Median from probability density function
$begingroup$
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
asked Dec 26 '18 at 13:14
Tjh ThonTjh Thon
112
$endgroup$
|
show 2 more comments
$begingroup$
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
asked Dec 26 '18 at 13:14
Tjh ThonTjh Thon
112
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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
asked Dec 26 '18 at 13:14
Tjh ThonTjh Thon
112
$endgroup$
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
asked Dec 26 '18 at 13:14
Tjh ThonTjh Thon
112
asked Dec 26 '18 at 13:14
Tjh ThonTjh Thon
112
edited Dec 26 '18 at 13:38
Tjh Thon
asked Dec 26 '18 at 13:14
Tjh ThonTjh Thon
112
asked Dec 26 '18 at 13:14
Tjh ThonTjh Thon
112
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
$endgroup$
add a comment |
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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.
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
$endgroup$
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.
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
answered Dec 27 '18 at 8:52
BruceETBruceET
35.6k71440
35.6k71440
add a comment |
add a comment |
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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
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