# C3_W1_Assignment - question on exercise 6

Could anyone help on the formula of exercise 6? I have a doubt on π(π₯πβ£πΆπ)=PDFgaussian(π₯π,ππΆπ,ππΆπ). Why P is PDF vs. PMF in this case? I donβt understand. From the course videos, what I understand is PDF could not be used to represent probability directly but we have to use PMF.

PDF are usually related to continuous random variables whereas PMF is to discrete random variable. In this case, the gaussian distribution is a continuous variable, therefore the name PDF makes sense. Note that it is also used PDF for discrete case in the notebook - this is just to make the notation simpler.

I think, though, that your concern is due to the fact that in a continuous random variable, the probability of a single value being observed is zero, but we can still ask how likely one value is to be observed. Take a Normal Distribution with mean 0 and standard deviation 1.

The probability that you observe the value 0 (note here we are talking about the value exactly zero) is 0, as the probability of you observe the value -600 is also 0. However, it is much more likely that you will observe (a value near of) zero than -600 in the case mentioned. This βlikelihoodβ is connected with the PDF.

Please, tell me if this is clear.

Cheers,
Lucas

Thank you so much Lucas. I got your points and understood now.