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简易版本Random Walk证明

2021-09-21 02:01:57  阅读:214  来源: 互联网

标签:Xi 1X Random sqrt Walk will 简易 steps ...+


作业,存档一下过程

1、Task

​ 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.

2、Relationship Conclusion

\[d = \sqrt{n} \]

​ Just a approximation of the result.

3、Evidence (Mathematics deduction)

To get to the conclusion of

\[d = \sqrt{n} \]

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

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

And we assume him walking on

​ West-East direction (x axis) for i steps

​ North-South direction (y axis) for k steps

\[n = i+k \]

Then we will have

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

If we see Xa/Ya represent the steps as -1/1 for opposite direction.

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

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

​ Each XaXa pair will be within the following types:

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

​ and the probability of these pairs will be the same because it's Random

​ On average will be 0,

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

​ Therefore,

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

​ the same procedure may be easily adapted to Y²

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

So, we can approximately deduce that

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

QED

标签:Xi,1X,Random,sqrt,Walk,will,简易,steps,...+
来源: https://www.cnblogs.com/Lamplight/p/15313522.html

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