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!).



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 $1$

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  • $\begingroup$ Thank you very much, however I still don't get what is N-1. $\endgroup$ – Benjamin Sauvé Jul 29 at 17:54
  • $\begingroup$ @BenjaminSauvé That is defined in terms of N-2 and so on $\endgroup$ – Superfast Jellyfish Jul 30 at 5:14
  • $\begingroup$ I just have trouble understanding what he is saying here: The expected value of D2N for N>1 can be obtained from DN−1. If, after N−1 steps, we have DN−1, then after N steps we have DN=DN−1+1 or DN=DN−1−1. $\endgroup$ – Benjamin Sauvé Jul 30 at 10:48
  • $\begingroup$ Expectation is additive. Meaning <a+b>=<a>+<b>. So if $D_n$ is the expectation of steps after n steps, it is expectation of n-1 steps plus whatever the step value is at nth step. Now those can be either +1 or -1. So that’s what the statement says. $\endgroup$ – Superfast Jellyfish Jul 30 at 12:33
  • $\begingroup$ but what do you mean by the expectation of steps after n steps? n is always going to be plus 1 regardless right? Assuming the Dn-1, after one step it would be D0? $\endgroup$ – Benjamin Sauvé Jul 31 at 1:32

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