How Does a Drunk's Random Walk Between Lamp Posts Model a Binomial Distribution?

[CAF123]

Homework Statement

A drunk lurches from one lamp post to the next on his way home. At each lamp post he pauses and is equally likely move towards or away from home. Suppose the posts are separated by a distance ##a## and find the mean and standard deviation of his displacement ##d## from the starting point, after ##N## steps.

Homework Equations

Binomial distribution may be incorporated

The Attempt at a Solution

I assume that each step the drunk makes is of length ##a##. He starts at some origin and then takes a total of N steps to reach a displacement d. Let ##r## be the number of steps to the right and let ##\ell## be the number of steps to the left. Then ##r + \ell = N##. At a given post, the probability that the drunk moves right is 1/2 = probability that the drunk moves left. The probability of getting ##r## right steps is then binomially distributed;$$P(\text{r right steps}) = {N \choose r} p^r (1-p)^{N-r},$$ and since ##p = 1/2##, this is the same as $${N \choose r} \frac{1}{2^N}$$ Similarly, $$P(\text{l left steps}) = {N \choose N-r} \frac{1}{2^N}$$ which is the same distribution.
I do not know how to proceed from here. I was also thinking that the expected value of displacement is given by the sum of all possible displacements multiplied by their corresponding probabilities, however, I am not sure how to obtain this in practice.

Many thanks.

 

[Dick]

Homework Statement

A drunk lurches from one lamp post to the next on his way home. At each lamp post he pauses and is equally likely move towards or away from home. Suppose the posts are separated by a distance ##a## and find the mean and standard deviation of his displacement ##d## from the starting point, after ##N## steps.

Homework Equations

Binomial distribution may be incorporated

The Attempt at a Solution

I assume that each step the drunk makes is of length ##a##. He starts at some origin and then takes a total of N steps to reach a displacement d. Let ##r## be the number of steps to the right and let ##\ell## be the number of steps to the left. Then ##r + \ell = N##. At a given post, the probability that the drunk moves right is 1/2 = probability that the drunk moves left. The probability of getting ##r## right steps is then binomially distributed;$$P(\text{r right steps}) = {N \choose r} p^r (1-p)^{N-r},$$ and since ##p = 1/2##, this is the same as $${N \choose r} \frac{1}{2^N}$$ Similarly, $$P(\text{l left steps}) = {N \choose N-r} \frac{1}{2^N}$$ which is the same distribution.
I do not know how to proceed from here. I was also thinking that the expected value of displacement is given by the sum of all possible displacements multiplied by their corresponding probabilities, however, I am not sure how to obtain this in practice.

Many thanks.

Your 'right steps' and 'left steps' distributions are counting the same thing. Just use one of them. If you take r steps right (and N-r left) where do you wind up?

 

[CAF123]

Hi Dick,

Your 'right steps' and 'left steps' distributions are counting the same thing. Just use one of them. If you take r steps right (and N-r left) where do you wind up?

This depends on the combination of right and left steps. The displacement is given by ##ka## where ##k## is an integer. If there are no left steps, then ##k = r = N##.

 

[Ray Vickson]

Hi Dick,

This depends on the combination of right and left steps. The displacement is given by ##ka## where ##k## is an integer. If there are no left steps, then ##k = r = N##.

You seem to be missing the point: given N and r, what is the displacement X? You need to figure this out because you are being asked to find the mean and standard deviation of X.

 

[CAF123]

You seem to be missing the point: given N and r, what is the displacement X? You need to figure this out because you are being asked to find the mean and standard deviation of X.

I see, so then ##X = ra - \ell a = ra - (N-r)a##. To obtain the mean it is a case of solving the equation ##X_{avg} = 2r_{avg} a - Na##. I can see where this method is going and in fact I have already solved it this way. I was wondering if there was a way of solving this explicitly with the distributions I have derived or via conditional probabilities.

 

[mfb]

To find the mean, you can simply use the symmetry of the problem. The standard deviation is more interesting.

I was wondering if there was a way of solving this explicitly with the distributions I have derived or via conditional probabilities.

You can find r_avg with the distributions. I don't see where conditional probabilities would occur, as all steps are independent.

 

[CAF123]

To find the mean, you can simply use the symmetry of the problem. The standard deviation is more interesting.

Since the drunk is equally likely to go backwards or forwards at each post, the mean should be zero.

You can find r_avg with the distributions.

##r_{avg}## is the expected number of right steps. This is just an expectation of a binomial distribution (regard each step as a trial and moving to the right an event - at each post the probability of the event is 1/2).

E[right steps] = r(1/2) + (N-r)(1/2) = N/2. How would I show this using my distributions?

I also want to try to use the explicit expression for the expectation that is ##E[X] = \sum_i i P_i##, where ##i## is a possible displacement and ##P_i## is the probability of that displacement. The summation would be over the entire sample space which is all possible displacements. Is it possible to find such a form?

I don't see where conditional probabilities would occur

I did see a method in my examples class, but the page was full with conditional probabilities that I preferred to focus on my own method. I think he was conditioning on the fact that we start at, say, ##x_o = 0## and then considered for example P(X=x_o+a|x_o=0).

 

[mfb]

Since the drunk is equally likely to go backwards or forwards at each post, the mean should be zero.

Right.

##r_{avg}## is the expected number of right steps. This is just an expectation of a binomial distribution (regard each step as a trial and moving to the right an event - at each post the probability of the event is 1/2).

E[right steps] = r(1/2) + (N-r)(1/2) = N/2. How would I show this using my distributions?

That is the easy way, indeed. You can calculate the probability of every r from 0 to N and then calculate the expectation value based on the distribution. I don't see a reason for that, but you can do it.

I also want to try to use the explicit expression for the expectation that is ##E[X] = \sum_i i P_i##, where ##i## is a possible displacement and ##P_i## is the probability of that displacement. The summation would be over the entire sample space which is all possible displacements. Is it possible to find such a form?

See above, that is nearly the same (just split it in odd and even i to make it easier).

I did see a method in my examples class, but the page was full with conditional probabilities that I preferred to focus on my own method. I think he was conditioning on the fact that we start at, say, ##x_o = 0## and then considered for example P(X=x_o+a|x_o=0).

Ah well... that is more complicated than necessary.