Why does the one-dimensional variant of a 2-d random walk not work?

Question

There is a two-dimensional random walk that one can find here which works perfectly in Octave. However, when I tried to write a one-dimensional random walk program, I got an error. Here is the program:

t=[];
x=[];
for i=1:100000
    J=rand;
    if J<0.5
        x(i+1)=x(i)+1;
        t(i+1)=t(i)+1;
    else
        x(i+1)=x(i)-1;
        t(i+1)=t(i)+1;
    end
end

plot(t,x)

Here is the error:

error: A(I): index out of bounds; value 1 out of bound 0

Thank you.

Answer 1

No need for a loop:

N = 100000;
t = 1:N;
x = cumsum(2*(rand(1,N)<.5)-1);
plot(t,x)

For the 2D case you could use the same approach:

N = 100000;
%// t = 1:N; it won't be used in the plot, so not needed
x = cumsum(2*(rand(1,N)<.5)-1);
y = cumsum(2*(rand(1,N)<.5)-1);
plot(x,y)
axis square