Suppose that $X_{1}, X_{2}, \ldots $ are i.i.d. symmetric random variables with the finite variance $\sigma^{2}$, let $N_{p}\in \text{Fs}(p)$ be independent of $X_{1},X_{2},\ldots$, and set $Y_{p}=\sum_{k=1}^{N_{p}}X_{k}$. Show that \begin{equation*} \begin{split} \sqrt{p}Y_{p}\overset{d}{\rightarrow} L(a)\qquad \text{as}\quad p\rightarrow 0. \end{split} \end{equation*} and determine $a$.
Attempt at solution:
$ \psi_{X}(t)=\mathrm{E}[e^{tX}]\\ \psi_{X}(\sqrt{(0)}t))=1\\ \frac{d}{dp}\psi_{X}(\sqrt{p}t))=\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)\\ g_{N_{p}}(t)=\sum_{k=1}^{\infty}t^{k}(1-p)^{k-1}p=\frac{p}{1-p}\sum_{k=1}^{\infty}t^{k}(1-p)^{k}\\ =\frac{p}{1-p}\sum_{k=1}^{\infty}(t(1-p))^{k}=\frac{p}{1-p}(\frac{1}{1-(t(1-p)}-\frac{1-(t(1-p)}{1-(t(1-p)})\\ =\frac{p}{1-p}(\frac{(t(1-p)}{1-(t(1-p)})=\frac{pt}{1-(1-p)t}\\ \sqrt{p}Y_{p}= \sqrt{p}\sum_{k=1}^{N_{p}}X_{k}=\sum_{k=1}^{N_{p}}(\sqrt{p}X_{k})\\ \psi_{\sqrt{p}Y_{p}}(t)=g_{N_{p}}(\psi_{X}(\sqrt{p}t))=\frac{p\psi_{X}(\sqrt{p}t)}{1-(1-p)\psi_{X}(\sqrt{p}t)}\\ \lim_{p\rightarrow 0}\sqrt{p}Y_{p}=\lim_{p\rightarrow 0}\frac{p\psi_{X}(\sqrt{p}t)}{1-(1-p)\psi_{X}(\sqrt{p}t)}\\ \text{indeterminate, use l'hopital}\qquad \frac{0}{0}\\ =\lim_{p\rightarrow 0}\frac{\frac{d}{dp}}{\frac{d}{dp}}\frac{p\psi_{X}(\sqrt{p}t)}{1-(1-p)\psi_{X}(\sqrt{p}t)}\\ =\lim_{p\rightarrow 0}\frac{p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t)}{-\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t)}\\ =\lim_{p\rightarrow 0}\frac{p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t)}{\frac{1}{2\sqrt{p}}(-t\psi_{X}'(\sqrt{p}t)+pt\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t))}\\ =\lim_{p\rightarrow 0}\frac{2\sqrt{p}(p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t))}{(p-1)t\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t)}\\ =\lim_{p\rightarrow 0}\frac{tp\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t)}{(p-1)t\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t)}\\ $
At this point I'm stuck, this limit goes to $\frac{0}{t\psi_{X}'(\sqrt{p}t)}=0$. I guess that I am supposed to use the fact that the distribution of $X_{k}$ is symmetric, but cannot figure out how.
I worked on it some more and got this solution: \begin{equation*} \begin{split} \varphi_{X}(t)=\mathrm{E}(e^{itX})\\ Y_{p}=\sum_{k=1}^{N_{p}}X_{k}=(\varphi_{X_{1}}(t))^{N_{p}} \end{split} \end{equation*} Since \begin{equation*} \begin{split} \varphi_{X}(t)=1+\sum_{k=1}^{n}X^{k}\cdot \frac{(it)^{k}}{k!}+\mathscr{o}(|t|^{n})\quad \text{as}\quad t\rightarrow 0 \end{split} \end{equation*} we get that \begin{equation*} \begin{split} \varphi_{X}(\sqrt{p} t)=1+\sum_{k=1}^{n}X^{k}\cdot \frac{(i\sqrt{p}t)^{k}}{k!}+\mathscr{o}(|\sqrt{p}t|^{n})\quad \text{as}\quad p\rightarrow 0 \end{split} \end{equation*} Thereby \begin{equation*} \begin{split} \sqrt{p}Y_{p}=\sum_{k=1}^{N_{p}}\sqrt{p}X_{k}=(\varphi_{X_{1}}(\sqrt{p}t))^{N_{p}}=(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\mathrm{E}X^{2}+\mathscr{o}(pt^{2}))^{N_{p}}\\ =(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))^{N_{p}} \end{split} \end{equation*} \begin{equation*} \begin{split} g_{N_{p}}(t)=\mathrm{E}(t^{N_{P}})=\sum_{k=1}^{\infty}t^{k}(1-p)^{k-1}p=\frac{p}{1-p}\sum_{k=1}^{\infty}t^{k}(1-p)^{k}\\ =\frac{p}{1-p}\sum_{k=1}^{\infty}(t(1-p))^{k}=\frac{p}{1-p}(\frac{1}{1-(t(1-p)}-\frac{1-(t(1-p)}{1-(t(1-p)})\\ =\frac{p}{1-p}(\frac{(t(1-p)}{1-(t(1-p)})=\frac{pt}{1-(1-p)t}\\ \end{split} \end{equation*} \begin{equation*} \begin{split} \varphi_{\sqrt{p}Y_{p}}(t)=g_{N_{p}}( \varphi_{X}(\sqrt{p} t))=\mathrm{E}((1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))^{N_{p}})\\ =\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{1-(1-p)(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\ =\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{1-(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\ =\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{-i\sqrt{p}t\mathrm{E}X+\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\ =\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{p(-i\frac{\sqrt{p}}{p}t\mathrm{E}X+\frac{t^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})))}\\ =\frac{1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})}{-i\frac{\sqrt{p}}{p}t\mathrm{E}X+\frac{t^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\ \end{split} \end{equation*} Since \begin{equation*} \begin{split} Z\in L(0,\sigma)\\ U=Z+\mu\\ \psi_{U}(t)=\psi_{Z+\mu}(t)=\mathrm{E}(e^{t(Z+\mu)})=\mathrm{E}(e^{\mu t} \cdot e^{tZ})=e^{\mu t}\mathrm{E}(e^{tZ})=e^{\mu t}\cdot \frac{1}{1-\frac{\sigma^{2}}{2}t^{2}}\\ U \in L(\mu,\sigma) \end{split} \end{equation*} \begin{equation*} \begin{split} \mathrm{E}(X)=0 \end{split} \end{equation*} and \begin{equation*} \begin{split} \lim_{p\rightarrow 0} \varphi_{\sqrt{p}Y_{p}}(t)= \lim_{p\rightarrow 0}g_{N_{p}}( \varphi_{X}(\sqrt{p} t))\\ =\lim_{p\rightarrow 0}e^{\mu t}\cdot \big(\frac{1+i\sqrt{p}t(0)-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})}{-i\frac{\sqrt{p}}{p}t(0)+\frac{t^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+(1+i\sqrt{p}t(0)-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\big)\\ =e^{\mu t}\cdot \frac{1}{1-\frac{\sigma^{2}}{2}t^{2}}\\ \sqrt{p}Y_{p}\rightarrow L(\sigma)\quad \text{as} \quad p\rightarrow 0. \end{split} \end{equation*}