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?
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I think you mean all odd-order central moments. Then you don't need the symmetric-about-$0$ assumption provided in the answers. – E2R0NS Oct 24 '24 at 17:42
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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.
Mike Spivey
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1The 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 $0$. – Dilip Sarwate Oct 14 '11 at 00:19
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@Dilip: I can certainly make my answer more precise on that count. – Mike Spivey Oct 14 '11 at 02:56
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@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. – Mike Spivey Oct 14 '11 at 16:13
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@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? – nikos Oct 15 '11 at 16:54
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@nikos: No. For instance, in your example $E[X^1] = m = 125$, not $0$. The higher-order odd moments will be more complicated, but in general they will not be equal to $0$ (or $m$, either, I believe). – Mike Spivey Oct 15 '11 at 17:04
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@Mike Spivey: I see...and if i have the following specific problem: let i be a discrete random variable and p(i), i=0,1,...,255 the corresponding histogram and it is given that: for i=0 to 20 the height of the histogram is $C/2$, for i=21 to 234 the height is $C$ and for i=235 to 255 the height is $C/2$. In that occasion is it true that all odd-order moments around the mean value m are all equal to zero. And if yes how can i prove it? – nikos Oct 16 '11 at 00:44
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@nikos: Yes; all odd-order moments around the mean value $m$ will be $0$. To prove it, do a variable transformation. If $X$ is your random variable, let $Y = X-m$. Then $Y$ is symmetric about $0$. Then, for $n$ odd, $E[(X-m)^n] = E[Y^n] = 0$, by the argument in my answer above. (And my apologies for misunderstanding your first comment; you did say "central moment," and I missed it.) – Mike Spivey Oct 16 '11 at 03:46
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I'm assuming here that we mean "symmetric about 0"; if the distribution is symmetric about some other value then this is obviously false. (Unless we are considering central moments, in which case we can assume without loss of generality that the center is 0.)
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 $E[X^n] = 0$.
Nate Eldredge
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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. – Dilip Sarwate Oct 14 '11 at 13:17
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@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. – kaka Jan 06 '14 at 23:44
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@kaka: the question is about symmetric distributions, so we necessarily have $E[X]=0$ if it exists. (Take $n=1$ in my answer!) – Nate Eldredge Jan 07 '14 at 03:26
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1@ Nate Eldredge: It is not necessary to have $E[X]=0$ for a symmetric distribution. – kaka Jan 07 '14 at 09:18
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2@kaka: we must not be using the same definition of "symmetric distribution". Mine is "X and -X are identically distributed". What is yours? – Nate Eldredge Jan 07 '14 at 14:59