Consider the set of probability distributions and partition this set into equivalence classes $$[F] = \{G: X_1, X_2 \overset{i.i.d.}\sim F,\, Y_1, Y_2 \overset{i.i.d.}\sim G \implies\mathrm{Law}(X_1 - X_2) = \mathrm{Law}(Y_1 - Y_2)\}$$ In other words, let two distributions be in the same equivalence class if the convolution of the first distribution with its negative is equal to the convolution of the second distribution with its negative. My question concerns what we can say about which distributions are in the same class as one another.

As an initial observation, it is clear that all members of a given location family belong to the same equivalence class. Therefore, let's restrict our attention to $L^1$ and make the normalization that we only consider distributions with mean 0. Again, this equivalence class isn't a singleton. Specifically, $-F \in [F]$, so for asymmetric distributions, each equivalence class has at least 2 members. My question then, is if $-F$ is the only other member of $[F]$. Or short of this, are there any known results that at least describe what $[F]$ looks like?

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    $\begingroup$ In symmetric distributions (where it might make sense to drop the requirement of $L^1$), this might be of interest $\endgroup$ Aug 24, 2019 at 4:36
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    $\begingroup$ @KaviRamaMurthy But if we're restricting to mean 0 (like I said in my post), there's only one degenerate distribution, so this isn't really a counterexample. $\endgroup$ Aug 24, 2019 at 4:46

1 Answer 1


Note that $G\in [F] \iff \forall t \in \mathbb R, \phi_F(t)\phi_F(-t)=\phi_G(t)\phi_G(-t)$.

Assume that $F$ and $G$ have moments of all orders and that $E(|X_1|^n)$ and $E(|Y_1|^n)$ don't grow too fast. Then $\phi_F$ and $\phi_G$ are analytic and $F$ and $G$ are characterized by their moments (i.e. the derivatives $\phi_F^{(n)}(0)$ and $\phi_G^{(n)}(0)$).

Note that the constraint $\forall t \in \mathbb R, \phi_F(t)\phi_F(-t)=\phi_G(t)\phi_G(-t)$ leaves a degree of freedom on all the $\phi_F^{(2n+1)}(0)$ and $\phi_G^{(2n+1)}(0)$. It's not hard to come up with vastly different $\phi_F^{(n)}(0)$ and $\phi_G^{(n)}(0)$ (i.e. vastly different $F$ and $G$) such that $\forall t \in \mathbb R, \phi_F(t)\phi_F(-t)=\phi_G(t)\phi_G(-t)$.

So $[F]$ is probably too complicated to be described explicitely.


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