# Proof of multivariate central limit theorem

$$\newcommand{\phi}{\varphi}$$ $$\newcommand{\eps}{\epsilon}$$

I'm using the book called 'A Course in Large Sample Theory' from Thomas S. Ferguson. During studying the proof of the central limit theory in the book, I don't understand something so I ask a question here.

The theorem states the following: Let $$X_1, X_2, \dots$$ be i.i.d. random vectors with mean $$\mu$$ and finite covariance matrix, $$\Sigma$$. Then $$\sqrt{n}(\bar{X}_n - \mu)\overset{d}{\to}N(0,\Sigma)$$ where $$\overset{d}{\to}$$ denotes the convergence in distribution.

The proof of this theorem is following: Since $$\sqrt{n}(\bar{X}_n - \mu) = (1/\sqrt{n})\sum_{1}^{n}(X_j - \mu)$$, we have \begin{align*} \phi_{\sqrt{n}(\bar{X}_n - \mu)}(t) &= \phi_{\sum_{1}^{n}(X_j - \mu)}(t/\sqrt{n}) \\ &=\phi(t/\sqrt{n})^n \end{align*} where $$\phi(t)$$ is the characteristic function of $$X_j - \mu$$. Then, since $$\phi(0) = 1, \dot{\phi}(0) = 0$$, and $$\ddot{\phi}(\eps)\to -\Sigma$$ as $$\eps\to 0$$, we have, applying Taylor's theorem, \begin{align} \phi_{\sqrt{n}(\bar{X}_n - \mu)}(t) &= \left(1 + \frac{1}{n}t'\int_{0}^1\int_0^1 v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right)^{n} \\ &\to \exp\left(\lim_{n\to\infty}t'\int_0^1\int_0^1v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right) \\ &= \exp(-(1/2)t'\Sigma t). \end{align} My first questions are following:

1. How do we know $$\ddot{\phi}(\eps)\to -\Sigma$$? I think it should be trivial since the book simply said we know it.
2. How does the first equality (after applying Taylor's theorem) hold?
3. How does the second ($$\dots = \exp(-(1/2)t'\Sigma t)$$) equality hold?

If there is already an answer for this question, I'm sorry for reposting it and please share the link in the commend, I'll close my post then. Any help regarding this question would be so helpful.

• Should there be an exponent of $n$ on $$1 + \frac{1}{n}t'\int_{0}^1\int_0^1 v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t$$?
– jld
Oct 20, 2020 at 21:39
• @jld Yes, that is correct. Sorry about confusion Oct 20, 2020 at 21:55

$$\newcommand{\E}{\operatorname{E}}\newcommand{\vp}{\varphi}\newcommand{\0}{\mathbf 0}$$This answers your first two questions. I'll update as I have time unless someone else answers the 3rd question before I do.
There aren't any regularity conditions mentioned here so I'll freely interchange limits with expected values (so e.g. I can exchange a derivative and expectation). I'll also take $$\E X = \0$$ without loss of generality. I'm also assuming $$\vp$$ refers to a characteristic function.
For your question 1, we have $$\vp(t) = \E(e^{it^TX})$$ so $$\dot{\vp}(t) = \E(iX \cdot e^{it^TX})$$ and $$\ddot{\vp}(t) = \E(-XX^T \cdot e^{t^TX}).$$ Then $$\lim_{t\to \0} \ddot{\vp}(t) = -\E\left(XX^T \cdot\lim_{t\to\0} e^{it^TX}\right) \\ = -\E(XX^T) = -\Sigma$$ since the mean is zero.
Then for question 2, we have the integral $$\int_0^1\int_0^1 v\ddot{\vp}(uv \cdot t/\sqrt n)\,\text du\,\text dv \\ = \E_X\left[-XX^T \int_0^1\int_0^1v e^{iuvt^TX/\sqrt n}\,\text du\,\text dv\right].$$ I'll first integrate w.r.t. $$u$$ to get $$\int_0^1 e^{iuvt^TX/\sqrt n}\,\text du = \frac{e^{ivt^TX/\sqrt n} - 1}{i vt^TX/\sqrt n}$$ so then canceling the $$v$$s I have $$\frac{\sqrt n}{i t^TX} \left( \int_0^1 e^{ivt^TX/\sqrt n}\,\text dv - 1\right) \\ = \frac{\sqrt n}{i t^TX}\left(\frac{e^{it^TX/\sqrt n}}{it^TX/\sqrt n} - \frac 1{it^TX/\sqrt n} - 1\right) \\ = -\frac{n}{(t^TX)^2}\left(e^{it^TX/\sqrt n} - 1 - i t^TX / \sqrt n\right).$$ This means $$1 + \frac{1}{n}t^T\left(\int_{0}^1\int_0^1 v\ddot{\vp}(uv\cdot t/\sqrt{n})\,\text d u\,\text dv\right)\cdot t \\ = 1 + \frac{1}{n}t^T\E_X\left[-XX^T \int_0^1\int_0^1v e^{iuvt^TX/\sqrt n}\,\text du\,\text dv\right]\cdot t \\ =1 + t^T\E_X\left[XX^T \frac{1}{(t^TX)^2}\left(e^{it^TX/\sqrt n} - 1 - i t^TX / \sqrt n\right)\right]\cdot t \\ = 1 + \E_X(e^{it^TX/\sqrt n}) - 1 - i t^T\E_X(X) / \sqrt n \\ = \E_X(e^{i t^TX/\sqrt n}).$$ Applying the exponent of $$n$$ gives the desired result there.