# Thread: Random walk with 1/(1+x) probability

1. ## Random walk with 1/(1+x) probability

You are walking along the x axis and you start at $x = 0$. At each time step you move to $x + 1$ with probability $p = \frac{1}{1+x}$ and to $x-1$ with probability $p = \frac{x}{1+x}$. What is the expected number of steps until you return to 0 for the first time?

2. ## Re: Random walk with 1/(1+x) probability

Should I post a solution or are people trying this? o.o

3. ## Re: Random walk with 1/(1+x) probability

I looked at it for 10 min and decided it was more than I wanted to spend time on.

4. ## Re: Random walk with 1/(1+x) probability

number of steps 1 with probability 0

number of steps 2 with probability 1/2!

number of steps <2n-1> with probability 0

number of steps <2n> with probability 1/(n+1)!

Does this make sense?

5. ## Re: Random walk with 1/(1+x) probability

If n is number of steps then
distance = (1+x)[n/(1+x)] + (1-x)[nx/(1+x)] = 0
gives n(1+x^2) = 0
n = 0 since 1+x^2 is not zero