<p>I think the step</p>

<blockquote>
  <p><em>"...where $P(a, t) = exp(-d(a,t)\Delta)$ is the discrete time analogue of $d(a,t)$..."</em></p>
</blockquote>

<p>is the problem.</p>

<p>In continuous time I guess we have</p>

<p>$$\dot m(a,t) = -d(a,t)m(a,t) \implies m(a,t) = m_0\exp \{-d(a,t)t\}$$</p>

<p>We then have, discretizing,</p>

<p>$$\frac {m(a,t+\Delta) - m(a,t)}{m(a,t)} = \frac {\exp \{-d(a,t)(t+\Delta)\}-\exp \{-d(a,t)t\}}{\exp \{-d(a,t)t\}}$$</p>

<p>$$=\exp \{-d(a,t)\Delta\}-1 \approx -d(a,t)$$</p>

<p>$$\implies d(a,t) \approx 1-\exp \{-d(a,t)\Delta\}$$</p>

<p>So it should be $P(a, t) = 1-\exp(-d(a,t)\Delta)$. Now $a$ changes exactly in the same way as $t$ does (at $t+\Delta$ those of age $a$ at $t$ will be of age $a+\Delta$). So $m$ changes in both arguments together.</p>

<p>$$m(a+\Delta, t+\Delta) = (1-P(a, t))m(a, t) = \exp(-d(a,t)\Delta m(a, t)$$</p>

<p>$$\approx [1-d(a,t)\Delta]m(a, t)$$</p>

<p>I understand the left hand side as "the people that were of age $a$ at time $t$, and are still alive at time $t+\Delta$, where they have become of age $a+\Delta$. Is this what you are after?</p>

CC BY-SA 3.0
2015-05-08T18:06:18.960Z
2015-05-08T19:41:33.510Z
2015-05-08T19:41:33.510Z
Alecos Papadopoulos
<p>We have a population of people with different age $a$, time is indexed with $t$. There is a rate at which people die, $d(a, t)$. For simplicity, ignore births. I want to compute the evolution of the distribution of ages over time.</p>

<p>Let $m(a, t)$ denote the mass of people at age $a$ and point in time $t$. I will start with a discrete time approximation and let $\Delta$ go to zero. At each discrete point in time,</p>

<p>$$ m(a+\Delta, t+\Delta) = (1-P(a, t))m(a, t)$$</p>

<p>where $P(a, t) = exp(-d(a,t)\Delta)$ is the discrete time analogue of $d(a,t)$. As I'm going to let $\Delta\to 0$, I can approximate $P$ with $(1-\Delta d)$:</p>

<p>$$ m(a+\Delta, t+\Delta) = \Delta d(a,t)m(a, t)$$</p>

<p><strong>Issue</strong> Already here I'm having trouble understanding what happened: $\Delta d(a,t)$ denotes the mass of people who died during $\Delta$ at age $a$ and time $t$. Shouldn't this be <strong>negatively</strong> affecting the mass of people alive? I was expecting something along the lines of </p>

<p>$$ m(a+\Delta, t+\Delta) = (1- \Delta d(a,t))m(a, t)$$</p>

CC BY-SA 3.0
2015-05-08T14:35:42.140Z
if you let delta go to zero,  you get $dm(t,d)= -\lambda(t,d)m(t,d)$. Is that what you want?
An old man in the sea.
CC BY-SA 3.0
2015-05-08T14:37:25.727Z
suggestion: You could improve the question by giving more details on what your objective, the properties of the functions on the RHS of the equation, etc.
An old man in the sea.
CC BY-SA 3.0
2015-05-08T14:45:31.583Z
@Anoldmaninthesea. I thought that it might be beneficial to simplify as much as possible, but as requested, I've rewritten the question.
FooBar
<mathematical-economics><continuous-time>
Computing the continuous time survival rate
FooBar
# Computing the continuous time survival rate

(Asked by: username_0 on 2015-05-08)

We have a population of people with different age $a$, time is indexed with $t$. There is a rate at which people die, $d(a, t)$. For simplicity, ignore births. I want to compute the evolution of the distribution of ages over time.

Let $m(a, t)$ denote the mass of people at age $a$ and point in time $t$. I will start with a discrete time approximation and let $\Delta$ go to zero. At each discrete point in time,

$$ m(a+\Delta, t+\Delta) = (1-P(a, t))m(a, t)$$

where $P(a, t) = exp(-d(a,t)\Delta)$ is the discrete time analogue of $d(a,t)$. As I'm going to let $\Delta\to 0$, I can approximate $P$ with $(1-\Delta d)$:

$$ m(a+\Delta, t+\Delta) = \Delta d(a,t)m(a, t)$$

**Issue** Already here I'm having trouble understanding what happened: $\Delta d(a,t)$ denotes the mass of people who died during $\Delta$ at age $a$ and time $t$. Shouldn't this be **negatively** affecting the mass of people alive? I was expecting something along the lines of

$$ m(a+\Delta, t+\Delta) = (1- \Delta d(a,t))m(a, t)$$

**Tags:** mathematical-economics, continuous-time
> **Comments:**
> **username_1**: if you let delta go to zero,  you get $dm(t,d)= -\lambda(t,d)m(t,d)$. Is that what you want?  
> **username_1**: suggestion: You could improve the question by giving more details on what your objective, the properties of the functions on the RHS of the equation, etc.  
> **username_0**: @Anoldmaninthesea. I thought that it might be beneficial to simplify as much as possible, but as requested, I've rewritten the question.  

---

## 1 answers

### Answer by username_2 on 2015-05-08. Score: 2 | Accepted answer

I think the step

> *"...where $P(a, t) = exp(-d(a,t)\Delta)$ is the discrete time analogue of $d(a,t)$..."*

is the problem.

In continuous time I guess we have

$$\dot m(a,t) = -d(a,t)m(a,t) \implies m(a,t) = m\_0\exp \{-d(a,t)t\}$$

We then have, discretizing,

$$\frac {m(a,t+\Delta) - m(a,t)}{m(a,t)} = \frac {\exp \{-d(a,t)(t+\Delta)\}-\exp \{-d(a,t)t\}}{\exp \{-d(a,t)t\}}$$

$$=\exp \{-d(a,t)\Delta\}-1 \approx -d(a,t)$$

$$\implies d(a,t) \approx 1-\exp \{-d(a,t)\Delta\}$$

So it should be $P(a, t) = 1-\exp(-d(a,t)\Delta)$. Now $a$ changes exactly in the same way as $t$ does (at $t+\Delta$ those of age $a$ at $t$ will be of age $a+\Delta$). So $m$ changes in both arguments together.

$$m(a+\Delta, t+\Delta) = (1-P(a, t))m(a, t) = \exp(-d(a,t)\Delta m(a, t)$$

$$\approx [1-d(a,t)\Delta]m(a, t)$$

I understand the left hand side as "the people that were of age $a$ at time $t$, and are still alive at time $t+\Delta$, where they have become of age $a+\Delta$. Is this what you are after?


---

