Is it possible to generate a 2D distribution function $f(x,y)$ with function supports specified as $[-a,a]$ and $[-b,b]$ for $x$ and $y$ respectively, such that it always has moments which are NON ZERO up-to a certain higher order say $p+q$?
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- $\begingroup$ Question modified after an answer was posted. $\endgroup$Did– Did2013-03-08 13:28:40 +00:00Commented Mar 8, 2013 at 13:28
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6 Every distribution whose suppport is bounded has moments of every order. Here, the support is included in $[-a,a]\times[-b,b]$ hence bounded.
If the support of the distribution is included in $[a,b]\times[c,d]$ with $0\lt a\lt b$ and $0\lt c\lt d$, then the $(i,j)$-moment is between $a^ic^j$ and $b^id^j$ for every nonnegative $i$ and $j$, in particular this moment is finite and positive.
- $\begingroup$ And can we ensure those moments are always non-zero even if the support is possitive i.e., for the supports [a,b] and [c,d]? $\endgroup$hAcKnRoCk– hAcKnRoCk2013-02-12 09:44:20 +00:00Commented Feb 12, 2013 at 9:44
- $\begingroup$ Not sure I get your question but in any case see revised version. $\endgroup$Did– Did2013-02-12 09:48:43 +00:00Commented Feb 12, 2013 at 9:48
- $\begingroup$ thanks for your answer. But every distribution whose support is bounded has moments of every order might not be true. As an example, consider a distribution which is symmetric. In this case, some of the 3rd order moments become zero. Is it not? $\endgroup$hAcKnRoCk– hAcKnRoCk2013-03-08 09:06:45 +00:00Commented Mar 8, 2013 at 9:06
- $\begingroup$ Example (and objection) unrelated to my post. Would you be saying "to have moments of any order" to mean "to have moments of any order, whose values are not zero"? $\endgroup$Did– Did2013-03-08 09:12:39 +00:00Commented Mar 8, 2013 at 9:12
- $\begingroup$ Yes. Sorry. So i should have phrased it 'moments of any order, but non-zero. I ve edited the question. $\endgroup$hAcKnRoCk– hAcKnRoCk2013-03-08 09:24:15 +00:00Commented Mar 8, 2013 at 9:24
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You could use the Moment Generating Function? Up to some conditions, i.e. when it exists, the MGF uniquely defines a distribution, by basically defining all its moments.
