Expected value and life

Question

Let $e_{x} = \int_{0}^{\infty} p_{x}(t) \ dt$ where $p_{x}(t)$ is the probability that a person aged $x$ will survive at least $t$ more years. Why is $e_{x} \leq e_{x+1}+1$?

We know that $e_{x} \geq e_{x+1}$ since the expected future lifetime of a younger person is greater than that of an older person. So maybe we can show that $e_{x}-e_{x+1} \leq 1$ or $$\int_{0}^{\infty} p_{x}(t) - p_{x+1}(t) \ dt \leq 1$$

Edit. We know that for a discrete random variable, $$\mathbb{E}(Y|X) = \int_{\Omega} Y(\omega) \mathbb{P}(dw|X=x)$$

$$= \frac{\int_{X=x} Y(\omega)\mathbb{P}(dw)}{\mathbb{P}(X=x)}$$

$$= \frac{\mathbb{E}(Y \mathbb{1}_{(X=x)})}{\mathbb{P}(X=x)}$$

What does $w$ above represent?

Answer 1

Let $n_x(d)$ be the expected number of people alive of age $x$ on date $d$. Then, independent of $d$, $$ n_{x+t}(d+t)=n_x(d)p_x(t)\tag{1} $$ Using $(1)$ with $x,d,t+1$, yields $$ p_x(t+1)=\frac{n_{x+t+1}(d+t+1)}{n_x(d)}\tag{2} $$ Using $(1)$ with $x+1,d+1,t$, yields $$ p_{x+1}(t)=\frac{n_{x+t+1}(d+t+1)}{n_{x+1}(d+1)}\tag{3} $$ Using $(1)$ with $x,d,t=1$, yields $$ n_{x+1}(d+1)=n_x(d)p_x(1)\tag{4} $$ Combining $(2)$, $(3)$, and $(4)$ gives $$ p_x(t+1)=p_x(1)p_{x+1}(t)\le p_{x+1}(t)\tag{5} $$ Therefore, we get $$ \begin{align} e_x &=\int_0^1 p_x(t)\,\mathrm{d}t+\int_0^\infty p_x(t+1)\,\mathrm{d}t\\ &\le1+\int_0^\infty p_{x+1}(t)\,\mathrm{d}t\\ &=1+e_{x+1}\tag{6} \end{align} $$

Answer 2

As the notation almost suggests, $$e_x=\int_0^\infty p_x(t)\,dt$$ is the expected additional life at age $x$.

For in general, if $W$ is a non-negative random variable defined on $[0,\infty)$, with cumulative distribution function $F_W(w)$, then $$E(W)=\int_0^\infty (1-F_W(w))\,dw.$$

If everybody who is age $x$ automatically survives to age $x+1$, then $e_x=e_{x+1}+1$. But perhaps not every person age $x$ makes it to $x+1$. Informally, any who die force $e_x$ below $e_{x+1}+1$.

If you want to calculate formally, let $d$ be the probability of dying by age $x+1$ given one has made it to $x$. (So $d=1-p_x(1)$.)

Then with probability $d$ one's additional life will be $\le 1$, so has expectation $\le 1$. And with probability $(1-d)$ one's additional life has expectation $e_{x+1}+1$. Thus $$e_x \le (d)(1) +(1-d)(e_{x+1}+1).$$ Expand. The right-hand side is $(1-d)e_{x+1} +1$, which is $\le e_{x+1}+1$.

Answer 3

The result to be proved is a particular case of the following fact:

The function $x\mapsto e_x+x$ is nondecreasing.

To show this, introduce the (random) lifetime $L$ and note that $p_x(t)=\mathrm P(L\geqslant x+t\mid L\geqslant x)$, hence $$ e_x=\mathrm E\left(\int_0^{\infty}\mathbf 1_{L\geqslant x+t}\,\mathrm dt\,\Big|\, L\geqslant x\right)=\mathrm E(L-x\mid L\geqslant x)=\mathrm E(L\mid L\geqslant x)-x. $$ Fix some $0\leqslant x\leqslant y$. Then, $$ (e_x+x)-(e_y+y)=\frac{\mathrm E(L:L\geqslant x)}{\mathrm P(L\geqslant x)}-\frac{\mathrm E(L:L\geqslant y)}{\mathrm P(L\geqslant y)}, $$ which has the sign of $$ v(x,y)=\mathrm E(L:L\geqslant x)\cdot\mathrm P(L\geqslant y)-\mathrm E(L:L\geqslant y)\cdot\mathrm P(L\geqslant x). $$ Adding and substracting $\mathrm E(L:L\geqslant y)\cdot\mathrm P(L\geqslant y)$ yields $$ v(x,y)=\mathrm E(L:y\geqslant L\geqslant x)\cdot\mathrm P(L\geqslant y)-\mathrm E(L:L\geqslant y)\cdot\mathrm P(y\geqslant L\geqslant x). $$ Hence $v(x,y)$ has the sign of $$ u(x,y)=\mathrm E(L\mid y\geqslant L\geqslant x)-\mathrm E(L\mid L\geqslant y). $$ The first term on the RHS is $\leqslant y$ and the second term on the RHS is $\geqslant y$, hence $u(x,y)\leqslant0$. QED.

Edit: For every random variable $X$ and disjoint events $A$ and $B$ with positive probabilities, $$ \mathrm E(X\mid A\cup B)-\mathrm E(X\mid B)=\left(\mathrm E(X\mid A)-\mathrm E(X\mid B)\right)\cdot\frac{\mathrm P(A)}{\mathrm P(A\cup B)}. $$ Hence, $\mathrm E(X\mid A\cup B)\geqslant\mathrm E(X\mid B)$ if and only if $\mathrm E(X\mid A)\geqslant\mathrm E(X\mid B)$.