If $\mu$ is an infinitely divisible probability measure on $[0,\infty)$, then the Lévy measure of $\mu$ is the vague limit of $n\mu^{*1/n}$

Question

If $\nu$ is a finite measure on $(\mathbb R,\mathcal B(\mathbb R))$, let $\nu^{\ast k}$ denote the $k$-fold convolution¹ of $\nu$ with itself for $k\in\mathbb N_0$, $$\exp(\nu)\mathrel{:=}\sum_{k=0}^\infty\frac{\nu^{\ast k}}{k!}$$ and $$\operatorname{CPoi}_\nu\mathrel{:=}\frac{\exp(\nu)}{\exp(\nu)(\mathbb R)}.$$ Moreover, let $$\mathcal L_\nu(t)\mathrel{:=}\int\nu({\rm d}x)e^{-tx}\;\;\;\text{for $t\in\mathbb R$}$$ denote the Laplace transform of $\nu$.

Let $\mu$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$. Remember that $\mu$ is called infinitely divisible if for all $k\in\mathbb N$, there is a probability measure $\nu$ on $(\mathbb R,\mathcal B(\mathbb R))$ with $\mu=\nu^{\ast k}$.

We can show that

1. $\mu$ is infinitely divisible;

and

2. There is a sequence $(\nu_n)_{n\in\mathbb N}$ of finite measures on $(\mathbb R,\mathcal B(\mathbb R))$ with $\operatorname{CPoi}_{\nu_n}\xrightarrow{n\to\infty}\mu$ weakly

are equivalent.

Now assume $\mu$ is a probability measure on $([0,\infty),\mathcal B([0,\infty)))$. By the Lévy-Khinchin formula,

3. $\mu$ is infinitely divisible;

and

4. There is a $\alpha\ge0$ and a $\sigma$-finite measure $\nu$ on $((0,\infty),\mathcal B((0,\infty)))$ with $$-\ln\mathcal L_\mu(t)=\alpha t+\int1-e^{-tx}\:\nu({\rm d}x)\;\;\;\text{for all $t\ge0$}\tag4$$

are equivalent.

Question: How can we show that if (3.) holds, then the $\nu$ from (4.) is equal to the vague limit² of $\left.n\mu^{\ast1/n}\right|_{(0,\:\infty)}$?

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¹ I.e. $\nu^{\ast 0}:=\delta_0$ is the Dirac measure on $(\mathbb R,\mathcal B(\mathbb R))$ at $0$ and if $k\in\mathbb N$, then $\nu^{\ast k}$ is the pushforward $\tau_k\left(\nu^{\otimes k}\right)$ of the $k$-fold product measure $\nu^{\otimes k}$ of $\nu$ with itself under the map $$\tau_k:\mathbb R^k\to\mathbb R\;,\;\;\;x\mapsto x_1+\dotsb+x_k.$$

² I.e. $$n\int_{(0,\:\infty)}f\:{\rm d}\mu^{\ast1/n}\xrightarrow{n\to\infty}\int f\:{\rm d}\nu\;\;\;\text{for all }f\in C_0((0,\infty))\tag5.$$

Answer 1

There are at least three ways to show that $n \mu^{*1/n}$ converges to $\nu$ vaguely in $\mathbb{R} \setminus \{0\}$. Let $X_t$ be the Lévy process such that $X_1$ has distribution $\mu$, and let $f$ be a smooth, compactly supported $f$ on $\mathbb{R} \setminus \{0\}$. It is sufficient to show that $t^{-1} \mathbb{E} f(X_t) \to \int f(x) \nu(dx)$ as $t \to 0^+$. Here are brief descriptions of the three methods.

1. Write $X_t = Y_t + Z_t$, where $Y_t$ is a compound Poisson process with Lévy measure $\nu$ restricted to $\mathbb{R} \setminus (-\varepsilon, \varepsilon)$ for a sufficiently small $\varepsilon > 0$ (such that $f = 0$ on $[-2\varepsilon, 2\varepsilon]$), and $Z_t$ is the "remaining part" of $X_t$ (the existence of such decomposition follows from the Lévy–Itô theorem — see Theorem 19.2 in Sato's book on Lévy processes). Then $$t^{-1} \mathbb{E} |f(Y_t)| \to \int f(x) \nu(dx)$$ as $t \to 0^+$, as can be easily proved by conditioning on the number of jumps (only the term with a single jump has positive contribution). Furthermore, $$t^{-1} \mathbb{E} |f(X_t) - f(Y_t)| \leqslant t^{-1} \mathbb{P}[|Y_t| < \varepsilon] \mathbb{P}[|Z_t| > \varepsilon] + t^{-1} \mathbb{P}[|Y_t| \geqslant \varepsilon] \|f'\|_\infty \mathbb{E} |Z_t|$$ and both terms can be verified to converge to zero as $t \to 0^+$.

2. Use an appropriate variant of Plancherel's theorem: if $\psi = -\ln \varphi_\mu$ is the characteristic (Lévy—Khintchine) exponent, then $$t^{-1} \mathbb{E} f(X_t) = t^{-1} \mathbb{E} (f(X_t) - f(0)) = \int \hat{f}(z) t^{-1} (e^{-t \psi(z)} - 1) dz \to -\int \hat{f}(z) \psi(z) dz $$ by the dominated convergence theorem. Using the explicit form of $\psi$, the facts that $f'(0) = f''(0) = 0$, and again an appropriate variant of Plancherel's theorem, we can show that $$-\int \hat{f}(z) \psi(z) dz = \int f(x) \nu(dx),$$ as desired.

3. My favourite one, using semigroup theory. Let $L$ be the generator of the transition semigroup $P_t$ of $X_t$. We have $$t^{-1} \mathbb E f(X_t) = t^{-1} (\mathbb E f(X_t) - f(0)) = t^{-1} (P_t f(0) - f(0)) \to L f(0) $$ as $t \to 0^+$. By the expression for $L$ and the fact that $f(0) = f'(0) = f''(0) = 0$, we have $$L f(0) = \int f(x) \nu(dx).$$

I am pretty much sure one can find the above arguments in the literature, but I do not have a reference off the top of my head.

Answer 2

Partial answer

Let

• $(b,\sigma)\in\mathbb R\times[0,\infty)$;
• $\nu$ be a $\sigma$-finite measure on $\mathbb R$ with $$\int1\wedge x^2\:\nu({\rm d}x)<\infty\tag a$$ and $\nu(\{0\})=0$;
• $\mu$ be a probability measure on $\mathbb R$ with $$\ln\varphi_\mu(t)={\rm i}tb-\frac{\sigma^2}2t^2+\int e^{{\rm i}tx}-1-1_{(-1,\:1)}(x){\rm i}tx\:\nu({\rm d}x)\tag b$$ for all $t\in\mathbb R$.

We can construct a real-valued random variable $Y$ on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $Y\sim\mu$ in the following way:

• Let $$I_k:=\left(-\frac1k,-\frac1{k+1}\right]\cup\left[\frac1k,\frac1{k+1}\right)$$ and $$\nu_k(B):=\nu(B\cap I_k)\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ for $k\in\mathbb N$
• Note that $$\nu(I_0)+\int_{(-1,\:1)}\nu({\rm d}x)=\int1\wedge x^2\:\nu({\rm d}x)<\infty\tag c.$$
• Let $(X_k)_{k\in\mathbb N_0}$ be a real-valued independent process on $(\Omega,\mathcal A,\operatorname P)$ with$^1$ $$X_k\sim\operatorname{CPoi}_{\nu_k}\;\;\;\text{for all }k\in\mathbb N_0\tag d.$$
• Note that $$\operatorname E[X_k]=\int_{I_k}\nu({\rm d}x)x\tag e$$ and $$\operatorname{Var}[X_k]=\int_{I_k}\nu({\rm d}x)x^2\tag f$$ for all $k\in\mathbb N_0$.
• It's easy to see that $$M_k:=\sum_{i=1}^k\left(X_i-\operatorname E\left[X_i\right]\right)\;\;\;\text{for }k\in\mathbb N$$ is a martingale with $$\operatorname E\left[M_k^2\right]=\sum_{i=1}^k\operatorname{Var}[X_i]\tag g\;\;\;\text{for all }n\in\mathbb N$$
• Let $\mathcal F^X_\infty:=\sigma(X_k:k\in\mathbb N)$.
• Then, $$\sup_{k\in\mathbb N}\operatorname E\left[M_k^2\right]=\int_{(-1,\:1)}\nu({\rm d}x)x^2<\infty\tag h$$ and hence $$M_k\xrightarrow{k\to\infty}M_\infty\;\;\;\text{almost surely}\tag i$$ for some real-valued $\mathcal F^X_\infty$-measurable square-integrable random variable $M_\infty$ on $(\Omega,\mathcal A,\operatorname P)$ by the martingale convergence theorem.
• Now let $Z$ be a real-valued standard normally distributed random variable on $(\Omega,\mathcal A,\operatorname P)$ independent of $(X_n)_{n\in\mathbb N_0}$
• It's easy to show that $$Y:=b+\sigma Z+X_0+M_\infty\sim \mu.$$

Now we know that there characteristic exponent of $\mu^{\ast1/n}$ is similarly given by $$\ln\varphi_{\mu^{\ast1/n}}(t)={\rm i}tb^{(n)}-\frac{\left|\sigma^{(n)}\right|^2}2t^2+\int e^{{\rm i}tx}-1-1_{(-1,\:1)}(x){\rm i}tx\:\nu^{(n)}({\rm d}x)\tag j,$$ where $b^{(n)}:=b/n$, $\sigma^{(n)}:=\sigma/\sqrt n$ and $\nu^{(n)}:=\nu/n$, for all $n\in\mathbb N$. Define $\left(\nu^{(n)}_k,X^{(n)}_k,M^{(n)}_k,M^{(n)}_\infty,Y^{(n)}\right)$ in the same way as before, but with $(b,\sigma,\nu)$ replaced by $\left(b^{(n)},\sigma^{(n)},\nu^{(n)}\right)$.

We then should be able to show that for all $\varepsilon_1,\varepsilon_2>0$, there are $k_0,n_0\in\mathbb N$ with $$p_{k_0}^{(n)}:=\operatorname P\left[\left|b^{(n)}+\sigma^{(n)}Z-\sum_{k=1}^{k_0}\operatorname E\left[X^{(n)}_k\right]\right|+\sum_{k>k_0}\left(X_k^{(n)}-\operatorname E\left[X_k^{(n)}\right]\right)>\varepsilon_1\right]\le\frac{\varepsilon_2}n\tag k$$ for all $n\ge n_0$.

Now let $$W^{(n)}:=\sum_{k=0}^{k_0}X_k^{(n)}\;\;\;\text{for }n\in\mathbb N.$$

In order to conclude that $n\mu^{\ast1/n}\to\nu$ vaguely, it is sufficient to show that $$n\operatorname P\left[Y^{(n)}\in(a,b]\right]\xrightarrow{n\to\infty}\nu((a,b])\tag l$$ for all $a,b\in\mathbb R$ with $a<b$ and $\nu(\{a\})=\nu(\{b\})=0$.

Using the simple estimate $\operatorname P\left[U\in(a+\varepsilon_1,b-\varepsilon_1]\right]-\operatorname P[|V|>\varepsilon_1]\le\operatorname P[U+V\in(a,b]]\le\operatorname P[U\in(a-\varepsilon_1,b+\varepsilon_1]]+\operatorname P[|V|>\varepsilon_1]$, for every random variables $U,V$, we obtain \begin{equation}\begin{split}&\operatorname P\left[W^{(n)}\in(a+\varepsilon_1,b-\varepsilon_1]\right]-p_{k_0}^{(n)}\\&\;\;\;\;\le\operatorname P[Y^{(n)}\in(a,b]]\\&\;\;\;\;\le\operatorname P[W^{(n)}\in(a-\varepsilon_1,b+\varepsilon_1]]+p_{k_0}^{(n)}\end{split}\tag m\end{equation} for all $n\ge n_0$. On the other hand, letting $J_{k_0}:=\bigcup_{k=0}^{k_0}I_k$, $$n\operatorname P\left[W^{(n)}\in B\right]=\nu(B\cap J_{k_0})\xrightarrow{k_0\to\infty}\nu(B)\;\;\;\text{for all }B\in\mathcal B(\mathbb R)\tag n.$$ So, unless I'm missing something, we should be able to conclude as long as we can justify that we can take $k_0$ as large as desired, while maintaining $(k)$.

I would highly appreciated if someone could fill this gap.

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$^1$ One issue, which I'm unable to resolve at the moment, is that $\operatorname{CPoi}_{\eta}$ is only well-defined, when $\eta$ is a finite measure on $\mathbb R$, but (unless I'm missing something) the assumptions don't imply that $\nu_k$ is finite for all $k\in\mathbb N_0$. Since $\nu$ is arising from the Lévy-Khinchin formula, we may be able (I don't know whether this is true) to assume that $\nu$ is locally finite. Under this assumption, $\nu_k$ would be finite for all $k\ne0$ (since the support of $\nu_k$ is contained in $(-1,1)$ for $k\ne0$).