Averaging a Random Walk with Numbers Spanning [-1,1]

조회 수: 1 (최근 30일)
Michael
Michael 2012년 11월 8일
댓글: Michael 2013년 12월 11일
Hello,
I need a random number generator whose values span [-1,1] (was using 2*rand-1) and I'm going to count the number of sign changes, then run that a whooole bunch of times and then average the sign changes.
I feel like I'm getting good results when I don't incorporate the '-1' part of the 2*rand-1... can anyone help me clear this issue up?
% Begin Loop for averaging sign changes
for j=1:10 % Number of iterations: better statistics the larger this is
clear count T starty e i
% Preallocate and such
steps=10; % Number of steps to be taken! should be large.. was testing
% small values
y_init=0;
y=y_init*ones(1,steps);
e=2*randn(1,steps)-1; % Numbers span [-1,1]
count=0;
%
% Begin loop to (repetitively) count sign changes in random walk
for i=1:(steps-1);
y(i+1)=y(i)+e(i+1)
end
%
% From MATLAB 'help rand'
% Example 1: Generate values from the uniform distribution on the
% interval [a, b].
% r = a + (b-a).*rand(100,1);
% newstep= -1 + 2.*rand(100,1)
%
zeros(j)=length(find(abs(diff(sign(y)))==2));
if j==1
time=[1:1:steps];
plotrandom=plot(time,y);
end
end
AVGnoBounces=mean(zeros)
Thanks for any help!!! -Mike
  댓글 수: 2
Walter Roberson
Walter Roberson 2012년 11월 8일
Naming a variable "zeros" is going to interfere with use of the zeros() function, and is going to confuse readers.
Michael
Michael 2013년 4월 3일
Sorry...

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채택된 답변

Matt Fig
Matt Fig 2012년 11월 9일
편집: Matt Fig 2012년 11월 9일
Wouldn't this be easier?
T = 2*rand(1,10)-1;
Num_change = sum(logical(diff(sign(T))))
  댓글 수: 1
Michael
Michael 2013년 12월 11일
So Sorry I haven't seen your result or thanked you!

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