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It is given that X has a pdf of

$f_X(x)$ = $\frac{sin(\pi r)}{\pi x}$$(x - \theta)^{-r}$$\theta^r$ for $x>\theta$, and

$f_X(x)$ = $0$ for $0

Question: Show that $f_X(x)$ is a pdf.

I am familiar with the properties that define a pdf, and however I am confused with how to go about showing the part of the definition that states: $\int f_X(x)$ $dx=1$

My attempt:

$\int \frac{sin(\pi r)}{\pi x}$$(x - \theta)^{-r}$$\theta^r$ $dx$

= $\frac{sin(\pi r)}{\pi}$ $\int $$(x - \theta)^{-r}$$\theta^r$$x^{-1}$ $dx$

= $\frac{sin(\pi r)}{\pi}$ $\int \left(\frac{\theta}{x - \theta}\right)^r$$x^{-1}$ $dx$

This is the part where I keep getting confused.

I know that I need to use the Gamma function identity: $\Gamma(a)\Gamma(1-a)$ = $\frac{\pi}{sin(\pi a)}$.

As well as, an integral representation of the Beta function: $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ (which I'm torn between $B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1} dt$ or $B(a,b)=\int_0^\infty \frac{t^{a-1}}{(t+1)^{a+b}} dt$)

I'd really appreciate some guidance on how to apply the representation of the Beta function to the function within the integral.

1 Answers 1

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We have \begin{align} \int_\theta^\infty f_X(x)dx &\ = \frac{\sin(\pi r)}{\pi}\int_\theta^\infty \frac{1}{x}(x - \theta)^{-r}\theta^r dx\\ &\ = \frac{1}{\Gamma(r)\Gamma(1-r)}\int_1^\infty \frac{1}{y} (y - 1)^{-r} d y \tag{$x = \theta y$}\\ &\ = \frac{\Gamma(1)}{\Gamma(r)\Gamma(1-r)} \int_0^1 (1 - z)^{-r} z^{r-1} dz \tag{$y = \frac{1}{z}$}\\ &\ = \frac{1}{B(r, 1-r)} \cdot B(r, 1 - r) = 1 \end{align}