I'm completely new to Baayesian statistics and am having some questioning about the following:

Given a probability distribution defined by the pdf

$\pi(x|\theta) = C(\theta)x^2 e^{-\theta x^{3}}$ for $x \in [0, \inf)$ and for all $\theta > 0$, determine the function $C(\theta)$

I also know that

$\pi(x|\theta) \propto x^2 e^{-\theta x^3}$

Now this is in a bayesian setting and I was considering Bayes formula

$\pi(\theta|x) = \frac{\pi(\theta)\pi(x|\theta)}{\pi(x)}$ and rewriting it as $\pi(x|\theta) = \frac{\pi(x)\pi(\theta|x)}{\pi(\theta)}$

Where $\pi(\theta)$ would then be the prior, $\pi(x|\theta)$ the "likelihood" and $\pi(\theta|x)$ the posterior. So as I understand it, finding $C(\theta)$ would be similar to finding the normalizing constant in Bayes formula seeing as $C(\theta) = 1/\pi(\theta)$ and $x^2 e^{-\theta x^3} = \pi(x)\pi(\theta|x)$, but following the same procedure with finding the marginal density would mean to find the integral

$\int_{0}^{\infty} x^2 e^{-\theta x^3} dx = \pi(\theta)$

But I don't know if it's even possible to find a closed form of this integral so it feels like I am on the wrong track?

  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – Silverfish Sep 10 '16 at 19:48
  • $\begingroup$ For finding closed form integrals you can always (well, not in closed-books exams...) try a computer algebra system with symbolic integration. WolframAlpha is available online, try wolframalpha.com/input/… $\endgroup$ – Silverfish Sep 10 '16 at 19:50

You just need to solve the integral $$ \int_{0}^{\infty}\pi(x|\theta)dx=\int_{0}^{\infty}C(\theta)x^2e^{-\theta x^3} dx=1 , \forall \theta > 0 $$

$$=C(\theta) \int_{0}^{\infty}x^2e^{-\theta x^3} dx=1 $$ Letting $-\theta x^3 =u$ then $\frac{\mathrm{d} u}{\mathrm{d} x} = -3\theta x^2$, this implies $$C(\theta) \int_{0}^{-\infty}x^2e^{u} \frac{du}{-3\theta x^2}=1$$ $$C(\theta)\frac{1}{3\theta }\int_{-\infty}^{0}e^{u} du=1$$ $$C(\theta)=3\theta$$

  • $\begingroup$ Ah, of course... Thank you! Need to do some repetition in calculus I notice. $\endgroup$ – Knut Sep 10 '16 at 21:01

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