Symmetrical distribution This is written in my book.
Since f(t) is symmetrical about the line t=0, all the moments of odd order about origin vanish
but can't find the explanation for it.
Please explain this
 A: If some random variable $X$ has a symmetric distribution, this means that $X$ and $-X$ have the same distribution.  This implies for any function $f$,  $E[f(X)] = E[f(-X)]$ if $E[f(X)]$ exists.  We can represent odd order moments by letting  $f(x) = x^{n}$ with $n$ being an odd integer.  If $X$ has an $n$-th moment it follows that $E[X^{n}]$ exists, thus
\begin{align*}
E[X^{n}] &= E[(-X)^{n}] \\ &= E[(-1)^{n} X^{n}] \\ &= E[-(X^{n})] \\ &= -E[X^{n}] \\
\implies & E[X^n] = 0
\end{align*}
hence all the moments of odd order for a symmetric distribution are $0$.
A: The symmetry mentioned means that $f(-t) = f(t)$ for all $t$. Then odd moments would involve summing (integrating) $t^nf(t)$ for odd values of $n$ (1,3,5, etc) over an interval centered at $t=0$. But that is an odd function of $t$, so the values at negative values of $t$ would be exactly the opposite of the values at the corresponding positive values of $t$, so the sum (integral) would be zero.
In other words, $$\int_{-\infty}^{\infty}t^nf(t)\;dt=0$$ for $n=1,3,5,\ldots$
