The convolution of two Borel measures $\mu$ and $\nu$ is given by $$(\mu * \nu)(E) = \int_{-\infty}^\infty \nu(E - x) \; \mu(d x).$$

I have been trying to figure out whether the following cancellation law holds for convolutions: $$\text{if } \nu \neq 0 \text{ and } \mu_1 * \nu = \mu_2 * \nu, \text{ then also } \mu_1 = \mu_2$$ Here I use $\mu = \nu$ to mean that $\mu(E) = \nu(E)$ for all Borel measurable $E$.

Intuitively, I feel that the cancellation property should hold, but I have seen examples of the cancellation property failing for very similar kinds of convolution, so I am unsure.

Does the cancellation property above hold for convolution of measures? And if not, can we recover it by restricting for example to probability measures and/or continuous measures?

  • $\begingroup$ $\mu$ and $\nu$ are only defined on measurable sets, so you can't really interpret it any way else $\endgroup$ – Jakobian Jul 16 '19 at 13:18
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    $\begingroup$ This is definitely not true for complex measures (even continuous): take any continuous compactly supported, nonzero functions $f,g$ with disjoint supports, and look at the (continuous) measures $\mu,\nu_1,\nu_2$ such that $\hat\mu=f$, $\hat\nu_1=g$ and $\hat \nu_2=2g$. Then $\mu*\nu_1=\mu*\nu_2=0$, but $\nu_1\neq2\nu_1=\nu_2$. $\endgroup$ – tomasz Jul 16 '19 at 13:32
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    $\begingroup$ Even simpler, consider $\mu_1(E) = 0$ for every measurable $E$. We either need stronger conditions on $\mu_1$, or consider the situation in general, where equality holds for any $\mu_1$. $\endgroup$ – Jakobian Jul 16 '19 at 13:36
  • $\begingroup$ @Jakobian: Yeah, but that is completely trivial, obviously the question only makes sense for nonzero measures. $\endgroup$ – tomasz Jul 16 '19 at 13:37
  • $\begingroup$ @Jakobian Right, $\mu_1$ must of course be non-zero, I will add that. $\endgroup$ – mrp Jul 16 '19 at 13:38

Definition: We say that $\mu$ is infinitely divisible, if for any $n$ there exists a measure $\mu_n$ such that $\mu = (\mu_n)^{*n}$, the $n$-th convolution power. In other words, for any $n$ there exists $n$-th root of $\mu$.

Theorem: Suppose that $\mu_1, \mu_2, \nu$ are of finite measure and $\nu\neq 0$ is infinitely divisible. It holds that $$\mu_1*\nu = \mu_2*\nu \implies \mu_1 = \mu_2.$$

Proof: Without loss of generality we may assume that $\mu_1, \mu_2, \nu$ are probability measures. This is because for any Borel measure $\mu$, $\mu \neq 0\iff \mu(R)>0$, and $$(\mu_1*\nu)(R) = \mu_1(R)\nu(R) = \mu_2(R)\nu(R) = (\mu_2*\nu)(R).$$ From which $\mu_1(R) = \mu_2(R)$. If $\mu_1(R) = 0$, then $\mu_1 = \mu_2 = 0$ as desired. If $\mu_1(R) \neq 0$, then we can divide both sides by $\mu_1(R)\nu(R) = \mu_2(R)\nu(R)$, and take new (this time probability) measures $$\mu_1' = \frac{\mu_1}{\mu_1(R)}, \mu_2' = \frac{\mu_2}{\mu_2(R)}, \nu' = \frac{\nu}{\nu(R)}.$$ Those new measures satisfy $\mu_1'*\nu' = \mu_2'*\nu'$ since convolution is bilinear. We will keep refering to those as $\mu_1, \mu_2, \nu$.

Consider $\varphi_1, \varphi_2, \phi$, the characteristic functions of $\mu_1, \mu_2,\nu$. We have $$\varphi_{\mu_1*\nu} = \varphi_1\phi = \varphi_2\phi = \varphi_{\mu_2*\nu},$$ hence $$\forall_{x\in R}\ (\varphi_1(x) = \varphi_2(x)\ \lor \phi(x) = 0). \tag{1} $$ It is because for any probability measures $\mu, \hat{\mu}$ we can consider independent random variables $X\sim \mu, Y\sim \hat{\mu}$, and it is well known that $X+Y\sim \mu*\hat{\mu}$, hence $\varphi_{\mu*\hat{\mu}} = \varphi_{\mu}\varphi_{\hat{\mu}}$.

From [1] it follows that $\phi$ has no zeros, since it's a characteristic function of infinitely divisible measure. Hence from $(1)$ we have that $\varphi_1 = \varphi_2$, and so $\mu_1 = \mu_2$, as had to be shown.

Examples of infinitely divisible measures are Normal distribution or Cauchy distribution. Distributions of the form $\delta_a$ are also infinitely divisible. Another example would be Poisson distribution.

It can also be shown that it is false in general for probability measures. Consider functions $$\phi(x) = (-|x|+1)1_{[0, 1]}(|x|),\ \varphi_1(x) = e^{-|x|},\\ \varphi_2(x) = e^{-|x|}1_{[0, 1)}(|x|)+e^{-1}(-|x|+2)1_{[1, 2]}(|x|) $$ We have that those are characteristic functions from Pólya criterion, and they satisfy $(1)$, which is equivalent to $\varphi_1\phi = \varphi_2\phi$, and so $\mu_1*\nu = \mu_2*\nu$ where $\mu_1, \mu_2, \nu$ are probability measures corresponding to $\varphi_1, \varphi_2,\phi$, and $\mu_1 \neq \mu_2$ because $\varphi_1\neq \varphi_2$.

There can be doubt about if $\varphi_2$ is convex for positive values because of the point $x = 1$, but it can be easily checked that $(e^{-x})'_{x=1} = -e^{-1}$, so it is indeed a convex function.

In fact, those characteristic functions are all integrable, hence they correspond to absolutely continuous random variables (in addition, symmetric).

[1] https://math.stackexchange.com/a/416436/476484


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