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.
Best Answer
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
the corresponding distributions have moments of all orders
$|g_1|=|g_2|$
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 $T_2$.