2$ if and only if $\Phi(\omega_1,\omega_2)=h(\omega_1+\omega_2)$.

Answer 1

I am not very familiar with the characteristic function and fourier transform, but lets give a try.

$\Rightarrow )$ Suppose $X_1 = X_2$. Then $$ \Phi(\omega_1, \omega_2) = E[e^{i(\omega_1X_1+\omega_2X_2)}] = E[e^{i(\omega_1 + \omega_2)X_1}] = h(\omega_1+\omega_2)$$

where $h$ is the common characteristic function of $X_1$ and $X_2$

$\Leftarrow )$ Suppose $\Phi(\omega_1, \omega_2) = h(\omega_1+\omega_2)$. Note that $X_1 = X_2 \iff X_1 - X_2 = 0$. Consider the characteristic function of $X_1 - X_2$: $$ E[e^{i\omega(X_1 - X_2)}] = E[e^{i(\omega X_1 - \omega X_2)}] = \Phi(\omega, -\omega) = h(\omega - \omega) = h(0) = 1$$ which is a constant function independent of $\omega$. Last equality follows from the fact that $h(0) = h(0 + 0) = \Phi(0, 0) = 1$. And only the degenerate distribution at $0$ will have this characteristic function.