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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 $Med(X)\ge 0$.

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  • $\begingroup$ The median is unique. So you can’t find two values. $\endgroup$ – mathcounterexamples.net Dec 26 '18 at 13:16
  • $\begingroup$ I solve the equation and it gives me 2 $x$ values, I haven't determined which $Med(X)$ is yet. $\endgroup$ – Tjh Thon Dec 26 '18 at 13:17
  • $\begingroup$ Which equation did you solve? Please update the question with what you did. $\endgroup$ – mathcounterexamples.net Dec 26 '18 at 13:18
  • $\begingroup$ I have edited my post. $\endgroup$ – Tjh Thon Dec 26 '18 at 13:24
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    $\begingroup$ Only the non-negative solution should be median since you used that part of the cdf for which $x\ge 0$; $F(x)$ cannot be $1/2$ for any $x<0$. $\endgroup$ – StubbornAtom Dec 26 '18 at 13:46
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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.

enter image description here

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