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Suppose that $Y_{\lambda}=^{d}P(\lambda)$. Prove that $[Y_{\lambda}-\lambda]/{\sqrt{\lambda}}\to^{d}N(0,1)$ when $\lambda \to \infty$ using characteristic functions. So $$\phi(t)=\sum_{k=0}^{\infty} \frac{e^{it\lambda}}{\sqrt{\lambda}}\left(e^{-\lambda}\frac{\lambda^k}{k!}-\lambda\right)$$ where $\phi(t)$ stands for characteristic function. So $$\phi(t)=\frac{1}{\sqrt{\lambda}}\sum_{k=0}^{\infty} {e^{it\lambda}}\left(e^{-\lambda}\frac{\lambda^k}{k!}-\lambda\right)=\frac{1}{\sqrt{\lambda}}[\sum_{k=0}^{\infty} {e^{it\lambda}}e^{-\lambda}\frac{\lambda^k}{k!}-\sum_{k=0}^{\infty}e^{it\lambda}\lambda]$$ Then $$\frac{1}{\sqrt{\lambda}}[\sum_{k=0}^{\infty} {e^{it\lambda}}e^{-\lambda}\frac{\lambda^k}{k!}-\sum_{k=0}^{\infty}e^{it\lambda}\lambda]=\frac{1}{\sqrt{\lambda}}\left[ {e^{\lambda(e^{it-1})}}-\lambda\sum_{k=0}^{\infty}e^{it\lambda}\right].$$ Stuck at this point.

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    $\begingroup$ Characteristic function of $P(\lambda)$ is $\exp(\lambda(e^{it}-1))$. Subtracting $\lambda$ and rescaling we get $\exp(\lambda(e^{\frac{it}{\sqrt\lambda}}-1))\exp^{-\frac {it \lambda }{\sqrt {\lambda} }}$. Now Taylor expand in the exponent. $\endgroup$ – A.S. Dec 7 '15 at 22:07
  • $\begingroup$ Up to second term I get: $it\sqrt{\lambda}+\lambda-\lambda(1+it/\sqrt{\lambda}-t^2/(2\lambda))+....$ So how do i assume the rest is zero? $\endgroup$ – Don Dec 7 '15 at 23:14
  • $\begingroup$ The rest goes to zero. $\endgroup$ – A.S. Dec 8 '15 at 9:06

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