4
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ $\mu$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.