Median from probability density function












2












$begingroup$


I have a probability density function:



$f(x) = begin{cases} frac 1 4xe^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



To add more detail, cdf is:



$F(x) = begin{cases} 1+frac{-1}2xe^{frac {-x}2}-e^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



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 $Med(X)ge 0$.






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    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






  • 2




    $begingroup$
    Only the non-negative solution should be median since you used that part of the cdf for which $xge 0$; $F(x)$ cannot be $1/2$ for any $x<0$.
    $endgroup$
    – StubbornAtom
    Dec 26 '18 at 13:46


















2












$begingroup$


I have a probability density function:



$f(x) = begin{cases} frac 1 4xe^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



To add more detail, cdf is:



$F(x) = begin{cases} 1+frac{-1}2xe^{frac {-x}2}-e^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



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 $Med(X)ge 0$.






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    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






  • 2




    $begingroup$
    Only the non-negative solution should be median since you used that part of the cdf for which $xge 0$; $F(x)$ cannot be $1/2$ for any $x<0$.
    $endgroup$
    – StubbornAtom
    Dec 26 '18 at 13:46
















2












2








2





$begingroup$


I have a probability density function:



$f(x) = begin{cases} frac 1 4xe^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



To add more detail, cdf is:



$F(x) = begin{cases} 1+frac{-1}2xe^{frac {-x}2}-e^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



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 $Med(X)ge 0$.






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$




I have a probability density function:



$f(x) = begin{cases} frac 1 4xe^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



To add more detail, cdf is:



$F(x) = begin{cases} 1+frac{-1}2xe^{frac {-x}2}-e^{frac {-x}2}, & xge0 \ 0, & x < 0end{cases}$



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 $Med(X)ge 0$.






probability probability-theory probability-distributions




probability probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 13:38







Tjh Thon

















asked Dec 26 '18 at 13:14









Tjh ThonTjh Thon

112




112












  • $begingroup$
    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






  • 2




    $begingroup$
    Only the non-negative solution should be median since you used that part of the cdf for which $xge 0$; $F(x)$ cannot be $1/2$ for any $x<0$.
    $endgroup$
    – StubbornAtom
    Dec 26 '18 at 13:46




















  • $begingroup$
    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






  • 2




    $begingroup$
    Only the non-negative solution should be median since you used that part of the cdf for which $xge 0$; $F(x)$ cannot be $1/2$ for any $x<0$.
    $endgroup$
    – StubbornAtom
    Dec 26 '18 at 13:46


















$begingroup$
The median is unique. So you can’t find two values.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:16




$begingroup$
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




$begingroup$
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




$begingroup$
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




$begingroup$
I have edited my post.
$endgroup$
– Tjh Thon
Dec 26 '18 at 13:24




2




2




$begingroup$
Only the non-negative solution should be median since you used that part of the cdf for which $xge 0$; $F(x)$ cannot be $1/2$ for any $x<0$.
$endgroup$
– StubbornAtom
Dec 26 '18 at 13:46






$begingroup$
Only the non-negative solution should be median since you used that part of the cdf for which $xge 0$; $F(x)$ cannot be $1/2$ for any $x<0$.
$endgroup$
– StubbornAtom
Dec 26 '18 at 13:46












1 Answer
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oldest

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$begingroup$

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






share|cite|improve this answer









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    1 Answer
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    0












    $begingroup$

    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






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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






        share|cite|improve this answer









        $endgroup$



        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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 8:52









        BruceETBruceET

        35.7k71440




        35.7k71440






























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