简易版本Random Walk证明

发布时间:2022-07-05 发布网站:脚本宝典
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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...+Ykend{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 =1end{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

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