The characteristic function determines the distribution

Statement

Let $X$ and $Y$ be real random variables with characteristic functions $\varphi_X$ and $\varphi_Y$. If $\varphi_X(t) = \varphi_Y(t)$ for all $t \in \mathbb{R}$, then $X$ and $Y$ have the same law.

Remarks

The proof uses the Lévy inversion formula. For real $a < b$ that are continuity points of the cumulative distribution function $F_X$, $$F_X(b) - F_X(a) = \lim_{T \to \infty} \frac{1}{2\pi} \int_{-T}^{T} \frac{e^{-ita} - e^{-itb}}{it} \, \varphi_X(t) \, dt.$$ If $\int |\varphi_X(t)| \, dt < \infty$, then $X$ has the continuous density $f_X(x) = \frac{1}{2\pi} \int e^{-itx} \varphi_X(t) \, dt$. The same result holds for random vectors in $\mathbb{R}^n$, with $\varphi_X(t) = E[e^{i \langle t, X \rangle}]$.

Explanations

L1 human

Two different distributions cannot have the same characteristic function. So, to identify a distribution, it is enough to compute its characteristic function and recognize it.