# Convergence in distribution of $\frac{\sqrt{n}(\bar{X}-\theta)}{\theta}$

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.

• 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 $\theta$. Commented Apr 28, 2020 at 21:09

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