# In Feynman's “Random Walker” probability example, why is $D^2$ better for illustrating “wandering”?

Taken from Volume 1, Chapter 6, Section 3 of the Feynman Lectures on Physics.

Feynman says that in describing random, equally-probable-backwards-or-forwards motion, that,

We might therefore ask what is his average distance travelled in absolute value, that is, what is the average of $|D|$. It is, howevermore convenient to deal with another measure of "progress", the square of the distance: $D^2$ is positive for either positive or negative motion, and is therefore a reasonable measure of such random wandering.

We can show that the expected value of $D^2_N$ is just $N$, the number of steps taken.....

He seems to provide a reason that applies just the same to using absolute value. From there, beginning with that next paragraph, he doesn't make any sense at all, so I think this is central to his point. Why does he use squares of the distances instead of absolute value? And how does $D^2_N = N$?

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The difference between absolute values and squaring is purely a matter of convenience: it's less of a headache to deal with squares, than with absolute values because of the sign of what's inside the absolute value. If you're evaluating an integral, as you would be if you're after expected value, then there's no need to consider when $D$ is positive or negative. Really you want the root mean square distance: $\sqrt{\mathbb{E}[D^2]}$ to give you a sense of distance.

You can write $D=X_1+X_2+\ldots+X_N$ for $N$ steps of the random walk, each $X_i$ being an independent random coin toss, with $X_i=1$ having probability $1/2$ (and $X_i=-1$ having probability $1/2$), so that $\mathbb{E}[X_i]=0$ and $\mathbb{E}[X_i^2]=1$. Then by linearity of expectation,

$$\mathbb{E}[D^2]=\mathbb{E}[X_1^2]+\ldots+\mathbb{E}[X_N^2]=N$$

where we noticed by independence: $\mathbb{E}[X_iX_j]=0$ for $i\neq j$. Now try figuring out $\mathbb{E}[|D|]$. It's considerably more annoying! You can't expand out the absolute value like we did when squaring. Linearity of expectation was key here as well. So now you have to worry about when the walk is positive or negative. That's a harder but doable combinatorial problem. Ultimately you won't gain much more information about your random walk than you do from the square.

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I understood most of the first paragraph.....I don't get how you arrive at all of those conclusions that follow that first paragraph so quickly, almost seeming like it needn't any proof or explanation whatsoever. What is i? What is E? And how does E[Xi^2] = 1 (whatever it is)? What is j?!? – bittenfig Mar 6 '14 at 22:06
@bittenfig: Each $X_i$ is a step in the random walk. In other words, the $i$'th step of the walk either takes the walk up 1, or down 1, with equal probability. $\mathbb{E}$ is the symbol for expectation. As a leading question, how familiar are you with probability theory? – Alex R. Mar 6 '14 at 22:17
Probably not too familiar. But the concepts in this section of the FLP don't seem foreign, it's just that they aren't explained too well. – bittenfig Mar 6 '14 at 22:22
@bittenfig: if it helps you, take a look at the wiki here: en.wikipedia.org/wiki/Random_walk – Alex R. Mar 6 '14 at 22:27
Will do. Thanks! – bittenfig Mar 6 '14 at 23:19