I'd like to show counterexamples for:
a. $med(X+Y) = med(X) + med(Y)$
b. $med(aX+b) = a\cdot med(X) + b$.
- Showing that "a" is not correct:
Let $X, Y$ be independent random variables with exponential distribution and correspondeing: $F(t) = 1-e^{-t}$ (for $t \ge 0$).
To find the median I solve $F(t)=1-e^{-t}=1/2$ and then I see $med(X)+med(Y)=2log(2)$.
Now I'd like to take a look at $F_{X+Y}(t)$ - how can I get this function?
- Showing that "b" is not correct:
For showing this I need $F_{aX}(t)$ and $F_{aX+b}(t)$, correct? Well then I should await some answers to "1.", since I am stuck there.
Part (a)
Your calculation of the sum of the medians of $X$ and $Y$ is correct and gives $2\ln2\approx1.3863$.
As you say the next step is to get the distribution for $X+Y$. The prob that $X+Y\le t$ is $\int_{r=0}^te^{-t}\int_{s=0}^{t-r}e^{-s}\ ds\ dr=\int_0^te^{-r}(1-e^{r-t}\ dr=1-e^{-t}-te^{-t}$.
You can either solve $1-e^{-t}-te^{-t}=\frac{1}{2}$ numerically to find $t=1.6784$, or you can note that the function is increasing and less than $\frac{1}{2}$ at $t=2\ln2$.
Part (b)
Hint: why would you expect this to be incorrect? You have $X\le m$ iff $aX+b\le am+b$.