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Let $f(x) = k(\sin x)^5(1-\sin x)^7$ if $0 \lt x \lt \pi/2$ and $0$ otherwise. Find the value of $k$ that makes $f(x)$ a density function.

I'm struggling to understand how this relates to the Beta distribution and how to proceed with this question.

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  • $\begingroup$ I'm no probabilist or statistician, here is my take, PDF of beta distribution is $\frac{x^{\alpha-1}(1-x)^{\beta-1}} {B(\alpha,\beta)}$. $x$ in your problem is $sin(x)$ and $\alpha = 6, \beta=8$. So inorder for $f(x)$ to be a PDF, $K$ should be $B(6,8)$ where $B(\alpha,\beta)$ is a beta function. $\endgroup$
    – forecaster
    Oct 12, 2018 at 18:08
  • $\begingroup$ is this a howeowrk, if so please tag it as self study in the future. $\endgroup$
    – forecaster
    Oct 12, 2018 at 18:27
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    $\begingroup$ @forecaster - you've left out the change of variable from the $x$ of the initial problem to the $x$ of the beta distribution in your formulation; the integral isn't with respect to $\sin x$ but with respect to $x$. $\endgroup$
    – jbowman
    Oct 12, 2018 at 18:27
  • $\begingroup$ good point @jbowman. Clearly I missed it. $\endgroup$
    – forecaster
    Oct 12, 2018 at 18:28
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    $\begingroup$ @Forecaster That's a good approximation to the exact value, $\frac{1013760}{27312128 - 8693685 \pi }.$ $\endgroup$
    – whuber
    Oct 12, 2018 at 22:00

1 Answer 1

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The integral of $f$ can be expressed as a Beta function times a hypergeometric function. This suggests $f$ is not the density of any particular Beta distribution, but that it is indeed related.

To evaluate $k$ it's simpler and more elementary to use the substitution $y = \sin^2(x),$ $\mathrm{d}y = 2\sin(x)\cos(x)\mathrm{d}x$ and observe $\cos(x) = (1-\sin^2(x))^{1/2}$ to evaluate integrals of the form

$$\int_0^{\pi/2} \sin^m(x)\cos^n(x)\mathrm{d}x = \frac{1}{2}\int_0^1 y^{(m-1)/2} (1-y)^{(n-1)/2}\mathrm{d}y = \frac{1}{2}B\left(\frac{m+1}{2}, \frac{n+1}{2}\right)$$

and then apply the Binomial Theorem to expand $(1-\sin(x))^7,$ producing

$$\eqalign{ \int_0^{\pi/2} f(x)\mathrm{d}x &= k \int_0^{\pi/2} \sin^5(x)(1-\sin(x))^7 \mathrm{d}x \\ &=k \sum_{j=0}^7 (-1)^j \binom{7}{j}\int_0^{\pi/2}\sin^{5+j}(x)\mathrm{d}x \\ &= \frac{k}{2} \sum_{j=0}^7 (-1)^j \binom{7}{j} B\left(\frac{5+j+1}{2}, \frac{1}{2}\right). }$$

The alternating binomial coefficients $\binom{7}{j}$ in this sum are $$\binom{7}{j} = (1,-7,21,-35,35,-21,7,-1)$$ while the Beta function values are $$B\left(\frac{5+j+1}{2}, \frac{1}{2}\right) = \left(\frac{16}{15},\frac{5 \pi }{16},\frac{32}{35},\frac{35 \pi }{128},\frac{256}{315},\frac{63 \pi }{256},\frac{512}{693},\frac{231 \pi }{1024}\right),$$

giving

$$1 = \int_0^{\pi/2} f(x)\mathrm{d}x = k \color{blue}{\frac{1}{2}\left(\frac{26672}{495} -\frac{17563}{1024}\pi\right)}.$$

Thus, $k$ is the reciprocal of the blue quantity.

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