Show $\mu(A)= \alpha \int_A x^2 dF(x)$ for a compound Poisson process

Question

The question is suppose that $N,Y_1,Y_2,....,Y_n$ are independent and $Y_n$ has the common distribution function $F$ and $N$ has the Poisson distribution with mean $\alpha$.

If $F$ has mean $0$ and finite variance, then the canonical measure $\mu$ defined in \begin{equation*} \phi(t)= \exp \int_{R^1} (e^{itx}-1-itx)\frac{1}{x^2}\mu(dx) \end{equation*}
is specified by $\mu(A)= \alpha \int_A x^2 dF(x)$.

I tried to use the Inversion formula to solve this question, but not able to get the desired answer.

Can anyone please give me some hints to solve this question?

The following part (c) is the statement of the question.

Answer 1

If we substitute $$\mu (dx) = \alpha x^2 dF(x)$$ in your expression for $\phi$, then we get $$\phi (t) = \exp \left[\alpha\int_{\mathbb{R}} (e^{itx} -1 -itx)dF(x)\right]$$ Since $F$ has mean zero, we have $\int_{\mathbb{R}} x dF(x) = 0$, so that $\phi$ simplifies to $$\phi (t) = \exp \left[\alpha \int_{\mathbb{R}} (e^{itx} - 1)dF(x)\right]$$ Notice that this agrees with the characteristic function of $S$. By Fourier inversion (i.e. since characteristic functions determine measures), we conclude that $\mu (A) = \alpha \int_A x^2 dF(x)$ is indeed the canonical measure.

Answer 2

Presumably you $\phi$ is the characteristic function of $$ S=\sum^N_{n=1}Y_n$$ where $S\sim\operatorname{Po}(\alpha)$, $(Y_j:j\in\mathbb{N})$ i.i.d., with common distribution $F$, and $$E[Y_1]=\int y\,F(dy)=0,\qquad\sigma^2:=E[Y^2_1]=\int y^2\,F(dy)<\infty$$ Then $$\begin{align} \phi(t)&=E[e^{itS}]=\sum^\infty_{n=0}E\Big[e^{it\sum^n_{j=1}Y_j}\mathbb{1}(N=n)\Big]=\sum^\infty_{n=0}\frac{e^{-\alpha}}{n!}\alpha^n\Big(\int e^{tyi} F(dy)\Big)^n\\ &=\exp\Big(-\alpha\Big(\int (e^{ity}-1)\,F(dy)\Big)\Big) \end{align}$$ Now, because $\int y F(dy)=0$, and $y\mapsto e^{ity}-1-ity$ vanishes at $0$, $$\begin{align} \int (e^{ity}-1)\,F(dy)&=\int(e^{ity}-1-ity)\,F(dy)\\ &=\int_{\mathbb{R}\setminus\{0\}}(e^{ity}-1-ity)\frac{y^2}{y^2}\,F(dy)\\ &=\int_{\mathbb{R}\setminus\{0\}}(e^{ity}-1-ity)\frac{1}{y^2}\,\mu(dy) \end{align} $$ where $\mu(dy)=y^2\mathbb{1}_{\mathbb{R}\setminus\{0\}}(y)\,F(dy)$. The assumption $\sigma^2<\infty$ implies that $0\leq\mu(\mathbb{R})=\sigma^2<\infty$.