The shifted exponential distribution

A random variable $X$ is distributed according to the exponential distribution if the cdf is $\Phi(x) = 1-\exp\left(-\lambda x\right)$ . We are interested in the shifted exponential distribution of $X-1$ . Using eq. (16) it is

$$G_\alpha = \log(1-\alpha)+\alpha\int_0^\infty\frac{e^{-\lambda x}}{1+\alpha(x-1)}\quad.$$

The integral is directly related to the incomplete gamma function $\Gamma(a,z)=\int_z^\infty t^{a-1}e^{-t} dt$ and a linear change of variables gives the result

$$G_\alpha = \log(1-\alpha)+\exp\left(\frac{1-\alpha}{\alpha}\lambda\right)\, \Gamma\left(0, \frac{1-\alpha}{\alpha}\lambda\right)\quad.$$

Note that $\Gamma[0,z]$ is related to the exponential integral function $\mathrm{Ei}(z) = -\int_{-z}^\infty\frac{e^{-t}}{t} dt$ . i.e. $\Gamma[0,z] = -\mathrm{Ei}(-z)$ . Figure 2.2 shows the graphs of $G_\alpha$ for different values of $\lambda$ . Since ${\mathsf{E}}(X-1) = \frac{1}{\lambda} -1$ it is $G^\prime(0) = \frac{1}{\lambda}-1$ and therefore $G_\alpha$ attains its maximum at $\alpha>0$ only for $\lambda<1$ , i.e. those shifted exponential distributions for which the mean is positive.

Fig. 1: $G_\alpha (\lambda )$ for the exponential distribution for different parameters $\lambda$ .

Unfortunately no closed form solution for $G = \max_\alpha G_\alpha$ is available and one has to resort to numerical techniques.

Fig.: $G(\lambda)=\max_\alpha G_\alpha(\lambda)$ for the exponential distribution.