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I read in a book that since $\frac{\sqrt{n}(\bar{X}-\theta)}{\theta}$ converges in distribution to the standard normal distribution, i.e., $N(0,1)$ using the Central Limit Theorem, that $\sqrt{n}(\bar{X}-\theta)$ converges in distribution to $N(0,\theta)$. Can someone please explain why this is true?

Note: it is also given that $E(X_i) = Var(X_i) = \theta$ and the question is about finding the limiting distribution of the mean of the Poisson distribution.

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  • $\begingroup$ Except if you are using a very nonstandard notation for a normal distribution, it should converge to $N(0,\theta^2)$, that is, its variance would be $\theta^2$ and its standard deviation $\theta$. $\endgroup$
    – Raoul
    Commented Apr 28, 2020 at 21:09

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Let $Y_n=\frac{\sqrt{n}(\bar{X}-\theta)}{\theta}$ and put $W_n=\sqrt{n}(\bar{X}-\theta)$. Then for $w\in\mathbb{R}$ $$ P(W_n\leq w)=P(\theta Y_n\leq w)=P(Y_{n}\leq w/\theta)\to P(Z\leq w/\theta)=P(\theta Z\leq w) $$ since $Y_{n}$ converges in distribution to a standard normal where $Z$ is a standard normal random variable. By the definition of convergence in distribution in terms of distribution functions it follows that $$ W_{n}\stackrel{d}{\to}\theta Z\sim N(0, \theta) $$ where $\theta$ is the standard deviation.

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  • $\begingroup$ Doesn't your answer show that $\sqrt{n}(\bar{X}-\theta)$ converges in distribution to $N(0, \frac{w}{\theta})$? $\endgroup$ Commented Apr 28, 2020 at 19:01

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