# Expected time to hit a line

probability

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.

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.