Median from probability density function

Question

I have a probability density function:

$f(x) = \begin{cases} \frac 1 4xe^{\frac {-x}2}, & x\ge0 \\ 0, & x < 0\end{cases}$

To add more detail, cdf is:

$F(x) = \begin{cases} 1+\frac{-1}2xe^{\frac {-x}2}-e^{\frac {-x}2}, & x\ge0 \\ 0, & x < 0\end{cases}$

Find the $Med(x)$.

Solving

$$\int_{-\infty}^{x} f(x) dx=1+\frac{-1}2xe^{\frac {-x}2}-e^{\frac {-x}2}=\frac 12$$

I found 2 values $3.3567$ and $-1.5361$, and my book said that the answer is $-1.5361$. This is confusing because I thought $Med(X)\ge 0$.

Answer 1

Your PDF is for the distribution $\mathsf{Gamma}(\text{shape}=2,\text{rate}=\frac 1 2).$ See the relevant Wikipedia page (or your textbook) for details. You seek $F_X^{-1}(\frac 1 2) = 3.356694.$

In R statistical software, the inverse CDF (quantile function) is denoted qgamma with appropriate arguments.

qgamma(.5,2,1/2)
[1] 3.356694

I agree with the comment of @StubbornAtom, that the median cannot be negative. The "answer" −1.5361 is simply wrong.

Here is a graph of the PDF, with the location of the median shown by a dotted red line.