What is the product of a PDF and a probability? The question is relatively simple. 
I have a PDF f(x) in hand, which represent the probability density for two network nodes has distance x. Also, I have a probability r(x), which has the physical meaning that, the probability for an observed communication pair, which has inter-distance x, can successfully exchange their information. 
In my derivation process, I met the product of f(x) and r(x), and I am wondering that, if the knowledge of f(x) and r(x) is enough, can we obtain the product's mean and variance? and also, what this the physical meaning of f(x) \times r(x)?
Thanks.
 A: Consider the probability that two network nodes can successfully exchange their information:   $$\Pr(\text{success})= E[\mathbb{1}_\text{success}] = \int_x r(x) \,f(x) \,dx.$$ 
Since the indicator random variable $\mathbb{1}_\text{success}$ is a Bernoulli random variable, you could say $$Var[\mathbb{1}_\text{success}] =\Pr(\text{success})(1-\Pr(\text{success})) = E[\mathbb{1}_\text{success}]-E[\mathbb{1}_\text{success}]^2 $$ $$= \int_x r(x) \,f(x) \,dx-\left(\int_x r(x) \,f(x) \,dx\right)^2.$$   
A: Henry has given the algebraic interpretation that I would as well.  I'm just following on with my own intuitive understanding of the expression.
I don't know that there's a generic meaning for $f(x)r(x)$ when $f(x)$ is a PDF and $r(x)$ is a probability that depends on $x$.  If you integrate it over the support of $f(x)$, then you get, in some sense, the expected value of $r(x)$, but the product on its own is really only an infinitesimal slice of that integration.
For my own part, I find it easier to conceptualize first in the discrete case.  Suppose there are only two distances, $x = 1$ and $x = 2$, with probability $p_1 = 1/4$ and $p_2 = 3/4$, respectively.  We also have $r(1) = 4/5$ and $r(2) = 2/5$.  Then we would say that the probability of successful communication is given by
$$
P = p_1r(1)+p_2r(2) = \frac{1}{4}\cdot\frac{4}{5}+\frac{3}{4}\cdot\frac{2}{5}
  = \frac{1}{2}
$$
In other words, the expected value of $r(x)$ (in the context of the distribution of distance) is $1/2$.  But just the product $(1/4)(4/5)$ has no straightforward interpretation that I can think of, other than as part of this expectation.
A: If we multiply the pdf with a probability, it means that the pdf is scaled. However, the scaled pdf is no longer a valid pdf unless we normalize it. 
