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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.

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    $\begingroup$ My updated answer is worth a read. $\endgroup$ – Gabriel Romon Sep 4 '19 at 17:03
  • $\begingroup$ @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! $\endgroup$ – shalop Sep 5 '19 at 1:30
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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

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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 $T_2$.

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