Consider the cdf $F(x)=1-e^{-x}-xe^{-x}, 0\leq x <\infty$, zero elsewhere. Find the median of this distribution,
The given CDF is a complicated and I am finding it difficult to find x for which P(X<x)=0.5. Is there any other approach I can try?
1
-
$\begingroup$ This is the CDF of a certain Gamma distribution. See en.wikipedia.org/wiki/Gamma_distribution#Median_calculation. $\endgroup$– StubbornAtomJun 16, 2020 at 19:16
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
As commented, this is a CDF of a $Gamma(2;1)$
To understad this it is enough to derive F obtainina
$f_X(x)=xe^{-x}$
that is $f_X(x)=\frac{\theta^n}{\Gamma(n)}x^{n-1}e^{-\theta x}$
with $\theta=1$ and $n=2$
As you can see in the link already posted, the median has not a closed form, so an alternative method with respect to the one described in the link to calculate the median, is a numerical one...it is not difficult, you can use, for example, the bisection method finding immediately $Me\approx 1.678$
