# Feynman's random walk: How does he get $⟨D^2_N⟩=N$?

In Feynman's lectures, section 6.3, I follow most of his argument about a random walk, but I miss one step. To summarize, he's discussing a one-dimensional random walk (eg, determined by coin flips), and his notation is

$N$: steps taken

$D_N$: net distance from start at step $N$ for a given trial

$⟨D^2_N⟩$: expected value for $D_N$ (mean square distance)

I follow him as he shows

$⟨D^2_1⟩=1$, and

$⟨D^2_N⟩=⟨D^2_{N−1}⟩+1$.

Then he says it follows that

$⟨D^2_N⟩=N$.

I suppose this last step should be easy, but it's the one I don't follow! I'd appreciate any help. TIA.

• What would $\langle D_2^2\rangle$ be? – lemon Jun 13 '16 at 18:39
• Thanks @lemon! Guess I'm rusty on recognizing recursion. – bostonquad Jun 13 '16 at 19:13

Well, $<D_2^2>=<D_1^2> +1 = 2$ and $<D_3^2>=<D_2^2> +1=3$. I'm guessing you probably follow the pattern now.