这是一份liverpool利物浦大学PHYS392的成功案例

$$
I\left(X_{1}: X_{2}\right)=\int_{X} d x_{1} d x_{2} p\left(x_{1}, x_{2}\right) \ln \left(\frac{p\left(x_{1}, x_{2}\right)}{p\left(x_{1}\right) p\left(x_{2}\right)}\right)
$$
and $I\left(X_{1}: X_{2}\right) \geq 0$ with equality iff $p\left(x_{1}, x_{2}\right)=p\left(x_{1}\right) p\left(x_{2}\right)$. Let’s look as an example to the bi-variate normal distribution: $N(\boldsymbol{x} \mid \boldsymbol{\mu}, \mathbf{\Sigma})$ :with covariance matrix
$$
\boldsymbol{\Sigma}=\left(\begin{array}{cc}
\sigma_{1}^{2} & \rho \sigma_{1} \sigma_{2} \
\rho \sigma_{1} \sigma_{2} & \sigma_{2}^{2}
\end{array}\right) \quad \text { and } \operatorname{det}[\boldsymbol{\Sigma}]=\sigma_{1}^{2} \sigma_{2}^{2}(1-\rho)^{2}
$$

PHYS392 COURSE NOTES ：

$$
\Phi_{X_{n}}(t)=E\left[e^{i t X}\right]=\frac{1}{2}\left(1+e^{2 i t}\right)
$$
and for $Y_{n}=X_{n} / 3^{n}$ :
$$
\Phi_{Y_{n}}(t)=\Phi_{X_{n}}\left(t / 3^{n}\right)=\frac{1}{2}\left(1+e^{2 i t / 3^{n}}\right)
$$
Being all $X_{n}$ statistically independent, we have that
$$
\Phi_{X}(t)=\prod_{n=1}^{\infty} \frac{1}{2}\left(1+e^{2 i t / 3^{n}}\right)=\prod_{n=1}^{\infty} \frac{1}{2} e^{i t / 3^{n}} \cos \left(t / 3^{n}\right)=e^{i t / 2} \prod_{n=1}^{\infty} \cos \left(t / 3^{n}\right)
$$