Question

Brownian Motion: Calculate the average distance travelled for a random walk of 10 steps, if the probability of going forward on each step is twice the probability of going backward.

Answer 1

Let the probability of going forward is P(f) and going backward is P(b).

We know that P(f) = 2 * P(b), and also P(f) + P(b) = 1

substituting for P(f), we get

P(b) = 1/3 and P(f) = 2/3

Total number of steps taken (N) = 10 steps

Let's say n₁ steps are taken forward and n₂ steps are taken backwards.

Hence, there are total ¹⁰Cn₁ ways of taking n₁ steps forward and n₂ steps backward.

We can write the probabilty of taking n₁ steps forward as using binomial probability distribution as we have only two alternatives:

P(n1) = ¹⁰C_n1 * P(f)ⁿ¹ * P(b)ⁿ²

But, since it is binomial distribution we can directly write its mean as

<n₁> = P(f) * N

= 2/3 * 10 = 6.33

<n₂> = P(b) * N

= 1/3 * 10 = 3.33

So, the average distance travelled will be equal to |<n₁> - <n₂>| = 3 steps