# Feynman's random walk (6-3)

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

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 $$1$$

• Thank you very much, however I still don't get what is N-1. – Benjamin Sauvé Jul 29 at 17:54
• @BenjaminSauvé That is defined in terms of N-2 and so on – Superfast Jellyfish Jul 30 at 5:14
• 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. – Benjamin Sauvé Jul 30 at 10:48
• 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. – Superfast Jellyfish Jul 30 at 12:33
• 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? – Benjamin Sauvé Jul 31 at 1:32