1
$\begingroup$

Let's say you are given that $E(X^n)$ = $\frac{n!}{((n+3!)/3!)}$ for a random variable $X$. So the first 4 moments are $\frac{1}{4}, \frac{1}{10}, \frac{1}{20}, \frac{1}{35}$, and so on. Is there any way to back into an approximation of the pdf from this?

It seems like the boundedness of $X$ would suggest uniqueness, but I am having difficulty finding whether we can actually figure out what the random variable might be, or at least learn something more about $X$.

$\endgroup$
3
  • $\begingroup$ Hi Donald, if you don't get any answers here, then after a few days pop over to stats.stackexchange.com, and then add a link here to the new question over there. $\endgroup$
    – David Roberts
    Commented Dec 2, 2012 at 21:39
  • $\begingroup$ Could you please verify your formulas? I don't see how you get those first moments from your formula. And could you give any motivation for the question? $\endgroup$ Commented Dec 2, 2012 at 22:10
  • $\begingroup$ Apologies, it is actually n!/((n+3)!/3!). These are the moments implied by a characteristic function whose corresponding distribution function I do not know. Thanks for the clarification. I've edited the question. $\endgroup$ Commented Dec 2, 2012 at 22:37

1 Answer 1

2
$\begingroup$

In general, it is not possible to uniquely recover the exact distribution function given the moments. An nice counter example is given on page 48 of Stoyanov's "Counterexamples in Probability and Statistics" (link below). It is possible when moment generating function is smooth and finite around the origin, and the exact answer for the inversion is given by the Mellin transform (http://en.wikipedia.org/wiki/Bromwich_integral). Also see the discussion here: https://stats.stackexchange.com/questions/32706/existence-of-the-moment-generating-function-and-variance. You mentioned in the comments that you determined these moments from the characteristic function. The characteristic function does uniquely determine the probability distribution function by its inverse Fourier transform. Working with the characteristic function itself should be easier.

Link to counterexample: http://books.google.com/books?id=irKSXZ7kKFgC&pg=PA46&lpg=PA46&dq=counterexamples+in+probability+moment+exist&source=bl&ots=tY_zdlzcoM&sig=Ed4mbVV7O8KAvzuERZKNR5LhVG0&hl=en&sa=X&ei=Ifi7UNOZJor48gTj-YDYDw&ved=0CCwQ6AEwAA#v=onepage&q=counterexamples%20in%20probability%20moment%20exist&f=false

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .