# What does $N(x|\mu, \sigma^2)$ mean?

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.

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

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