Feynman's random walk (6-3)

Question

How does Feynman get to $D^2_{N-1}$?

The expected value of $D^2_N$ for $N>1$ can be obtained from $D_{N−1}$. If, after $N−1$ steps, we have $D_{N−1}$, then after $N$ steps we have $D_N=D_{N−1}+1$ or $D_N=D_{N−1}−1$. And could someone help me with this sentence?

In a number of independent sequences, we expect to obtain each value one-half of the time, so our average expectation is just the average of the two possible values. The expected value of $D^2_N$ is then $D^2_{N−1}+1$. In general, we should expect for $D^2_{N−1}$ its “expected value” $⟨D^2_{N−1}⟩$ (by definition!).

https://www.feynmanlectures.caltech.edu/I_06.html

Answer 1

What he’s saying is the following

$$\langle D^2_N\rangle= P_+\Big(\langle D^2_{N-1}\rangle +1\Big)+ P_-\Big(\langle D^2_{N-1}\rangle+1\Big)$$

Where he says that probability of each branch (plus $P_+$ or minus $P_-$) is equal to half. As they are equally likely. This means that the above equation evaluates to $$\langle D^2_N\rangle= \langle D^2_{N-1}\rangle+1$$

You can use this definition recursively down to $D^2_1$ which has to be defined which in this case is $1$