some small changes
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@ -122,7 +122,7 @@ More formally:
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For every fixed $n$, $Y_n$ and $Z_n$ are independent.
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\end{claim}
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\begin{subproof}
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This is obvious, but well prove it carefully here.
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This is obvious, but we will prove it carefully here.
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\begin{IEEEeqnarray*}{rCl}
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&&(\bP \otimes \bP) [Y_n \in (a,b) , Z_n \in (a',b') ]\\
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&=& (\bP\otimes\bP) \left( (\omega, \omega') : X_n(\omega) \in (a,b) \land X_n(\omega') \in (a',b') \right)\\
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@ -50,32 +50,38 @@ So far we have dealt with the average behaviour,
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\[
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\frac{\overbrace{X_1 + \ldots + X_n}^{\text{i.i.d.}}}{n} \to \bE(X_1).
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\]
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We now want to understand \vocab{fluctuations} from the average behaviour,
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We now want to understand fluctuations from the average behaviour,
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i.e.\[
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X_1 + \ldots + X_n - n \cdot \bE(X_1).
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\]
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% TODO improve
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The question is, what happens on other timescales than $n$?
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An example is
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\[
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\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} } \xrightarrow{n \to \infty} hv \cN(0, \Var(X_i)) (\ast)
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\]
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Why is $\sqrt{n}$ the right order? (Handwavey argument)
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\begin{equation}
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\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }
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\xrightarrow{n \to \infty} \cN(0, \Var(X_i))
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\label{eqn:lec09ast}
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\end{equation}
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Why is $\sqrt{n}$ the right order?
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Handwavey argument:
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Suppose $X_1, X_2,\ldots$ are i.i.d. $\cN(0,1)$.
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Suppose $X_1, X_2,\ldots$ are i.i.d.~with $X_1 \sim \cN(0,1)$.
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The mean of the l.h.s.~is $0$ and for the variance we get
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\begin{IEEEeqnarray*}{rCl}
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\Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) &=& \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right)\\
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&=& \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
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\Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} })
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&=& \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right)\\
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&=& \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
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\end{IEEEeqnarray*}
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For the r.h.s.~we get a mean of $0$ and a variance of $1$.
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So, to determine what $(\ast)$ could mean, it is necessary that $\sqrt{n}$
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So, to determine what \eqref{eqn:lec09ast} could mean, it is necessary that $\sqrt{n}$
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is the right scaling.
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To define $(\ast)$ we need another notion of convergence.
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To make \eqref{eqn:lec09ast} precise,
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we need another notion of convergence.
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This will be the weakest notion of convergence, hence it is called
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\vocab{weak convergence}.
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This notion of convergence will be defined in terms of characteristic functions of Fourier transforms.
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This notion of convergence will be defined in terms of
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characteristic functions of Fourier transforms.
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\subsection{Characteristic Functions and Fourier Transform}
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