Probability density with finite moment?

[bjj_99]

Homework Statement

From Hoel, Port, & Stone, Chapter 4, Exercise 9: Construct an example of a density that has a finite moment of order r but has no higher finite moment. Hint: Consider the series \sum_{k=1}^{\infty} k^{-(r+2)} and make this into a density.

Btw, this is for my own self-study. I don't even know if schools use this book anymore (my edition is from 1971).

Homework Equations

The r-th moment for a discrete random variable is defined as \sum_{k}k^r f(k) where f(k) is a probability density function (i.e., \sum_{k}f(k) = 1 )

The Attempt at a Solution

If we were to take the Hint and pretend f is a probability density defined by f(k) = k^{-(r+2)}, then I see that a moment of order > r would result in something greater than a harmonic series, which is divergent. But I don't see how to turn f into a probability density function. I know \sum_{k=1}^{\infty} k^{-2} = \pi / 6, but I'm pretty sure there are no "convenient" solutions for \sum_{k=1}^{\infty} k^{-(r+2)} where r is an arbitrary positive integer. If there were, than I could just normalize the series to 1 by dividing it, e.g., \frac{6}{\pi} \sum_{k=1}^{\infty} k^{-2} qualifies as a probability density for r=0.

Any help would be much appreciated! A hint or two would be ideal. All of the exercises in this book have been relatively straightforward up to this point, so I feel like this question should not be that difficult. Perhaps I am just misunderstanding the question.

 

[bjj_99]

Ah, figured it out. I was looking for a nice solution when the raw form would suffice... just divide the series above by itself. So the probability density function would be:
<br /> \frac{1}{\sum_{k=1}^{\infty} k^{-(r+2)}} \sum_{k=1}^{\infty} k^{-(r+2)}
<br /> = \sum_{k=1}^{\infty} \frac{k^{-(r+2)}}{\sum_{k=1}^{\infty} k^{-(r+2)}}

This converges to 1 for r >= 0. And moments of order > r would diverge.