Let's say X is Zeta distributed: $$ p( X = k ) = \frac{k^{-s}}{\sum_{n=1}^\infty k^{-s}} $$
Then, the MRLF (Mean Residual Life Function) of the Zeta distribution is:
\begin{align} e(u) &= E[X - u \mid x > u ] \\ &=\sum_{i=u}^\infty i \frac{i^{-s}}{\sum_{j=u}^\infty j^{-s}} - u \\ &= \frac{\sum_{i=u}^\infty i^{1-s}}{\sum_{j=u}^\infty j^{-s}} - u \\ &= \frac{\zeta(s-1, u)}{\zeta(s, u)} - u \end{align}
Two observations:
- Plotting the MRLF suggests that it is linear. This is somewhat expected because the Zeta distribution is kind of a discrete version of the Pareto distribution, whose MRLF is linear.
- It looks like the slope of the MRLF is some nice function of $s$ (estimated):
I was wondering if there is a way of showing that these two observations are correct.
You are just saying that as $n\to \infty$
$$\frac{n^{1-s}-(n+1)^{1-s}}{s-1}-n^{-s}=\int_n^{n+1} (x^{-s}-n^{-s})dx= \int_n^{n+1} \int_n^x s t^{-s-1}dtdx=O(s n^{-s-1})$$
So that for $s > 1$ as $u \to \infty$ $$\sum_{n\ge u} n^{-s} \sim\frac{u^{1-s}}{s-1}$$ and for $s > 2$ $$e(u)=\frac{\sum_{n\ge u} n^{1-s}}{\sum_{n\ge u} n^{-s}}-u\sim \frac{\frac{u^{2-s}}{s-2}}{\frac{u^{1-s}}{s-1}}-u = u (\frac{s-2}{s-1}-1)$$