My Latex skills are horrible, and I'm unable to type my problem out in its full glory on this interface, but I believe I'll be able to sufficiently communicate my concerns to you all.
A problem in a textbook that I'm reading asked the reader to prove the central limit theorem for sample means using characteristic functions.
The characteristic function I got for the sample mean is $(E(e^{itX/n}))^n$ where $X$ is a random variable with mean $\mu$ and variance $\sigma^2$.
Once you expand $e^{itX/n}$ using Taylor series, apply expectation, raise the result to the $n^{th}$ power, and then take limits, the limit I get is $e^{it\mu}$ which is clearly incorrect.
Can someone explain to my why this doesn't work?
Thank you!
Best Answer
Let $\mu_{n} = \frac{S_{n} - E[S_{n}]}{\sqrt{D[S_{n}]}}$. Now, to prove the central limit theorem, we have to show that the characteristic function of $\mu_{n} \rightarrow e^{-t^2 / 2}$. Let $a = E[\xi_{i}], \; \sigma^{2} = D[\xi_{k}]$. Now, let's transform $\mu_{n}$: $$ \mu_{n} = \frac{\sum_{i = 1}^{n} \xi_{k} - E[\xi_{k}]}{\sqrt{\sum_{i = 1}^{k} D[\xi_{k}]}} = \sum_{i = 1}^{n} \frac{\xi_{k} - a}{\sqrt{n} \cdot \sigma} = \sum_{i = 1}^{n} \frac{\eta_{k}}{\sqrt{n}} $$ Here $\eta_{k} = \frac{\xi_{k} - a}{\sigma}$. Now, note that $E[\eta_{k}] = 0, D[\eta_{k}] = 1 \rightarrow E[\eta_{k}^2] = 1$ So, the characteristic function of $\eta_{k}$ is $\phi_{\eta_{k}}(t) = 1 - \frac{t^2}{2} + o(t^{2})$ So, $$\phi_{\mu_{n}}(t) = E \left[\exp \left(i \cdot t \cdot \sum_{i = 1}^{n} \frac{\eta_{k}}{\sqrt{n}}\right) \right ] = (because \; \eta_{k} \; are \; independent) = \prod\limits_{k = 1}^{n} \phi_{\eta_{k}} \left(\frac{t}{\sqrt{n}} \right) = \left(1 - \frac{t^2}{2n} + o(t^2 / n) \right)^{n} \rightarrow e^{-\frac{t^2}{2}}$$ So $\mu_{n} \xrightarrow[]{d} N (0, 1)$