# Expected number of steps for reaching $K$ in a random walk

Assuming steps are $+1/-1$ with a $50/50$ probability. What is the expected step count for reaching $10, 100$ or $K$?

• Answer: +Infinity. – Did Feb 17 '13 at 20:26
• But even though it takes infinite time on average, it still happens sooner or later with probability 1. – hmakholm left over Monica Feb 17 '13 at 20:50

Let's study the expected time it takes for the symmetric random walk to reach $+1$, assuming it begins at $0$. Let $S$ denote the number of steps we take until we reach $+1$ that is $S_1 = \min\{n > 0; X_n = +1\}$ where $X_n$ is your symmetric random walk. Note that $S_1$ is an odd number.

$$P(S_1 = 1) = \frac{1}{2}\\ P(S_1 = 3) = \frac{1}{2^3} \\ P(S_1 = 5) = 2 \frac{1}{2^5}\\ P(S_1 = 2n+1) = C_n \frac{1}{2^{2n+1}}$$

where $C_n$ is the number of non positive paths of length $2n$ begin at $0$ and end at $0$, this is given by $C_n = \frac{1}{n+1} {2n\choose n}$ see https://en.wikipedia.org/wiki/Catalan_number

Now we calculate $$E[S_1] = \sum_{k=0}^\infty (2k + 1) C_k \frac{1}{2^{2k+1}}$$

Now assymptotically, $$C_n \sim \frac{4^n}{n^{3/2}\sqrt{\pi}}$$

therefore $$C_k \frac{1}{2^{2k+1}} \sim \frac{1}{2 k^{3/2}\sqrt{\pi}}$$

and therefore

$$(2k + 1) C_k \frac{1}{2^{2k+1}}\sim \frac{2k+1}{2 k^{3/2}\sqrt{\pi}} > \frac{1}{k}$$

therefore the sum diverges and $E[S_1] = \infty$. Now for $S_{10} = \min\{n > 0, X_n = 10\}$ we have $S_{10} \geq S_1$ therefore $$E[S_{10}]\geq E[S_1] = \infty$$

• Thank you for this. I assumed the expected amount of steps would be a bit smaller. – ubershmekel Aug 31 '15 at 4:45
• Let's say we were trying to figure out P(S2=2) , P(S2 = 4), P(S2 = 6), etc., how could we modify this answer to get there? – Blake Steines May 22 '19 at 22:22

Let $$x$$ be the expected number of steps to go from $$0$$ to $$1$$. Then, half the time you'll get there in one step, and the other half of the time you'll take one step and then need to cover twice the distance (because now you need to go from $$-1$$ to $$1$$). Therefore

$$x = \frac{1}{2}(1) + \frac{1}{2}(1 + 2x)$$

which simplifies to

$$x = x + 1$$

which means that

$$x = \infty$$

So if going from $$0$$ to $$1$$ in expectancy takes infinity steps, then going from $$0$$ to any other number also takes infinity steps.