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I have a random sum of independent random variables, i.e. $S_N=X_1+X_2+\ldots+X_N$, where $$P(X=1)=P(X=-1)=0.5$$ which I think would make $X_i$ belong to a Rademacher distribution. If $N\sim \mathrm{Po}(\lambda)$ and is independent, how can I show that $$\frac{S_N}{\sqrt{\lambda}}\rightarrow N(0,1)$$ as $\lambda$ goes to infinity?

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  • $\begingroup$ You do not need the $S_N$ to be iid but the $X_i$ to be iid. $\endgroup$
    – Jimmy R.
    Oct 6, 2016 at 8:10
  • $\begingroup$ Did you try to compute the characteristic function of $S_N/\sqrt{\lambda}$? $\endgroup$
    – Did
    Oct 6, 2016 at 8:10
  • $\begingroup$ @Did Yes, but I get caught in the Cos(z) , the CF of Rademacher, and can never get it to converge to N(0,1). $\endgroup$
    – litmus
    Oct 6, 2016 at 8:13
  • $\begingroup$ Which function did you get? $\endgroup$
    – Did
    Oct 6, 2016 at 8:25
  • $\begingroup$ @Did I got $\varphi_{\frac{S_N}{\sqrt{\lambda}}}=...=e^{\lambda(e^{i*cos(\frac{t}{\sqrt{\lambda}})}-1)}$, does this converge to $N(0,1)$ as $\lambda$ goes to inf? $\endgroup$
    – litmus
    Oct 6, 2016 at 8:39

1 Answer 1

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Recall the characteristic function of the standard normal distribution: $$\int_{-\infty}^\infty e^{izx}\frac1{\sqrt{2\pi}} e^{-\frac12 x^2}\,\mathsf dx = e^{-\frac12 z^2}. $$ Now, the characteristic function of $X_1$ is $$\varphi_X(z) = \frac12\left(e^{iz} +e^{-iz}\right) = \cos z,$$ and the generating function of $N$ is $$g_N(z) = \sum_{n=0}^\infty z^n\frac{\lambda^n}{n!}e^{-\lambda} = e^{\lambda(z-1)}. $$ Hence the characteristic function of $S_N/\sqrt\lambda$ is \begin{align} \varphi_{S_N/\sqrt\lambda}(z) &= g_N(\varphi_X(z/\sqrt\lambda))\tag1\\ &= g_N(\cos(z/\sqrt\lambda))\\ &= \exp\left(\lambda\left(\cos(z/\sqrt\lambda)-1\right)\right)\\ &=\exp\left(\lambda\left(1-\frac12\frac{z^2}\lambda + O\left(\frac{z^4}{\lambda^2}\right)-1 \right) \right)\\ &\stackrel{\lambda\to\infty}\longrightarrow \exp\left(-\frac12 z^2\right). \end{align}

To justify $(1)$, we have $$ \varphi_{S_N/\sqrt\lambda}(z) = \mathbb E\left[e^{iz\left(S_N/\sqrt{\lambda}\right)} \right] = \mathbb E\left[ e^{i\left(z/\sqrt\lambda\right)S_n}\right] =\varphi_{S_N}\left(z/\sqrt\lambda\right) $$ and for any $\theta\in\mathbb C$, \begin{align} \varphi_{S_N}(\theta) &= \mathbb E\left[ e^{i\theta S_N}\right]\\ &= \mathbb E\left[ \mathbb E\left[ e^{i\theta S_N} \mid N\right]\right]\\ &= \sum_{n=0}^\infty \mathbb E\left[ e^{i\theta S_N} \mid N=n\right]\mathbb P(N=n)\\ &= \sum_{n=0}^\infty \mathbb E\left[e^{i\theta S_n} \right]\mathbb P(N=n)\\ &= \sum_{n=0}^\infty \prod_{k=1}^n \mathbb E\left[e^{i\theta X_k} \right]\mathbb P(N=n)\\ &= \sum_{n=0}^\infty \mathbb E\left[e^{i\theta X_1} \right]^n\mathbb P(N=n)\\ &= g_N\left(\mathbb E\left[e^{i\theta X_1}\right]\right)\\ &= g_N\left(\varphi_X(\theta)\right). \end{align}

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  • $\begingroup$ Two points to modify. First, $\varphi_N(z)=\exp(\lambda(e^{iz}-1))$, not $\varphi_N(z)=\exp(\lambda(iz-1))$. Second, if the $\varphi$s are really characteristic functions, then $\varphi_{S_N/\sqrt\lambda}(z) \ne \varphi_N(\varphi_X(z/\sqrt\lambda))$, rather $\varphi_{S_N/\sqrt\lambda}(z) \ne g_N(\varphi_X(z/\sqrt\lambda))$ where $g_N(s)=E(s^N)$ is the generating function of $N$, not its characteristic function. Correcting all this should bring you back to the formulas in my comments on main (and explain the "mystery" of your factor $i$...). $\endgroup$
    – Did
    Oct 6, 2016 at 9:57
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    $\begingroup$ Are you able to show the crucial step $$\varphi_{S_N/\sqrt\lambda}(z) = g_N(\varphi_X(z/\sqrt\lambda))\ ?$$ $\endgroup$
    – Did
    Oct 7, 2016 at 19:57

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