# Characteristic functions agree in a neighborhood of zero

Suppose I have two random variables $$X,Y$$ with finite mean. Let's say that their characteristic functions agree in a small neighborhood of zero (this means that there is some $$\epsilon>0$$ such that $$\Bbb E[e^{itX}] = \Bbb E[e^{itY}]$$ for all $$|t|<\epsilon$$). Can I conclude that $$X \stackrel{d}{=} Y$$?

Remarks: If I remove the condition of finite mean then certainly the answer is no. This is a trivial consequence of Polya's criterion (see Theorem 3.3.22 in Durrett's book). On the other hand, if I instead impose the stronger moment condition that $$\Bbb E[e^{\lambda|X|}]< \infty$$ for some small $$\lambda >0$$, then $$X$$ and $$Y$$ certainly must have the same distribution (because then the characteristic functions extend to analytic functions on some domain of $$\Bbb C$$ which contains the entire real line).

Hence the real underlying question is: what is the minimal number of moments needed by $$X,Y$$ so that $$X \stackrel{d}{=} Y$$ under the given assumptions on characteristic functions? I suspect that the mgf condition is the minimal one (and tried to construct a counterexample using the lognormal distribution), but I could not prove it. If that's wrong then my next guess is that two moments or one moment would suffice, hence the original question.

• My updated answer is worth a read. – Gabriel Romon Sep 4 '19 at 17:03
• @GabrielRomon thanks so much for that answer. The Fourier analytic approach is short so I accepted that, but a purely probabilistic approach is most certainly valuable as well! – shalop Sep 5 '19 at 1:30

Say $$X,Y$$ are absolutely continuous random variables so the question becomes

Given $$f,g \ge 0, \|f\|_{L^1}=\|g\|_{L^1}=1$$ whose Fourier transform $$\hat{f},\hat{g}$$ agree on $$(-\epsilon,\epsilon)$$ is there a simple condition implying that $$f= g$$ ?

I'm affraid that when we can't assume $$\hat{f}-\hat{g}$$ is analytic then the answer is no.

Take $$\hat{\phi} \in C^\infty_c$$ real even and supported on $$|t|> r$$, thus $$\phi$$ is Schwartz real and even, take $$F\ge 0$$ Schwartz such that $$F- \phi\ge 0$$ (*) and let $$f = \frac{F}{\|F\|_{L^1}} \ge 0, \quad g = \frac{F-\phi}{\|F\|_{L^1}} \ge 0, \quad \|f\|_{L^1}=1, \quad \|g\|_{L^1} = \hat{g}(0)=1, \qquad \hat{f}-\hat{g}=\frac{\hat{\phi}}{\|F\|_{L^1}}$$

(*) To construct $$F$$ we can use that for $$\varphi \in C^\infty_c$$ and $$u$$ continuous rapidly decreasing then $$\varphi \ast u$$ is Schwartz

Example 18 in Ushakov's Selected Topics in Characteristic Functions gives two different characteristic functions $$g_1, g_2:\mathbb R\to \mathbb R$$ such that

1. the corresponding distributions have moments of all orders

2. $$|g_1|=|g_2|$$

The equality of absolute values implies that $$g_1=g_2$$ in a neighborhood of $$0$$ (hence the moments are equal).

After inspection I believe there are several typos/mistakes in their example, so I will deviate quite a bit from what's written in the book.

$$\bullet$$ Let $$(a_n)_{n\geq 1}$$ be a sequence of positive reals such that $$\sum_{n=1}^\infty a_n <\infty$$. Let $$A=\sum_{n=1}^\infty a_n$$.

Let $$X_1,X_2$$ be i.i.d with distribution $$\mathcal U([-1,1])$$. Recall that the c.f. of this distribution is $$t\mapsto \frac{\sin(t)}{t}$$. Let $$(Z_n)_{n\geq 1}$$ be i.i.d with the same distribution as $$X_1+X_2$$ and set $$Z=\sum_{n=1}^\infty a_n Z_n$$. Since $$|Z_n|\leq 2$$, the series $$\sum_{n\geq 1} a_n Z_n$$ is absolutely convergent, hence $$Z$$ is well-defined.
By dominated convergence, $$\phi_Z(t) = E(\lim_n e^{it\sum_{k=1}^n a_k Z_k})=\lim_n E( e^{it\sum_{k=1}^n a_k Z_k})=\lim_n \prod_{k=1}^n \left(\frac{\sin(a_kt)}{a_kt}\right)^2 = \prod_{n=1}^\infty \left(\frac{\sin(a_nt)}{a_nt}\right)^2$$

Note that $$\phi_Z$$ is $$\geq 0$$, even and integrable (since it is $$\leq \frac{1}{a_1^2t^2}$$). Integrability implies that the distribution of $$Z$$ is absolutely continuous, let $$f_Z$$ denotes its density. Note that the support of $$f_Z$$ is a subset of $$[-2A,2A]$$. Since $$f_Z(x)=\frac{1}{2\pi} \int e^{-itx} \phi_Z(t) dt = \frac{\int \phi_Z}{2\pi} \int e^{itx} \frac{\phi_Z(-t)}{\int \phi_Z} dt=\frac{\int \phi_Z}{2\pi} \int e^{itx} \frac{\phi_Z(t)}{\int \phi_Z} dt$$

we conclude that $$\frac{2\pi}{\int \phi_Z} f_Z = \phi_Y$$ where $$Y$$ has density $$\displaystyle t\mapsto \frac{\phi_Z(t)}{\int \phi_Z}$$. This implies that the support of $$\phi_Y$$ is a subset of $$[-2A,2A]$$.

$$\bullet$$ Now comes the tricky part. Note that \begin{aligned} \phi_Y(t) + \frac 1{2i}(\phi_Y(t+4A)-\phi_Y(t-4A)) &= \int \frac{\phi_Z(x)}{\int \phi_Z}(\sin(4Ax)+1)e^{itx} dx \\ &= \int f_{T_1}(x) e^{itx} dx \\ &=\phi_{T_1}(t) \end{aligned} where $$T_1$$ is a r.v. with density $$\frac{\phi_Z(x)}{\int \phi_Z}(\sin(4Ax)+1)$$ (this function integrates to $$1$$ because $$\phi_Z$$ is even).

Similarly, \begin{aligned} \phi_Y(t) - \frac 1{2i}(\phi_Y(t+4A)-\phi_Y(t-4A)) &= \int \frac{\phi_Z(x)}{\int \phi_Z}(1-\sin(4Ax))e^{itx} dx \\ &= \int f_{T_2}(x) e^{itx} dx \\ &=\phi_{T_2}(t) \end{aligned} where $$T_2$$ is a r.v. with density $$\frac{\phi_Z(x)}{\int \phi_Z}(1-\sin(4Ax))$$.

By the remark on the support of $$\phi_Y$$,we have for $$t\in[-2A,2A]$$: $$\phi_{T_1}(t) = \phi_{T_2}(t)=\phi_Y(t)$$ and when $$|t|>2A$$ we have $$|\phi_{T_1}(t)| = |\phi_{T_2}(t)|$$ so that $$|\phi_{T_1}(t)| = |\phi_{T_2}(t)|$$ everywhere.

$$\bullet$$ Besides, for any $$n\geq 1$$, $$f_{T_1}(x) = O\left(\frac{1}{x^{2n}} \right)$$ hence $$T_1$$ has moments of every order, and similarly for $$T_2$$.