Let X be a continuous random variable with probability density function
$$f_X(x)=\begin{cases}
(2+x), & \text{$-2<x<-1$} \\
(2-x), & \text{$1<x<2$} \\
0, & \text{elsewhere}
\end{cases}$$
Find $m=\pi_{0.5}$ of $X$. Is it unique?
$$\int_{-2}^{-1} (2+x) dx+\int_{1}^{2} (2-x) dx$$
we get $2x+2x=4x$.
Then set $f_X(x)=4x=\frac{1}{2}$
$4*(\pi_{0.5})=\frac{1}{2}$ get $\frac{1}{8}$.
Is it right? And for uniqueness, what does it want? Am I supposed to graph and tell it?
Best Answer
By definition, a median of $X$ is any real number $m$ which satisfies $$ \mathbb P(X\leqslant m) \geqslant\frac12,\quad \mathbb P(X\geqslant m)\geqslant\frac12. $$ It is clear that $\mathbb P(X\leqslant -1) = \frac12$, and since $f(x)=0$ on $(-1,1)$, we also have $\mathbb P(X\leqslant x) = \frac12$ for $-1\leqslant x\leqslant 1$. By symmetry, we have $\mathbb P(X\geqslant x)=\frac12$ for $-1\leqslant x\leqslant 1$. So the medians of $X$ are given by the closed interval $[-1,1]$.