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**1. Hello gentlemen! I've studying the Feymann lectures, and I came across an interesting question, from reading the chapter 6, about probability.**

In this chapter, Feymann deduces the binomial distribution, and I am ok with that.

But, in the 3rd section, when adresses to the random walk problem, he registered something that was very cumbersome for me to understand.

Let me try to explain.

He deal with two variables when solving the binomial distribution problem: n, the number of tosses of a "fair" coin; and k, the number of heads thrown. In the end, he gets:

[itex]P(k,n)=\frac{\left( \stackrel{n}{k} \right)}{2^n}.[/itex]

Ok, I got the point.

But later, he starts to solve the random walk problem, and introduces another variable, D, which is the net distance traveled in N steps.

Ok.

Stablishing a relation to the binomial distribution problem, D is just the difference between the number of heads and the number of tails, as heads stands for a forward step and tais for backward steps.

So,

[itex]D = N_H - N_T,[/itex]

and

[itex]N = N_H + N_T.[/itex]

So,

[itex]D = 2N_H - N.[/itex]

All right. Nothing difficult up to now.

But then, comes the magic.

He afirms that "We have derived earlier an expression for the expected distribution for D. Since N is just a constant, we have the corresponding distribution for D."

I got the impression that he is trying to pass the idea that as D is in a linear relation with [itex]N_H[/itex], they must have the same distribution.

In this chapter, Feymann deduces the binomial distribution, and I am ok with that.

But, in the 3rd section, when adresses to the random walk problem, he registered something that was very cumbersome for me to understand.

Let me try to explain.

He deal with two variables when solving the binomial distribution problem: n, the number of tosses of a "fair" coin; and k, the number of heads thrown. In the end, he gets:

[itex]P(k,n)=\frac{\left( \stackrel{n}{k} \right)}{2^n}.[/itex]

Ok, I got the point.

But later, he starts to solve the random walk problem, and introduces another variable, D, which is the net distance traveled in N steps.

Ok.

Stablishing a relation to the binomial distribution problem, D is just the difference between the number of heads and the number of tails, as heads stands for a forward step and tais for backward steps.

So,

[itex]D = N_H - N_T,[/itex]

and

[itex]N = N_H + N_T.[/itex]

So,

[itex]D = 2N_H - N.[/itex]

All right. Nothing difficult up to now.

But then, comes the magic.

He afirms that "We have derived earlier an expression for the expected distribution for D. Since N is just a constant, we have the corresponding distribution for D."

I got the impression that he is trying to pass the idea that as D is in a linear relation with [itex]N_H[/itex], they must have the same distribution.

**2. In my opinion, Feymann did not express himself clearly, as he gives the impression that D and k have the same distribution because they have a linear relation between them. But, this is far from obvious, at least in my opinion. I cannot imagine that if tho variables are related by a linear relation, they will have same probability distribution.**

I think both D and k have same distribution because they are basically Dichotomous variables. I would like to receive your advices. Thank you in advance.

I think both D and k have same distribution because they are basically Dichotomous variables. I would like to receive your advices. Thank you in advance.

P.S.: sorry for my poor english.