 Solution

Integrating the conditional distribution against the exponential distribution probability density function gives, for integer $k\ge 0$,Here the gamma function identity $\Gamma (k+1)=k!$ evaluates the integral. Henceon $\{0,1,2,\dots \}$. Put $r=1/(1+\lambda )$ and $\rho =1-r$. The geometric series gives $F(k)=1-r^{k+1}$. Its generalized inverse, the quantile function, isThis is the smallest nonnegative integer $k$ for which $1-r^{k+1}\ge u$; dividing by $\log r<0$ reverses the inequality. Draw $U$ from the uniform distribution on $(0,1)$ and use this formula. The commonly used expression $\lfloor \log (1-U)/\log r\rfloor$ has the same probability distribution, but differs from the exact generalized inverse when $U$ is precisely a jump value of the cumulative distribution function. Such values have probability zero for an ideal continuous uniform draw.