I am stuck at solving the following problem (at what I believe is the last step):
Determine which distributions on the non-negative reals, if any, with mean $\mu$ are such that $2\mu$ is a median.
My thoughts so far:
Let's call the distribution in question $X$. Given that $X$ takes values in the non-negative reals, we can apply the Markov inequality $\left(P(X\geq t)\leq\frac{E[X]}{t}, \mbox{ for } t\geq 0\right)$. Taking $t = \mu$ in this inequality we get that $$ P(X\geq 2\mu)\leq \frac{\mu}{2\mu} = \frac{1}{2} $$
However, given that $2\mu$ is the median, we have that $P(X\geq 2\mu)\geq \frac{1}{2}$. Hence we get that $P(X\geq 2\mu) = \frac{1}{2}$. Therefore we have equality in the Markov inequality. Hence $X\in\left\{ 0,2\mu\right\}$.
- How do I determine where $X$ takes the value $0$ and where $X$ takes the value $2\mu$ (if there is any such way)?
- From 1, how do I get the pdf of $X$?
Let $p:=P(X=2\mu)$ so that $1-p=P(X=0)$.
Then $\mu=(1-p)\cdot0+p\cdot2\mu=2\mu p$ implying that $p=\frac12$.
Btw, from $X\in\{0,2\mu\}$ a.s. it follows immediately that $X$ is discrete, hence has a PMF and not a PDF.