Let steps of equal length be taken along a line. Let be the probability of taking a step to the right, the probability of taking a step to the left, the number of steps taken to the right, and the number of steps taken to the left. The quantities , , , , and are related by
(1) |
and
(2) |
Now examine the probability of taking exactly steps out of to the right. There are ways of taking steps to the right and to the left, where is a binomial coefficient. The probability of taking a particular ordered sequence of and steps is . Therefore,
(3) |
where is a factorial. But this is simply a binomial distribution, so the mean number of steps to the right is
(4) |
and the mean number of steps to the left is
(5) |
Similarly, the variance is given by
(6) |
and the root-mean-square deviation is
(7) |
Consider now the distribution of the distances traveled after a given number of steps,
(8) |
as opposed to the number of steps in a given direction. The above plots show for and three values , , and , respectively. Clearly, weighting the steps toward one direction or the other influences the overall trend, but there is still a great deal of random scatter, as emphasized by the plot below, which shows three random walks all with .
Surprisingly, the most probable number of sign changes in a walk is 0, followed by 1, then 2, etc.
For a random walk with , the probability of traveling a given distance after steps is given in the following table.
steps | 0 | 1 | 2 | 3 | 4 | 5 | |||||
0 | 1 | ||||||||||
1 | 0 | ||||||||||
2 | 0 | 0 | |||||||||
3 | 0 | 0 | 0 | ||||||||
4 | 0 | 0 | 0 | 0 | |||||||
5 | 0 | 0 | 0 | 0 | 0 |
In this table, subsequent rows are found by adding half of each cell in a given row to each of the two cells diagonally below it. In fact, it is simply Pascal's triangle padded with intervening zeros and with each row multiplied by an additional factor of 1/2. The coefficients in this triangle are given by
(9) |
(Papoulis 1984, p. 291). The moments
(10) |
of this distribution of signed distances are then given by
(11) |
|||
(12) |
|||
(13) |
|||
(14) |
so the mean is , the skewness is , and the kurtosis excess is
(15) |
The expectation value of the absolute distance after steps is therefore given by
(16) |
|||
(17) |
This sum can be done symbolically by separately considering the cases even and odd. First, consider even so that . Then
(18) |
|||
(19) |
|||
(20) |
|||
(21) |
But this sum can be evaluated analytically as
(22) |
Writing , plugging back in, and simplifying gives
(23) |
where is the double factorial.
Now consider odd, so . Then
(24) |
|||
(25) |
|||
(26) |
|||
(27) |
|||
(28) |
But this sum can be evaluated analytically as
(29) |
Writing , plugging back in, and simplifying gives
(30) |
|||
(31) |
|||
(32) |
Both the even and odd solutions can be written in terms of as
(33) |
or explicitly in terms of as
(34) |
|||
(35) |
The first few values of for , 1, ... are therefore 0, 1, 1, 3/2, 3/2, 15/8, 15/8, 35/16, 35/16, ... (OEIS A086116 and A060818; Abramowitz and Stegun 1972, Pr
Let steps of equal length be taken along a line. Let be the probability of taking a step to the right, the probability of taking a step to the left, the number of steps taken to the right, and the number of steps taken to the left. The quantities , , , , and are related by
(1) |
and
(2) |
Now examine the probability of taking exactly steps out of to the right. There are ways of taking steps to the right and to the left, where is a binomial coefficient. The probability of taking a particular ordered sequence of and steps is . Therefore,
(3) |
where is a factorial. But this is simply a binomial distribution, so the mean number of steps to the right is
(4) |
and the mean number of steps to the left is
(5) |
Similarly, the variance is given by
(6) |
and the root-mean-square deviation is
(7) |
Consider now the distribution of the distances traveled after a given number of steps,
(8) |
as opposed to the number of steps in a given direction. The above plots show for and three values , , and , respectively. Clearly, weighting the steps toward one direction or the other influences the overall trend, but there is still a great deal of random scatter, as emphasized by the plot below, which shows three random walks all with .
Surprisingly, the most probable number of sign changes in a walk is 0, followed by 1, then 2, etc.
For a random walk with , the probability of traveling a given distance after steps is given in the following table.
steps | 0 | 1 | 2 | 3 | 4 | 5 | |||||
0 | 1 | ||||||||||
1 | 0 | ||||||||||
2 | 0 | 0 | |||||||||
3 | 0 | 0 | 0 | ||||||||
4 | 0 | 0 | 0 | 0 | |||||||
5 | 0 | 0 | 0 | 0 | 0 |
In this table, subsequent rows are found by adding half of each cell in a given row to each of the two cells diagonally below it. In fact, it is simply Pascal's triangle padded with intervening zeros and with each row multiplied by an additional factor of 1/2. The coefficients in this triangle are given by
(9) |
(Papoulis 1984, p. 291). The moments
(10) |
of this distribution of signed distances are then given by
(11) |
|||
(12) |
|||
(13) |
|||
(14) |
so the mean is , the skewness is , and the kurtosis excess is
(15) |
The expectation value of the absolute distance after steps is therefore given by
(16) |
|||
(17) |
This sum can be done symbolically by separately considering the cases even and odd. First, consider even so that . Then
(18) |
|||
(19) |
|||
(20) |
|||
(21) |
But this sum can be evaluated analytically as
(22) |
Writing , plugging back in, and simplifying gives
(23) |
where is the double factorial.
Now consider odd, so . Then
(24) |
|||
(25) |
|||
(26) |
|||
(27) |
|||
(28) |
But this sum can be evaluated analytically as
(29) |
Writing , plugging back in, and simplifying gives
(30) |
|||
(31) |
|||
(32) |
Both the even and odd solutions can be written in terms of as
(33) |
or explicitly in terms of as
(34) |
|||
(35) |
The first few values of for , 1, ... are therefore 0, 1, 1, 3/2, 3/2, 15/8, 15/8, 35/16, 35/16, ... (OEIS A086116 and A060818; Abramowitz and Stegun 1972, Pr
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