简易版本Random Walk证明

作业，存档一下过程

1、Task 问题

​ Imagine a drunken man who, starting out leaning against a lamp post in the middle of an open space, takes a series of steps of the same length: 1 meter . The direction of these steps is randomly chosen from North, South, East or West. After n steps, how far (*d*), generally speaking, is the man from the lamp post? Note that d is the Euclidean distance of the man from the lamp-post. Deduce the relationship.
想象一个醉汉从一个路灯出发，每一步都迈出去一米的距离，每次随机选择东南西北四个方向之一。N步以后，他离路灯多远？推断此时距离d和步数n的关系。

2、Relationship Conclusion 结论

\[d = \sqrt{n} \]

​ Just a approximation of the result.
只是一个约等于的推测。

3、Evidence (Mathematics deduction) 证明（数学推导）

To get to the conclusion of
为了得到以下关系

\[d = \sqrt{n} \]

We consider the drunken man walking in a coordinate system and the lamp spot as the origin,

then we will get his position as (x,y)

and the distance will be
我们想象这个醉汉在一个坐标系中行走，路灯作为原点。他的位置为（x,y），此时他离路灯的距离d：

\[d=\sqrt{x^2+y^2} \]

And we assume him walking on
假设他在：

​ West-East direction (x axis) for i steps
东西方向走了i步

​ North-South direction (y axis) for k steps
南北方向走了k步

\[n = i+k \]

Then we will have
此时可以得出

\[\begin{cases} X = X1+X2+X3...+Xi \\Y= Y1+Y2+Y3...+Yk\end{cases} \]

If we see Xa/Ya represent the steps as -1/1 for the opposite direction.
当我们将Xa、Ya = -1/1 以代表ta每一步走的具体方向时，可以得到以上表达式

\[X^2=(X_1+X_2+X_3...+X_i)^2\\ \ \ \ \ \ =X_1^2+X_1X_2+X_1X_3+...+X_1X_i\\ \ \ \ \ \ \ \ \ \ \ +X_2^2+X_1X_2+X_2X_3+...+X_2X_i\\ \ \ \ \ \ \ \ \ \ \ +X_3^2+X_1X_3+X_2X_3+...+X_3X_i\\ ...\\\ \ \ \ \ \ \ \ \ \ +Xi^2+X_1X_i+X_2X_i+...+X_{i-1}X_i\\ =(X1^2+X2^2+X3^2...+Xi^2)+2(X_1X_2+X_1X_3+X_1X_4...+X_{i-1}X_i) \]

\[\because Xa = -1 \ or\ 1, \\Xa^2 = 1\\ \therefore (X1^2+X2^2+X3^2...+Xi^2) = 1*i =i \]

​ Each XaXa pair will be within the following types:
每一对XaXa属于以下四种之一

\[\begin{cases} 1,\ \ \ -1\ =-1 \\1,\ \ \ \ \ \ 1\ \ \ =1\\-1,\ \ \ 1\ \ \ =-1\\-1,\ -1\ =1\end{cases}\\ \]

​ and the probability of these pairs will be the same because it's Random
且因为他走的方向随机，所以每种情况的可能性相等。

​ On average will be 0,
平均来看 XaXa = 0

\[\therefore 2(X_1X_2+X_1X_3+X_1X_4...+X_{i-1}X_i) =0 \]

​ Therefore,
因此

\[X^2=(X1^2+X2^2+X3^2...+Xi^2)+2(X_1X_2+X_1X_3+X_1X_4...+X_{i-1}X_i)=i+0 =i \]

​ the same procedure may be easily adapted to Y²
同理可得Y^2

\[Y^2=(Y1^2+Y2^2+Y3^2...+Yk^2)+2(Y_1Y_2+Y_1Y_3+Y_1Y_4...+Y_{k-1}Y_k)=k+0 =k \]

So, we can approximately deduce that
综上所述，我们可以得到

\[d = \sqrt{X^2 +Y^2} =\sqrt{i+k}=\sqrt{n} \]

QED
证毕