Consider a typical random walk problem, where the probability to go right is $R$ and the probability to go left is $L$, where $R+L=1$. The particle can move 1 unit in each step, and starts at zero.

Let the particle move $n$ steps and write down its location away from zero.

Now repeat this a very large number of times.

What is the mean location of the particle from zero, and what is the standard deviation of the particle's location?

  • $\begingroup$ Do you know Binomal distribution and what happens in the limit ? $\endgroup$ – Chinny84 Oct 21 '15 at 9:49
  • $\begingroup$ I am familiar with this distribution but don't know what you mean by limit. $\endgroup$ – E Be Oct 21 '15 at 9:51
  • $\begingroup$ What does the Binomal distribution become in the limit the number of trials becomes large? $\endgroup$ – Chinny84 Oct 21 '15 at 9:52
  • 1
    $\begingroup$ It becomes a Normal distribution, according to the Central limit theorem. $\endgroup$ – E Be Oct 21 '15 at 10:05

You want the mean and variance of the sum of the random variables $X_1, X_2, ... , X_n$ which i.i.d with $P(X_i=1) = R$ and $P(X_i = -1) = 1-R$.

The moments can be calculated by direct summation:

\begin{equation} \begin{split} \langle\sum_{i=1}^n X_i\rangle &= \sum_{i=1}^n\langle X_i\rangle\\ &= \sum_{i=1}^n (2R-1)\\ &= n(2R-1) \end{split} \end{equation}


\begin{equation} \begin{split} \langle\Big(\sum_{i=1}^n X_i\Big)^2\rangle &= \langle\sum_{i=1}^n X_i\sum_{j=1}^n X_j\rangle\\ &= \sum_{i=1}^n\langle X_i^2\rangle + \sum_{i\neq j}\langle X_iX_j \rangle\\ &= \sum_{i=1}^n\langle X_i^2\rangle + \sum_{i\neq j}\langle X_i\rangle \langle X_j \rangle\\ &= n + n(n-1)(2R-1)^2 \end{split} \end{equation}

So the standard deviation is

$$\sqrt{\langle\Big(\sum_{i=1}^n X_i\Big)^2\rangle - \langle\sum_{i=1}^n X_i\rangle^2} = 2\sqrt{nR(1-R)}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.