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

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 $\phi_{X_j - \mu}$. 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.

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

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.

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

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 $\phi_{X_j - \mu}$. 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, $$ \begin{align} \phi_{\sqrt{n}(\bar{X}_n - \mu)}(t) &= \left(1 + \frac{1}{n}t'\int_{0}^1\int_0^1 v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right) \\ &\to \exp\left(\lim_{n\to\infty}t'\int_0^1\int_0^1v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right) \\ &= \exp(-(1/2)t'\Sigma t). \end{align}$$ 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.

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

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 $\phi_{X_j - \mu}$. 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.

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

I'm using the book called 'A Course in Large Sample Theory' from Champman and Hall. 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 $\phi_{X_j - \mu}$. 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, $$ \begin{align} \phi_{\sqrt{n}(\bar{X}_n - \mu)}(t) &= \left(1 + \frac{1}{n}t'\int_{0}^1\int_0^1 v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right) \\ &\to \exp\left(\lim_{n\to\infty}t'\int_0^1\int_0^1v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right) \\ &= \exp(-(1/2)t'\Sigma t). \end{align}$$ 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.

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

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 $\phi_{X_j - \mu}$. 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, $$ \begin{align} \phi_{\sqrt{n}(\bar{X}_n - \mu)}(t) &= \left(1 + \frac{1}{n}t'\int_{0}^1\int_0^1 v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right) \\ &\to \exp\left(\lim_{n\to\infty}t'\int_0^1\int_0^1v\ddot{\phi}(uv\cdot t/\sqrt{n})dudv\cdot t\right) \\ &= \exp(-(1/2)t'\Sigma t). \end{align}$$ 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.