Interpreting distance in random walk I've recently started reading about the random walk, from different sources across the internet, and there is this small detail that I'm not being able to wrap my head around.
Suppose we have, a symmetric random walk in $1$ dimension. The walker has an equal probability of going to the left or to the right, and he starts at the origin. The first obvious question is, what is the expected position of the walker after $n$ steps. The answer is obviously $0$. This is because, in the normal distribution of positions, $0$ has the highest probability. However, even though getting $0$ is the most probable or rather average outcome of all the simulations, it is not the most likely outcome.
This is because, with more steps, the variance increases, and the bell curve spreads out. Among individual positions, $0$ still has the highest probability, but the probability of not being $0$ increases. This is characterized by the standard deviation.
In symmetric walk, the standard deviation $\sigma$ and the root-mean-squard distance $x_{rms}$ are the same thing. Moreover we have $x_{rms} \propto \sqrt{n} \space\space$ as the distribution spreads out more and more.
As this happens, the likelihood of landing away from the center, increases, and so the walker stops at the distance away from zero.
Many books interpret the root mean squared distance as the most likely distance. This doesn't make sense to me. Yes, the root mean squared distance gives us a measure or an estimate of how far from the mean, the walker would stop. However, it surely isn't the most likely position.
For example, if $n=100$, we have $x_{rms}=10$. This is often interpreted as the most likely distance is positive or negative $10$ from the mean. However, shouldn't the interpretation be more like, the most likely position is between $10$ and $-10$? Shouldn't this be the correct interpretation ?
A higher value of $x_{rms}$ should be interpreted as a higher likelihood of landing away from the mean, shouldn't it ? I don't know why most books interpret this as the most probable position.
If I'm wrong, can someone give me the correct intuitive physical explanation of what $x_{rms}$ actually represents? To me, it is just an abstract measure of how far from the mean, the walker is expected to land.
 A: $x_{rms}$ is a kind of average (positive) distance away from the starting point.
Let's say the experiment was done 5 times with 100 steps each time.
A list of displacements at the end of 100 steps might be -5, -28, 32, 6, -12
These are squared and that makes them positive, an average found, then square rooted - r.m.s stands for root-mean-square.
In the example above of it's the square root of $$\frac{25+784+1024+36+144}{5}$$
i.e. 20.06
but for a larger sample it would work out as 10.
So $x_{rms}$ is not the most likely distance from the start, but a kind of average distance from the start that you would expect from a large number of trials.
If you wanted to pick a single place that was most likely after 100 steps it would be the starting point.  If there were only two steps, L= left, R= right, the 4 possibilities are LL, LR, RL, RR and the starting point is most likely.
A: I think the confusion here is about what is meant by most likely.
Most likely value to be measured is the value corresponding to the maximum of the probability distribution (its mode). This can be a very small value, but still greater than the probability of finding any other value.
Note further that for a continuous distribution the probability of hitting any specific point is zero - the meaningful probabilities are thsoe of hitting a finite interval. So we may ask, where are we most likely to find the walker after $n$ steps? The answer is within interval $[-\sigma, +\sigma]$, where $\sigma$ is the standard deviation of the statistical distribution: see here for the normal distribution.
