Is it true that for every symmetrical distribution all odd-order moments are equal to zero?

If yes, how would I be able to prove such a thing?


I'm going to assume you mean (1) symmetric about $0$, and (2) distributions for which the odd-order moments actually exist. Under these assumptions, yes.

In the continuous case, the $n$th moment for a distribution with pdf $f(x)$ exists and is $$\int_{-\infty}^{\infty} x^n f(x) \,dx,$$ exactly when the integral converges. So, what does it mean for the integral to converge?

Remember that we define $\int_{-\infty}^{\infty} x^n f(x) \,dx$ to be $$ \int_{-\infty}^0 x^n f(x) \, dx + \int_0^{\infty} x^n f(x) \, dx.$$ (Well, we could choose some point other than $0$ to split the integral, but $0$ works fine.) Convergence of $\int_{-\infty}^{\infty} x^n f(x) \,dx$ is equivalent to the convergence of both of these integrals. Existence of the $n$th odd-order moment, then, is equivalent to $\int_{-\infty}^0 x^n f(x) \, dx$ and $\int_0^{\infty} x^n f(x) \, dx$ both being finite.

Suppose $\int_0^{\infty} x^n f(x) \, dx = A$. If we let $u = -x$, then $$\int_{-\infty}^0 x^n f(x) \, dx = \int_{\infty}^0 (-u)^n f(-u) \, (-du) = (-1)^n \int_0^{\infty} u^n f(-u) \, du = - \int_0^{\infty} u^n f(u) \, du = -A,$$ where the first step is legitimate because the integral converges and the third step follows because $n$ is odd and $f$ is symmetric about the origin.

Thus, if the $n$th odd-order moment exists, it is equal to $\int_{-\infty}^{\infty} x^n f(x) \,dx = -A + A = 0.$

The discrete case for $n$ odd is analogous.

  • $\begingroup$ The Wikipedia article referenced restricts itself to integrals over finite intervals $(-A, A)$. So please considering editing your answer a little to make it more clear that the $n$-th order moment, $n$ odd, exists exactly when $\int_0^{\infty} x^n f(x) dx$, $n$ odd, has finite value (or converges); and if the moment exists, it has value $0$. $\endgroup$ – Dilip Sarwate Oct 14 '11 at 0:19
  • $\begingroup$ @Dilip: I can certainly make my answer more precise on that count. $\endgroup$ – Mike Spivey Oct 14 '11 at 2:56
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    $\begingroup$ And what about the neither-absolutely-continuous-neither-discrete case? $\endgroup$ – Did Oct 14 '11 at 5:51
  • $\begingroup$ @Didier: I skipped that, didn't I? I suppose to handle all cases the answer would need to be rewritten in measure-theoretic terms. My rustiness with the measure-theory approach to probability, though, makes me concerned that I would make too many mistakes were I to try to revise my answer along those lines. So I think I'll leave my answer as is and acknowledge (via this comment) that I've skipped the case you mention. $\endgroup$ – Mike Spivey Oct 14 '11 at 16:13
  • $\begingroup$ @Mike Spivey: thanks for your answer:) if it's not symmmetric around 0, if e.g. i want to consider the odd-order central moments around a mean value m, with p(i) being a function symmetric around m=125 for i 0-256 (discrete case). does it still hold that these odd-order central moments are all equal to zero? $\endgroup$ – nikos Oct 15 '11 at 16:54

I think it's easier than arguments about integrals and sums. Saying a random variable $X$ has a symmetric distribution means that $X$ and $-X$ have the same distribution. This means that for any function $f$, if $E[f(X)]$ exists then so does $E[f(-X)]$ and they are equal. So take $f(x) = x^n$ with $n$ odd. If $X$ has an $n$th moment, then $E[X^n]$ exists and so $$E[X^n] = E[(-X)^n] = E[(-1)^n X^n] = E[-(X^n)] = -E[X^n]$$ which means $E[X^n] = 0$.

  • $\begingroup$ What does "If $X$ has a $n$th moment, then $E[X^n]$ exists" mean? I was taught that the $n$th moment of $X$ was defined as the value of $E[X^n]$ if $E[X^n]$ exists, and was undefined otherwise. The reason I requested @MikeSpivey to revise his answer was to confront this notion of existence of $E[X^n]$, which is obvious to those well-versed in the measure-theoretic approach to probability, but is confusing to beginning students as the OP nikos might well be. See also the discussion here. $\endgroup$ – Dilip Sarwate Oct 14 '11 at 13:17
  • $\begingroup$ @Nate Eldredge : Your solution is nice. But how to prove it for $E[X]\neq0$. Since distributions of $X$ and $-X$ are equal only for zero-mean distribution. $\endgroup$ – kaka Jan 6 '14 at 23:44
  • $\begingroup$ @kaka: the question is about symmetric distributions, so we necessarily have $E[X]=0$ if it exists. (Take $n=1$ in my answer!) $\endgroup$ – Nate Eldredge Jan 7 '14 at 3:26
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    $\begingroup$ @ Nate Eldredge: It is not necessary to have $E[X]=0$ for a symmetric distribution. $\endgroup$ – kaka Jan 7 '14 at 9:18
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    $\begingroup$ @kaka: we must not be using the same definition of "symmetric distribution". Mine is "X and -X are identically distributed". What is yours? $\endgroup$ – Nate Eldredge Jan 7 '14 at 14:59

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