4
$\begingroup$

I am supposed to show that $f(x) = \sum_{k=1}^{K}\pi_k N(x|\mu_k, \sigma_{k}^2)$ complies with the properties of a density function but I have no idea how to do this since I am not sure what $N(x|\mu_k, \sigma_{k}^2)$ means.

I know $X \sim N(\mu, \sigma^2)$ means that the random variable X follows a normal distribution with mean $\mu$ and variance $\sigma^2$. I'm just not sure how $x|\mu_k$ changes things.

This is probably a very silly question but your help will be appreciated.

$\endgroup$
  • $\begingroup$ It's worth nothing that this is not standard notation, so feel vindicated in your confusion. $\endgroup$ – Cliff AB Feb 19 at 3:30
2
$\begingroup$

Here, it means the normal PDF: $$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$

The $\mu,\sigma^2$ in given side means that you can treat them as known quantities.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ooh ok I see. Thought there was something more to it. Thank you! $\endgroup$ – Susan-l3p Feb 18 at 20:28
3
$\begingroup$

$N(x|\mu, \sigma)$ combines the two notations: $x \sim N(\mu, \sigma)$ and $p(x| \mu, \sigma)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma$.

| cite | improve this answer | |
$\endgroup$

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.