Consider a projectile moving in a 2d plane in the +x direction toward a vertical line. Think of that line as a ruler. During each interval of time, it has .5 chance to shift up a unit and a .5 chance to shift down a unit along the y axis on its path. What is the probability distribution of the projectile landing on each point of that ruler? (it starts at 0)

So my method was to take the number of possible arrangements of 1 and -1 into n slots (n is the total number of time intervals) and divide it by the total number of possible arrangements.

To land on point m on the ruler, you must get $m+(\frac{n-m}{2})$ ups and $\frac{n-m}{2}$ downs. this gives a distribution of


However the integral of this function is 2 rather than 1. I'm guessing that each m represents both the positive and negative value of m, and I am double counting.


1 Answer 1


Be careful about how you account for the space of allowed values. For even $n$, only even final positions are possible. For odd $n$, only odd final positions are allowed. To show this, note that position starts at zero, and its parity flips every time step. In fact, your multiplicity term $\binom{n}{\frac{n-1}{2}}$ isn't even well defined when $n$ and $m$ do not have the same parity. (You could use gamma functions, but that definition wouldn't correspond to your counting problem.)

To show that your distribution is normalized, You need to prove the following:

$$1=\begin{cases} \sum\limits_{m=-n\\m\ \text{even}}^n\binom{n}{\frac{n-m}{2}}2^{-n} & n\ \text{even} \\ \\ \sum\limits_{m=-n\\m\ \text{odd}}^n\binom{n}{\frac{n-m}{2}}2^{-n} & n\ \text{odd} \end{cases}$$

Both sums can be reindexed into a binomial expansions of $(1+1)^p$. For the first one, try $n=2k$ and $m=2l$.

  • $\begingroup$ This is definiteley the reason thank you $\endgroup$ Sep 12, 2017 at 15:27

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