Let be an odd prime. A probability distribution on is symmetric and unimodal if and only if for all such that and
Exercise 9 asks us to show that if and are two symmetric unimodal distributions then their convolution is again symmetric and unimodal. The symmetry is easy: for any we have
First proof of unimodality
Let . It suffices to show that if then . In fact we shall prove that is a sum of terms of the form where and . Each such term is positive.
It is clear that one contribution to comes from
If we increase the indices of the terms, we get , and so on, until we reach . At this point the indices of the terms begin to increase, until we reach . The next term is , which vanishes, and then we continue until . (Note the change of sign.) Then finally the indices of the terms decrease until we reach . In this process we see each twice. A rather fiddly calculation shows that if then the coefficient of is
which is as required. The case should be similar, but I haven’t worked out the formula.
A better proof
A very slick proof of unimodality for random variables taking values in was given by Purkayastha in 1998. The argument can be adapted to our case.
Let be a random variable distributed so that . Then
A standard result states that if is any random variable taking non-negative real values then . Applying this result to the random variable we see that
For each there exists an such that
It therefore suffices to prove that for each fixed , the function
is a decreasing function of , for . (Note that and are computed in .) Now finally we have
which is always positive provided .
Note that the probability distribution of the position of a particle after steps of the symmetric random walk in which the steps have equal probability is certainly not unimodal, because if is even then the particle must be at an even position, and similarly if is odd. (Exercise 10 gives a monotonicity result for these positions.)
The result on convolutions can however be applied to the random walk where the steps all have equal probability . It would be interesting to know how quickly this walk approaches the uniform distribution compared to the walk with steps . On the one hand, one might expect the convergence to be slower, since at each time there is a good chance we will stand still. On the other hand, one would not expect parity to lead to any significant extra departure from randomness.