An ant is standing at the origin $(0,0).$ It starts moving in the $xy$-plane such that at each step, it moves northwards (in $+y$ direction) or eastwards (in $+x$ direction) with equal probabilities. The length of the first step is $1$ unit and the length of each subsequent step is half of its previous step.
after taking the first step, it can be at $(1,0)$ or $(0, 1)$ with equal probabilities. The length of the second step would be $0.5$ and third $0.25$.
$(x_i,y_i)$ is the position of the ant after $i$ steps have been taken. Thus, $$(x_0,y_0) = (0, 0)$$
Also,
$$(x, y) = \lim_{i \rightarrow ∞} (x_{i}, y_{i})$$
What will be the locus of $(x,y)$?
I started by calculating certain coordinates which will be $(0,1) (1,0)$
$$(0.5,1) (0,1.5) (1.5,0) (1,0.5)$$
$$(0.75,1) (0.5,1.25) (0.25,1.5) (0,1.75) (1.75,0) (1.5,0.25) (1.25,0.5) (1,0.75)$$
The probable pattern is that $x_n$ can be summation of $\frac{1}{2}$ until infinity or less than that and same goes for $y$. I plotted the points to get a start but didn't help much. What should be a better approach?
When he introduced the question, I remembered that this question already had an equivalent in mathematics. It's called the "Wiener process".This is a random walk.Let's take the one-dimensional version of this. A version where we can only go to -x and +x. Let's call the distance traveled "d" and look at the absolute distance that can be traveled. $d^2$ is the more plausible measure of randomization to deal with positive values.Let's express the expected value as $\langle{d_n^2}\rangle$ after n steps. One step later $d^2$ is always one so $\langle{d_1^2}\rangle = 1$. For $n > 1$, the expected value of $d_n^2$ can be obtained from $d_{n-1}$ Starting from the previous equation
$d_n = d_{n-1}\pm1$ and $d_n^2 = d_{n-1}^2\pm2d_{n-1}+1$.
With a large number of independent sequences, we expect to get one value half the time and the other value half the time, so the average expectation is the average of these two possible values. In that case $\langle{d_n^2}\rangle = \langle d_{n-1}^2\rangle+1.$
$\langle{d_1^2}\rangle = 1$, so $\langle{d_n^2}\rangle = n$.
Returning to your question, let's add y to this random walk. Suppose that in addition to each random choice of direction, the length of each stride also changes undetectable. If we call the length of a step s, s can take any value, but will most often be "closer" to 1, so $\langle s^2\rangle = 1$.
We derive x as before, only the equation becomes,
$\langle{d_n^2}\rangle = \langle{d_1^2}\rangle + \langle s^2\rangle = \langle{d_1^2}\rangle + 1$
As before we have $\langle{d_n^2}\rangle = n$
In short, the answer to the question is $\sqrt{\langle d^2\rangle} = \sqrt n$.