I am working on this problem:
Suppose that an observed value of X is equally likely to come from a continuous distribution for which the pdf is f or from one for which the pdf is g. Suppose that f(x) > 0 for 0 < x < 1 and f(x) = 0 otherwise, and suppose also that g(x) > 0 for 3 < x < 6 and g(x) = 0 otherwise. Determine (a) the mean and (b) the median of the distribution of X.
For the median, my intuition would be to just call it 0.5*F(0.5) + 0.5*G(0.5), treating it as a weighted probablity. But theoretically it's not clicking for me how to do that with the expectation/why that would work. Would I treat it as a conditional expectation? My thought would be to integrate x*f(x) from 0 to 1 and x*g(x) from 3 to 6, multiplying both antiderivatives by 0.5, but I don't know how to take it farther than that without actual equations. Can someone give me a thorough explanation of what the mechanics behind this problem are? Thanks!
You could model $X$ as $X = S \Phi + (1-S) \Gamma$, where $S$ takes the values $0,1$ with equal probability, $\Phi, \Gamma$ have pdfs $f,g$ respectively, and $S,\Phi,\Gamma$ are independent.
Then $E X = ES E \Phi + E (1-S) E \Gamma$, hence $EX = {1 \over 2} (\int x (f(x)+ g(x)) dx )$, that is, the average of the expectations of $\Phi,\Gamma$.
The median is any number $c$ such that $P[X \le c] = {1 \over 2}$, hence in this case we see that $P[X \le c] = {1 \over 2}$ for any $c \in [1,3]$.