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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?

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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 $Me\approx 1.678$

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