How to prove that $\mathring{e}_x \le \mathring{e}_{x+1} + 1$ - AMLCR Exercise 2.10a

Question

I'm going through Actuarial Mathematics for Life Contingent Risks, 3rd. ed Exercise 2.10.

Part a. is the following question:

Show that $\mathring{e}_x \le \mathring{e}_{x+1} + 1$.

Based on the definition from the book as well as the distributional property given, I know that

$$ \mathring{e}_x = \int_0^\infty {}_tp_x dt \quad \quad \quad (2.23) $$ This can also be re-written as

$$ \mathring{e}_x = \int_0^\infty {}_tp_x dt = \sum_{j=0}^\infty \int_j^{j+1} {}_tp_x dt $$

Now, using the definition above, and the fact that

$$ {}_tp_x = \frac{S_0(x+t)}{S_0(x)} = \frac{{}_{x+t}p_0}{{}_xp_0} $$

Trying to connect this to ${}_{t}p_{x+1}$, where $p_x = {}_1p_x$,

$$ {}_tp_{x+1} p_x = {}_{t+1}p_x $$ or $$ {}_{t-1}p_{x+1} p_x = {}_tp_x $$

Expressing $\mathring{e}_{x+1}$ as the integral definition,

$$ \mathring{e}_{x+1} = \int_0^\infty {}_tp_{x+1} dt = \int_0^\infty \frac{{}_{t+1}p_x}{p_x} dt = \frac{1}{p_x} \int_0^\infty {}_{t+1}p_x dt = \frac{1}{p_x} \biggr( \int_0^\infty p_{t+x} {}_tp_x dt \biggr) $$

This is where I'm stuck, since I am not sure how to extract the bounds of $\mathring{e}_{x+1} + 1$ . Is there a certain trick I can use in order to relate $\mathring{e}_x$ and $\mathring{e}_{x+1}$? I'm a bit stuck and don't know if there is a clever trick or another definition I should use in order to arrive at the inequality $\mathring{e}_x \le \mathring{e}_{x+1} + 1$.

Answer 1

Intuitively, with curtate life expectancies, you will either survive for a year or you won't. So $e_x$ is either 1 year (with probability $q_x$) or it is $1+e_{x+1} $ (with probability $p_x$). So

$$e_{x} =1 \, (1-p_x) + (1+e_{x+1}) p_x = 1+e_{x+1} p_x \leqslant 1+e_{x+1}$$

With exact life expectancies: $$\overset{\circ}{e}_{x} =\int_0^\infty {}_tp_x \, dt = \int_0^1 {}_tp_x \, dt + \int_1^\infty {}_tp_x \, dt $$ $$=\int_0^1 {}_tp_x \, dt + \int_0^\infty {}_tp_{x+1} \, dt $$ $$ \leqslant 1+ \overset{\circ}{e}_{x+1} $$

Answer 2

Let $f(z) = \mathring{e}_{x+z} + z$; it suffices to show that $f$ is monotone increasing. We have \begin{align*} f'(z) &= \left(\frac{d}{dz} \int_{0}^{\infty}\frac{S_0(x+z+t)}{S_0(x+z)}dt\right) + 1 \\ &=\left(\int_{0}^{\infty}\frac{-f_0(x+z+t)S_0(x+z) + S_0(x+z+t)f_0(x+z)}{S_0^2(x+z)}dt\right) + 1 \\ &= \frac{f_0(x+z)}{S_0^2(x+z)}\int_0^\infty S_0(x+z+t) dt \\ &\ge 0 \end{align*}