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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?

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Hint: first argue that $p_x(t+1)\leq p_{x+1}(t)$ –  Julian Rosen Jul 25 '12 at 18:36
Less than 2 hours. –  Did Jul 25 '12 at 22:13
Replaced by approximately 5 hours, which is only marginally better. –  Did Jul 25 '12 at 23:48
It is not true that $e_x\ge e_{x+1}$. Consider a hypothetical population where half the people die at age 30 and the other half at age 60, and try $x=29.5$. This phenomenon actually happens in real life due to infant mortality. –  Rahul Jul 26 '12 at 2:14
thomas: You see... –  Did Jul 27 '12 at 21:24
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3 Answers

up vote 1 down vote accepted

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 $$ \begin{align} e_x &=\int_0^1 p_x(t)\,\mathrm{d}t+\int_0^\infty p_x(t+1)\,\mathrm{d}t\\ &\le1+\int_0^\infty p_{x+1}(t)\,\mathrm{d}t\\ &=1+e_{x+1}\tag{6} \end{align} $$

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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 $\le e_{x+1}+1$.

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I was thinking you would answer this question. –  Graphth Jul 25 '12 at 18:40
@Andre Nicolas: I dont think you have to consider everybody. $e_x$ can be for just a single person and the argument would work. For example, if I am alive this year and make it to next year, then my $e_{x+1} = e_{x}-1$. –  thomas Jul 25 '12 at 22:12
@thomas: I wanted to use the informal large sample interpretation for the intuitive version of the argument. The formal computation is the one that calculates expectation by summing over the two conditions died, did not. –  André Nicolas Jul 25 '12 at 22:18
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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 $\mathrm E(X\mid A)\geqslant\mathrm E(X\mid B)$.

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How did you come up with $v(x,y)$? –  thomas Jul 25 '12 at 23:01
See Edit. $ $ $ $ –  Did Jul 25 '12 at 23:08
We know that $E(Y|X=x) = \int_{\Omega} Y(\omega)P(dw|X=x)$. Do you know what $w$ is in this context? –  thomas Jul 25 '12 at 23:22
Sorry but neither the content of your comment nor how it should be related to your post and/or to my answer, are clear to me. Please explain. –  Did Jul 25 '12 at 23:26
Since the Edit is unrelated to the original question, you should not include it in this question and, if interested, you should ask it as another question. (Furthermore, its content is dubious, for example, what is $x$? and, is $w=\omega$?) –  Did Jul 25 '12 at 23:47
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