Consider a particle in $\mathbb{R}^2$ that starts at $(0,0)$. This particle moves at a speed of 1 unit per second, and every second this particle moves randomly 1 unit directly north, south, east or west, with an equal probability of $\frac{1}{4}$.

What is the average time the particle will hit the line that passes through the points $(1,0)$ and $(0,1)$, i.e, $y = 1 -x$?

I have noticed that after one step there are two scenarios: a) the particle goes north (east) and hits the particle at the point $(1,0)$ ($(0,1)$) – this happening with probability $\frac{1}{2}$; b) the particle goes south (west) and its now at the point $(0,-1)$ ($(-1,0)$). From this scenario, it can occur that: b.1) the particle hits the line two steps after. For example, $(0,0)$ $\to$ $(-1,0)$ $\to$ $(-1,1)$ $\to$ $(0,1)$ – this occurring with probability $\frac{1}{2}$ (if I'm not mistaken); b.2) the particle goes back to the starting point – $(0,0)$ – and the "chain" starts over.

Having these scenarios, I then proceeded to calculate the expected value of the variable $N$, which I defined as the number of steps until the particle hits the $y = 1 – x$ line.

$$ E[N] = 2 \times \frac{1}{4} + \frac{1}{2}\Bigg[2 \times \frac{1}{4} + \frac{1}{4} (1 + E[N]) \Bigg] $$

Solving the equation for $E[N]$, I managed to obtain that $E[N] = 1$

Is this correct or am I missing something? If so, I would appreciate if you could give some tips.

## Best Answer

If you look at the distance to the line $L$, i.e. the process $Y_k = \text{dist}(X_k, L)$, then you should see that $Y_{k+1}=Y_k+1$ with probability $1/2$ and $Y_{k+1}=Y_k-1$ with probability $1/2$. Consequently, we are back to the usual one-dimensional problem consisting in computing how long it takes for a symmetric random walk started at zero, i.e. $S_k = 0$, to reach $S_k=1$. The expected time it takes is infinite.