1
$\begingroup$

The kernel of the normal distribution is typically written as : $ p(\theta | \mu, \sigma^2) \propto \exp\Big(- \dfrac{(\theta-\mu)^2}{2\sigma^2} \Big)$ where the normalizing constant is: $C = \dfrac{1}{\sqrt{2\pi}\sigma}$.

I have some lecture notes where this is simplified to:

$ p(\theta | \mu, \sigma^2) = \dfrac{1}{\sqrt{2\pi}\sigma} \exp\Big(- \dfrac{(\theta-\mu)^2}{2\sigma^2} \Big)$

$ p(\theta | \mu, \sigma^2) = \dfrac{1}{\sqrt{2\pi}\sigma} \exp\Big(- \dfrac{\theta^2-2\mu\theta + \mu^2}{2\sigma^2} \Big)$

$ p(\theta | \mu, \sigma^2) = \dfrac{1}{\sqrt{2\pi}\sigma} \exp\Big( \dfrac{-\mu^2}{2\sigma^2} \Big) \exp\Big(- \dfrac{\theta^2 - 2\mu \theta}{2\sigma^2} \Big)$

So the kernel simply becomes

$ p(\theta | \mu, \sigma^2) \propto \exp\Big(- \dfrac{\theta^2 - 2\mu \theta}{2\sigma^2} \Big)$ with normalizing constant: $C= \dfrac{1}{\sqrt{2\pi}\sigma} \exp \Big( \dfrac{-\mu^2}{2\sigma^2} \Big)$

I'm confused about the part where the exp is splitten in to two. How is: $ \exp \Big( \dfrac{ab}{c} \Big) = \exp \Big( \dfrac{a}{c} \Big) \exp \Big( \dfrac{b}{c} \Big) $ ??

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ It's not. In your block of three equations, look at line 2 carefully and compare it to your question. Write out what $a$ is and what $b$ is. Is $ab/c$ really what's in line 2? $\endgroup$
    – jbowman
    Feb 2, 2018 at 16:32
  • 2
    $\begingroup$ The relevant property is $\exp(a + b) = \exp(a)\exp(b)$ $\endgroup$
    – Matthew Gunn
    Feb 2, 2018 at 16:51
  • $\begingroup$ Ah.. I think I get it. So exp((a+b)/c)) = exp(a/c)*exp(b/c) $\endgroup$
    – Said Hassan
    Feb 2, 2018 at 16:54

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.