I've come across many articles that describe standard deviation to be a measure of how much spread out our distribution is. In other words, it is a measure of how far away from the mean, we expect a data point to land.
Now, imagine we have a distribution of random variables, like in the case of a symmetric random walk. The mean of the distribution is obviously $0$. So we can write $\langle x_i \rangle = 0 $.
In this case, let $\sigma$ be the standard deviation. Then we have the following : $$\sigma^2 = \langle x_i^2 \rangle – \langle x_i\rangle^2$$
In our example, we can say :
$$\sigma = \sqrt{\langle x_i^2\rangle}$$
The term on the right is often referred to as root mean squared distance in random walk and other problems of the sort. It is often interpreted as, if we repeat the experiment many many times, and take the squares of the final outcome, and average it, and then take the square root, our average outcome is $\sigma.$
What I know is, the likelihood of getting a data point is maximum between $+\sigma$ and $-\sigma$ i.e. within one standard deviation of the mean with a likelihood of around $70 $ percent. However, I still don't understand why many authors, interpret the root mean squared distance to be the average distance where we find the final answer. To me, the root mean squared distance is actually the boundary of the region (strip) where we are most likely to obtain our results.
However, even though we are likely to obtain our data within the root mean squared distance, or the standard deviation, we do find many data points outside of this range. If we take an average of all these distances where we happen to find the range, it does lie around the standard deviation.
Is this alternate interpretation of standard deviation correct? Normally standard deviation marks the boundary of the region in which, we are most likely to obtain the outcomes.
However, it is also the average distance from the mean, where we expect our outcomes to be. This is usually the description of the root mean squared position, but in this instance, it is equivalent to the standard deviation.
In random walk, most of our walkers, about $70$ percent of them, are found within the region bounded by $\pm \sigma.$ These walkers have taken more steps in one direction compared to the other. Since $\sigma \approx \sqrt{n}$, the difference in steps for 70 percent of walkers is less than $\sqrt{n}$. However, for the rest of the walkers, about 30 percent of them, this difference is more than $\sqrt{n}$ and can be enormously large. So, we have 70 percent of walkers with a small difference in steps and a small no. of walkers with a large difference in steps. The average difference thus happens to lie around the standard deviation, and so it is often interpreted as the average distance where we happen to find our walker.
Is this reasoning correct?
Best Answer
Re to "how can we interpret the standard deviation as a distance". As I wrote in a comment above, the following can be interpreted as a distance between the random variables $X$ and $Y$; $$ d_2(X, Y)=\sqrt{E[\lvert X-Y\rvert^2]}. $$ The subscript $2$ stands for $L^2$ (ell-two), which is a reference to functional analysis. This $d_2$ is exactly a generalization of the usual distance between points, that is $$ d_{\mathbb R^3}( (x_1, x_2, x_3), (y_1, y_2, y_3)) =\sqrt{ \sum_{j=1}^3 \lvert x_j- y_j\rvert^2}.$$ As you can see, $d_2$ is essentially the same thing as $d_{\mathbb R^3}$, only with an $E$ operator in place of a $\sum$.
With this interpretation, the standard deviation of a random variable $X$ with expectation $\mu$ reads $$ \sigma=d_2(X, \mu),$$ that is, standard deviation is the distance of $X$ from its mean.