0
$\begingroup$

I'm working through an example which can be found here (p. 36), if someone is interested.

I have an integral of the form: $$P(x|\mu)=\int d\sigma P(x|\mu, \sigma)P(\sigma)=\int d\sigma \frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{(x-\mu)^2}{2\sigma^2})P(\sigma)$$

where $P(x|\mu, \sigma)$ is a likelihood and $P(\sigma)$ a prior distribution.

Now $\beta=\frac{1}{\sigma^2}$ is defined and the distribution over $\beta$ is defined to be a Gamma distribution of the form: $$f(\beta) = \frac{a^v}{\Gamma(v)} \beta^{v-1} e^{-a\beta}$$

When substituting $\sigma$ for $\beta$ the following expression is obtained, which I am struggling to derive myself: $$\int d\beta \frac{a^v}{\sqrt{2\pi}\Gamma(v)} \exp(-\beta\frac{(x-\mu)^2}{2}) \beta^{v-\frac{1}{2}} e^{-a\beta}$$

I tried converting $P(\sigma)$ to $P(\beta)$ using the change of variable technique and used u-substitution to change the variable of integration to no avail. Can someone lead me through the individual steps?

$\endgroup$
  • $\begingroup$ At which page is the example? $\endgroup$ – callculus Mar 4 at 16:00
  • $\begingroup$ @callculus p.36 (edited the question) $\endgroup$ – Dahlai Mar 4 at 16:02
2
$\begingroup$

Since $\sigma = \beta^{-1/2}$, we have $d\sigma = -\frac{1}{2} \beta^{-3/2} \, d\beta$ and $$f_\sigma (\sigma) = f_\beta (\sigma^{-2}) \cdot 2 \sigma^{-3} = \frac{a^v}{\Gamma(v)} (\sigma^{-2})^{v-1} e^{-a \sigma^{-2}} (2\sigma^{-3});$$ it follows that $$\begin{align*} \int_{\sigma = 0}^\infty \frac{1}{\sqrt{2\pi} \sigma} e^{-(x-\mu)^2/(2\sigma^2)} f_\sigma(\sigma) \, d\sigma &= \int_{\beta = \infty}^0 \frac{a^v}{\Gamma(v)} \frac{\beta^{1/2}}{\sqrt{2\pi}} e^{-(x-\mu)^2 \beta/2} \beta^{v-1} e^{-a \beta} 2\beta^{3/2} \cdot \frac{1}{2} \beta^{-3/2} \, d\beta \\ &= \frac{a^v}{\Gamma(v) \sqrt{2\pi}}\int_{\beta=0}^\infty e^{-((x-\mu)^2/2 + a)\beta} \beta^{v-1/2} \, d\beta \end{align*}$$

as claimed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.