C3_W1_Assignment - question on exercise 6

Could anyone help on the formula of exercise 6? I have a doubt on 𝐏(π‘₯π‘˜βˆ£πΆπ‘–)=PDFgaussian(π‘₯π‘˜,πœ‡πΆπ‘–,πœŽπΆπ‘–). Why P is PDF vs. PMF in this case? I don’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.

Hi @Francis_Ouyang!

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€œlikelihood” is connected with the PDF.

Please, tell me if this is clear.


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