The position of a particle in one dimension changes in discrete steps. With each step it moves to the right, however, the length of the step is drawn from a uniform distribution from the interval \(\left[ {{\rm{λ }}\,{\rm{ - }}\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{w,}}\,{\rm{λ }}\,{\rm{ + }}\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{w}}} \right] \) , where λ and w are positive constants. If X denotes the distance from the starting point after N steps, the standard deviation \(\sqrt {\left\langle {{X^2}} \right\rangle \, - {{\left\langle X \right\rangle }^2}} \) for large values of N is
Understanding Particle Position in Random Walk
The problem describes a particle moving in one dimension through discrete steps. Each step is taken to the right, and the length of each step is a random variable drawn from a uniform distribution. We are asked to find the standard deviation of the particle's position after \(N\) steps for large \(N\).
Step Length Distribution
Let \(s_i\) be the length of the \(i\)-th step. The length \(s_i\) is drawn from a uniform distribution over the interval \(\left[ \lambda - \frac{1}{2}w,\ \lambda + \frac{1}{2}w \right]\).
For a uniform distribution over an interval \([a, b]\), the mean (expected value) and variance are given by:
- Mean: \(\frac{{a} + {b}}{{2}}\)
- Variance: \(\frac{{{{\left( {{b} - {a}} \right)}}^{2}}}{{{12}}}\)
Mean and Variance of a Single Step
For a single step \(s_i\) with \(a = \lambda - w/2\) and \(b = \lambda + w/2\):
The mean expected step length is:
$$\langle s_i \rangle = \frac{(\lambda - w/2) + (\lambda + w/2)}{2} = \frac{2\lambda}{2} = \lambda$$
The variance of a single step length is:
$$\text{Var}(s_i) = \frac{\left(\left(\lambda + \frac{w}{2}\right) - \left(\lambda - \frac{w}{2}\right)\right)^2}{12} = \frac{w^2}{12}$$
Total Distance after N Steps
The total distance from the starting point after \(N\) steps is the sum of the lengths of individual steps:
$$X = \sum_{i=1}^N s_i$$
Since the steps are independent, the expected value of \(X\) is the sum of the expected values of the individual steps:
$$\langle X \rangle = \left\langle \sum_{i=1}^N s_i \right\rangle = \sum_{i=1}^N \langle s_i \rangle = \sum_{i=1}^N \lambda = N\lambda$$
The variance of \(X\) is the sum of the variances of the individual steps because the steps are independent:
$$\text{Var}(X) = \text{Var}\left(\sum_{i=1}^N s_i\right) = \sum_{i=1}^N \text{Var}(s_i) = \sum_{i=1}^N \frac{w^2}{12} = N \frac{w^2}{12}$$
Recall that variance is defined as \(\left\langle {{X^2}} \right\rangle \, - {{\left\langle X \right\rangle }^2}\). So, \(\text{Var}(X) = \left\langle {{X^2}} \right\rangle \, - {{\left\langle X \right\rangle }^2}\).
Calculating Standard Deviation
The standard deviation is the square root of the variance:
Standard Deviation \( = \sqrt{\left\langle {{X^2}} \right\rangle \, - {{\left\langle X \right\rangle }^2}} = \sqrt{\text{Var}(X)}\)
Substituting the variance we found:
Standard Deviation \( = \sqrt{N \frac{w^2}{12}}\)
We can simplify this expression:
$$\sqrt{N \frac{w^2}{12}} = \sqrt{N} \times \sqrt{\frac{w^2}{12}} = \sqrt{N} \times \frac{w}{\sqrt{12}} = \sqrt{N} \times \frac{w}{2\sqrt{3}} = \frac{w}{2} \sqrt{\frac{N}{3}}$$
Thus, for large values of \(N\), the standard deviation of the distance \(X\) is \(\frac{{w}}{{2}}\,{\rm{ \times }}\sqrt {\frac{{N}}{3}} \).
Important Questions from Waves
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