The expected value
E[
Xt] with restarts after
t time units is implicitly given in [
9].
$$ E[X_{t}] = \frac{1}{1-\exp{(-(\lambda t)^{k})}}\left( t\exp{(-(\lambda t)^{k})}+\frac{1}{\lambda} \gamma(1+\frac{1}{k}, (\lambda t)^{k})\right), $$
(6)
where
\(\gamma (s, x) = {{\int \limits }_{0}^{x}}t^{s-1}e^{-t} \mathrm {d} t\) is the lower incomplete gamma function. To determine the derivative of
E[
Xt], the derivative of the lower incomplete gamma function
\({\gamma (1+\frac {1}{k}, (\lambda t)^k)}\) is considered. Using the definition of
γ and the chain rule yields:
$$ \begin{array}{@{}rcl@{}} \frac{\mathrm{d}}{\mathrm{d}{}t} {\gamma(1+\frac{1}{k}, (\lambda t)^k)}{} &=& k \cdot \lambda \cdot {\exp{(-(\lambda t)^k)}}{} \cdot \left( (\lambda t)^{k}\right)^{\frac{1}{k}} \cdot (\lambda t)^{k-1} \end{array} $$
(7)
$$ \begin{array}{@{}rcl@{}} & =& k \cdot \lambda \cdot {\exp{(-(\lambda t)^k)}}{} \cdot (\lambda t)^{k} \end{array} $$
(8)
We now calculate the derivative of
E[
Xt] w.r.t.
t and denote it by
g(
t). The derivative
g(
t) is then given by:
$$ \frac{\exp{(-(\lambda t)^{k})}}{1-\exp{(-(\lambda t)^{k})}}\left( 1-\frac{\exp{(-(\lambda t)^{k})} k(\lambda t)^{k} + k (\lambda t)^{k-1} \gamma(1+\frac{1}{k},(\lambda t)^{k})}{1-\exp{(-(\lambda t)^{k})}}\right). $$
We show that
g(
t) > 0 for all
t > 0. This inequality can be transformed to:
$$ \frac{1-\exp{(-(\lambda t)^{k})}}{k} - (\lambda t)^{k} \exp{(-(\lambda t)^{k})} - (\lambda t)^{k-1} \gamma\left( 1+\frac{1}{k}, (\lambda t)^{k}\right)> 0. $$
(9)
First, this inequality is observed for the limit
\(t \rightarrow 0\). The limit of the term
\(-(\lambda t)^{k-1} \gamma (1+\frac {1}{k},(\lambda t)^{k})\) has to be examined closely for
k < 1. The limit of all other terms is clearly zero. Due to L’Hospital’s rule, we have:
$$ \lim\limits_{t\rightarrow 0}- \frac{\gamma(1+\frac{1}{k},(\lambda t)^{k})}{(\lambda t)^{1-k}} = \lim\limits_{t\rightarrow 0}-\frac{\exp(-(\lambda t)^{k}) k (\lambda t)^{2k}}{1-k}=0. $$
Thus, the left side of inequality (
9) approaches zero in the limit. We now show that the derivative of the left side of inequality (
9) is positive for all
t > 0. This inequality can be transformed to
$$ \frac{1-\exp{(-(\lambda t)^{k})}}{k (\lambda t)^{k-1}} - \exp{(-(\lambda t)^{k})}\lambda t - \gamma\left( 1+\frac{1}{k}, (\lambda t)^{k}\right)> 0. $$
(10)
The derivative
h(
t) of the left side of inequality (
10) is given by:
$$ h(t)=\frac{(1-\exp{(-(\lambda t)^{k})})(1-k)\lambda}{k (\lambda t)^{k}} $$
It can be easily verified that for
t > 0,
h(
t) evaluates to:
$$ h(t) \begin{cases} > 0, \text{ for } k < 1, \\ = 0, \text{ for } k = 1, \\ < 0, \text{ for } k > 1. \end{cases} $$
(11)
Thus, for
k < 1 we have
h(
t) > 0 which implies
g(
t) > 0 for all
t. This property means that
E[
Xt] is strictly monotonically increasing. On the other hand,
k > 1 implies that
E[
Xt] is strictly monotonically decreasing. □