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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?

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  • $\begingroup$ 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
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    $\begingroup$ It becomes a Normal distribution, according to the Central limit theorem. $\endgroup$ – E Be Oct 21 '15 at 10:05
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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}

and

\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

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

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