I've just started studying probability, so I beg your pardon if my question is silly.
Consider a pdf $\large p_{f}(x)$, for a continuous random variable, that represents the distribution of images of all possible human faces. Now, this distribution is made up of many types of unique features, for instance faces with glasses, women, men, beard, no beard, and so on... For simplicity, let's say I'm only interested in "beard" with its own pdf $\large p_{b}$ and "no beard" with its own pdf $\large p_{nb}$.
Now, no human face can exist out of these two distributions, this implies the pdf $\large p_{f}(x)$ can be written only with $\large p_{b}(x)$ and $\large p_{nb}(x)$.
So how can I write $\large p_{f}(x)$ in terms of $\large p_{b}(x)$ and $\large p_{nb}(x)$ only?
If I'm talking about a discrete RV, then ofc it's simply the addition of pmf of these "beard" and "no beard" distributions. Which is not the case. So how can I add these two pdfs?
Thanks for your time.
I asked the same question in multiple ways (here, here and here) so that someone would get my doubt. And finally, thanks to jlammy who provided the perfect answer to the question.
What I was looking for, was the mixture distribution, here its explained really really well. The combination of density functions can be written as their weighted sum. Where the weights are their corresponding proportions in the population.
So in my above case, the $\large p_{f}(x)$ can simply be written as $\large p * p_{b}(x) + (1 - p) * p_{nb}$, where $\large p$ denotes the proportion of bearded humans, and $\large (1 - p)$ denotes the proportion of non bearded humans.