| |
\[
\begin{aligned}
M_n &= {X_1 + \cdots + X_n \over n} & \text{where } X\text{'s are iid's} \\
\end{aligned}
\] |
We have the familiar CLT as \(n \rightarrow \infty\):
| |
\[
\begin{aligned}
\bf{P}\left( \frac{M_n - \mu}{\sigma \sqrt{n}} \le \frac{\epsilon}{\sigma \sqrt{n}} \right) &= \bf{P}\left( z \le \frac{\epsilon}{\sigma \sqrt{n}} \right)
\approx \mathbb{\phi} \left(\frac{\epsilon}{\sigma \sqrt{n}} \right) \\
\end{aligned}
\] |
But how about the absolute value of z ?
| |
\[
\begin{aligned}
\bf{P}\left( \big|\frac{M_n - \mu}{\sigma \sqrt{n}}\big| \le \frac{\epsilon}{\sigma \sqrt{n}} \right) &= \bf{P}\left( |z| \le \frac{\epsilon}{\sigma \sqrt{n}} \right)
\approx 1 - 2\bigg[1 - \mathbb{\phi} \left(\frac{\epsilon}{\sigma \sqrt{n}} \right)\bigg] = 2 \cdot \mathbb{\phi}\left(\frac{\epsilon}{\sigma \sqrt{n}} \right) - 1 \\
\end{aligned}
\] |
In case we know the range \([a, b]\) of \(X\) but not the variance, we can make use of the upper bounds. That is,
| |
\[
\begin{aligned}
\displaystyle \sigma^2 \le \frac{(b-a)^2}{4} \\
\end{aligned}
\] |
which can be verified via the use of Chebyshev's Inequality. A proof can be found at p.268 of
Introduction to Probability, 2nd Edition.
Have you ever wondered why
| |
\[
\begin{aligned}
\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dx = 1
\end{aligned}
\] |
? That is, why
| |
\[
\begin{aligned}
\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\,dx = \sqrt{2\pi}
\end{aligned}
\] |
? Let
| |
\[
\begin{aligned}
\color{blue}{\mathbf{I}} &\color{blue}{= \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\,dx} \\
\mathbf{I}\cdot\mathbf{I} &= \left(\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\,dx \right)\left(\int_{-\infty}^{\infty} e^{-\frac{y^2}{2}}\,dy\right) \\
\mathbf{I}^2 &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x^2+y^2)}\,dx \,dy \\
\end{aligned}
\] |
The double integral is easier to resolve in polar coordinate. In general,
| |
\[
\begin{align*}
\iint_D g(x,y)\,dA = \iint_{D^*} g(\mathbf{T}(r,\theta))\big| \det D \mathbf{T}(r,\theta)\,\big|\, dr\, d\theta \\
\end{align*}
\] |
In this particular case,
| |
\[
\begin{align*}
g(x,y) &= e^{-\frac{1}{2}(x^2+y^2)} \\
\mathbf{T}(r,\theta) &= (r\cos{\theta}, r\sin{\theta}) = (x, y) \\
g(\mathbf{T}(r,\theta)) &= e^{-\frac{1}{2}((r\cos{\theta})^2+(r\sin{\theta})^2)} = e^{-\frac{1}{2}r^2} \\
D \mathbf{T}(r,\theta) &= \begin{bmatrix}\frac{d}{dr}r\cos{\theta} & \frac{d}{d\theta}r\cos{\theta}\\\frac{d}{dr}r\sin{\theta} & \frac{d}{d\theta}r\sin{\theta}\end{bmatrix}
= \begin{bmatrix}\cos{\theta} & -r\sin{\theta}\\\sin{\theta} & r\cos{\theta}\end{bmatrix} \\
\big| \det D \mathbf{T}(r,\theta)\,\big| &= \begin{vmatrix}\cos{\theta} & -r\sin{\theta}\\\sin{\theta} & r\cos{\theta}\end{vmatrix} = r \\
\end{align*}
\] |
Therefore
| |
\[
\begin{aligned}
\mathbf{I}^2 &= \int_0^{2\pi} \int_0^{\infty} e^{-\frac{1}{2}r^2} r\,dr\,d\theta \\
\end{aligned}
\] |
Let \(u = \frac{1}{2}r^2\), so \(du = r\,dr\)
| |
\[
\begin{aligned}
\mathbf{I}^2 &= \int_0^{2\pi} \int_0^{\infty} e^{-u} \,du\,d\theta = \int_0^{2\pi} -e^{-u} \bigg|_{u=0}^{\infty} \,d\theta = \int_0^{2\pi}d\theta = 2\pi \\
\therefore \color{blue}{\mathbf{I}} &\color{blue}{= \sqrt{2\pi}} \\
\end{aligned}
\] |
\(\Box\)
References:
