Moments of Z (about 0) are
$E(Z^n)=\int\limits_0^{\pi/2}\frac{2}{\pi}sin^{2n}\theta d\theta=\prod\limits_{j=0}^{n-1}\frac{2j+1}{2j+2}$
Would any stats expert explain me how the R.H.S is arrived at?
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asked Apr 11, 2016 at 16:36
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$\begingroup$ Substituting $x=\sin^2(\theta)$ gives $(1/\pi)\int_0^1 x^{n-1/2}(1-x)^{-1/2}dx$. The integral is the normalizing factor for a Beta distribution and therefore equals $B(n+1/2, 1/2).$ This holds true even when $n$ is non-integral, thereby generalizing the result. $\endgroup$– whuber ♦Commented Apr 11, 2016 at 19:16
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1 Answer
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Your question is more a math question than a stats question. Define for positive integers $m$,
$$\begin{align*} I(m) &= \int \sin^m \theta \, d\theta \\ &= \int \sin^{m-1} \theta \sin \theta \, d\theta \\ &= -\sin^{m-1} \theta \cos \theta + (m-1) \int \cos^2 \theta \sin^{m-2} \theta \, d\theta \\ &= -\sin^{m-1} \theta \cos \theta + (m-1)(I(m-2) - I(m)). \end{align*}$$ Consequently, $$m I(m) = -\sin^{m-1} \theta \cos \theta + (m-1)I(m-2),$$ which is the familiar reduction formula for powers of sine. It follows that $$\begin{align*} J(2n) &= \int_{\theta = 0}^{\pi/2} \sin^{2n} \theta \, d\theta \\ &= \frac{1}{2n} \left[-\sin^{2n-1} \theta \cos \theta \right]_{\theta = 0}^{\pi/2} + \frac{2n-1}{2n} J(2n-2) \\ &= \frac{2n-1}{2n} J(2n-2), \end{align*}$$ and since $J(0) = \pi/2$ trivially, we immediately have $$\frac{2}{\pi} J(2n) = \prod_{k=1}^n \frac{2k-1}{2k},$$ which is equivalent to your product.
