<p>Here's my best guess. I haven't checked to thoroughly if this is right, but maybe it will help.</p>

<h1>Evolution of population density</h1>

<p>I understand the model as follows. $f(a,t)$
is the density of people of age $a$ at time $t$. Suppose at time $t=0$, the density of the population is $f_0(a)$. To model the aging process as well as the mortality rate, the density $f$ must evolve over time so that it satisfies the condition that you derived. Thus, $m$ must satisfy the following partial differential equation
$$
\frac{\partial f}{\partial t} + \frac{\partial f}{\partial a} + d(a, t) f(a,t) = 0
$$
and the initial condition
$$
f(a,0) = f_0(a).
$$
I believe that PDEs of this form can have relatively straightforward solutions depending on the functional form chosen for $d$. Here are some cases.</p>

<p><strong>Case 1: Constant mortality rate, $d(a,t) = d_0$.</strong></p>

<p>Suppose a constant mortality rate, $d(a,t) = d_0$. Then (using Mathematica),</p>

<pre><code>DSolve[
 {D[f[a, t], t] + D[f[a, t], a] + d0 f[a, t] == 0, 
  f[a, 0] ==  f0[a]},
 f[a, t], {a, t}]
</code></pre>

<p>results in </p>

<pre><code>{{f[a, t] -&gt; E^(-a d0 + d0 (a - t)) f0[a - t]}}
</code></pre>

<p>So, as we can see, this gives a simple solution
$$
f(a,t) = \exp\{-a d_0 + d_0 (a-t)\} \cdot f_0(a-t)
$$
<strong>Case 2: Log Mortality, $d(a,t) = \log(a+1)$.</strong></p>

<p>Here we have</p>

<pre><code>DSolve[
 {D[f[a, t], t] + D[f[a, t], a] + Log[a+1] f[a, t] == 0, 
  f[a, 0] ==  f0[a]},
 f[a, t], {a, t}]
</code></pre>

<p>which results in </p>

<pre><code>{{f[a, t] -&gt; (1 + a)^(-1 - a) E^t (1 + a - t)^(1 + a - t) f0[a - t]}}
</code></pre>

<p>This also gives a simple solution
$$
f(a,t) = (1+a)^{-a-1} e^t (1+a-t)^{1+a-t} \cdot f_0(a-t)
$$</p>

<p><strong>Case 3: The general case</strong></p>

<p>For the case where we do not yet specify $d(a,t)$,</p>

<pre><code>DSolve[
 {D[f[a, t], t] + D[f[a, t], a] + d[a, t] f[a, t] == 0, 
  f[a, 0] ==  f0[a]},
 f[a, t], {a, t}]
</code></pre>

<p>gives us (in TeXForm for clarity)</p>

<p>$$
\begin{align*}
\left\{\left\{f(a,t)\to \\
   \text{f0}(a-t) \cdot \\
   \exp \left(\int_1^a
   -d(K[1],K[1]-a+t) \, dK[1]-\int_1^{a-t} -d(K[1],K[1]-a+t) \,
   dK[1]\right)\right\}\right\}
\end{align*}
$$</p>

CC BY-SA 3.0
2015-05-18T08:54:23.237Z
As stated in the comments of the question above, they are (Monte-Carlo) approaches that are pretty efficient on the generic case. Why would you assume log or constant MR?
VicAche
CC BY-SA 3.0
2015-05-18T18:07:10.290Z
I used those two cases as simple examples. I've put the general case in now. I don't quite see how you would use MC methods. I'll have to look into it more. I understand that Feynman-Kac gives an equivalence between certain PDEs and stochastic processes and that this allows us to solve these PDEs by approximated a conditional expectation with MC methods. However, as @FooBar mentioned, I'm having trouble seeing how to fit this problem into that framework.
jmbejara
CC BY-SA 3.0
2015-05-18T18:13:08.803Z
Also, as for efficiency, I believe that the strength of Feynman-Kac is that you can avoid Monte Carlo methods. As I understand it, solution by MC  methods is reliable but not particularly efficient. Shreve, in Stochastic Calculus for Finance II (p.268), says  "the Euler methods described (an MC method)... for determining this function converges slowly and gives the function value for only one pair (t,x). Numerical algorithms for solving [the PDE] converge quickly in the case of one dimensional $x$ being considered here... and given the function g(t,x) for all values simultaneously." (abridged)
jmbejara
CC BY-SA 3.0
2015-05-18T18:19:26.023Z
Your answer is much more convincing in its current form. You get the upvote :)
VicAche
CC BY-SA 3.0
2015-05-18T18:20:33.840Z
tho you don't quite answer the question I believe. You're giving the integral we need to compute, not a proposed method (except "using mathematica") to compute it.
VicAche
CC BY-SA 3.0
2015-05-18T18:37:59.703Z
You're right. I'm not sure where to go from there, but I figured this is maybe a helpful start. All I can suggest is to look into various quadrature methods. It would be nice to know more about the particular functional forms in order to see how to proceed. Also, this gives us a solution for $f$ and not for $\partial_t F$ as the OP requested. But, I suppose giving $f$ is helpful anyway.
jmbejara
CC BY-SA 3.0
2015-05-18T07:11:38.123Z
2015-05-18T17:56:10.397Z
2015-05-18T17:56:10.397Z
jmbejara
<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>Denote the mass of people at or below age $a$ by $F(a,t)$ </p>

<p>$$ F(a,t) = \int_0^{a} m(\tilde a,t) d\tilde a $$</p>

<p><strong>Ultimately, I am after some Kolmogorov forward equation</strong>, that is, the solution for </p>

<p>$$ \partial_t F(a,t)$$</p>

<p><strong>My approach</strong> Let $f(a, t)$ denote the density 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>$$ f(a+\Delta, t+\Delta) = (1-P(a, t))f(a, t)$$</p>

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

<p>$$ f(a+\Delta, t+\Delta) = (1- \Delta d(a,t))f(a, t)\\
\frac{f(a+\Delta, t+\Delta) -f(a,t)}{\Delta} = -d(a,t))f(a, t)\\
(\partial_t + \partial_a)f(a,t) = \lim_{\Delta\to 0}\frac{f(a+\Delta, t+\Delta) -f(a,t)}{\Delta} = -d(a,t))f(a, t)\\
$$</p>

<p>I can integrate both sides w.r.t. a and get</p>

<p>$$ \partial_t F(t, a) = - f(t, a) - \int q(t, a) f(t, a) da \\
\partial_t F(t, a) = - \partial_a F(t, a) - \int q(t, a) \partial_a F(t, a) da $$</p>

<p>I know that $\partial_a q(t, a) = q(t, a) (1-q(t, a))$. However, that doesn't really help me with solving the integral. Is there perhaps another angle to attack this problem? Or did I miss something?</p>

CC BY-SA 3.0
2015-05-11T16:28:39.750Z
What is $\partial_t F(a,t)$? Is that notation for $\frac{\partial F}{\partial t} (a, t)$?
jmbejara
CC BY-SA 3.0
2015-05-11T16:29:40.607Z
it is notation for $\frac{\partial F(a, t)}{\partial t}$; the left hand side is the total derivative.
FooBar
CC BY-SA 3.0
2015-05-11T16:32:03.120Z
If you know $m$, why can't you just compute $\frac{\partial F(a, t)}{\partial t} = \int_0^a m_t(\tilde a,t) d \tilde a$ with Leibniz's rule?
jmbejara
CC BY-SA 3.0
2015-05-11T16:35:23.497Z
Right, that would be a (quite computationally intensive) numerical approximation. I could do that, but if all else fails.
FooBar
CC BY-SA 3.0
2015-05-13T15:53:55.763Z
You can look at http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula
VicAche
CC BY-SA 3.0
2015-05-13T15:55:09.807Z
which is usually estimated using Monte-Carlo/Quasi-MC methods. To my knowledge, this is how we solve it
VicAche
CC BY-SA 3.0
2015-05-13T21:32:49.790Z
@VicAche I don't think my formula qualifies for Feynman-Kac: I have the final integral term $\int q \partial_a F da'$ that does not appear in your reference.
FooBar
CC BY-SA 3.0
2015-05-13T22:30:55.810Z
Will look into it :(
VicAche
CC BY-SA 3.0
2015-05-14T14:41:11.633Z
http://quant.stackexchange.com/questions/10359/is-there-an-intuitive-explanation-for-the-feynman-kac-theorem maybe you should ask your question there if this is not enough, actually. They seem far more confident that I am about the magic :)
VicAche
CC BY-SA 3.0
2015-05-18T07:12:29.537Z
Is there a difference between $m(a,t)$ and $f(a,t)$ used here?
jmbejara
<mathematical-economics><continuous-time>
Compute evolution of a distribution over time
FooBar
# Compute evolution of a distribution over time

(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.

Denote the mass of people at or below age $a$ by $F(a,t)$

$$ F(a,t) = \int\_0^{a} m(\tilde a,t) d\tilde a $$

**Ultimately, I am after some Kolmogorov forward equation**, that is, the solution for

$$ \partial\_t F(a,t)$$

**My approach** Let $f(a, t)$ denote the density 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,

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

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

$$ f(a+\Delta, t+\Delta) = (1- \Delta d(a,t))f(a, t)\\
\frac{f(a+\Delta, t+\Delta) -f(a,t)}{\Delta} = -d(a,t))f(a, t)\\
(\partial\_t + \partial\_a)f(a,t) = \lim\_{\Delta\to 0}\frac{f(a+\Delta, t+\Delta) -f(a,t)}{\Delta} = -d(a,t))f(a, t)\\
$$

I can integrate both sides w.r.t. a and get

$$ \partial\_t F(t, a) = - f(t, a) - \int q(t, a) f(t, a) da \\
\partial\_t F(t, a) = - \partial\_a F(t, a) - \int q(t, a) \partial\_a F(t, a) da $$

I know that $\partial\_a q(t, a) = q(t, a) (1-q(t, a))$. However, that doesn't really help me with solving the integral. Is there perhaps another angle to attack this problem? Or did I miss something?

**Tags:** mathematical-economics, continuous-time
> **Comments:**
> **username_1**: What is $\partial_t F(a,t)$? Is that notation for $\frac{\partial F}{\partial t} (a, t)$?  
> **username_0**: it is notation for $\frac{\partial F(a, t)}{\partial t}$; the left hand side is the total derivative.  
> **username_1**: If you know $m$, why can't you just compute $\frac{\partial F(a, t)}{\partial t} = \int_0^a m_t(\tilde a,t) d \tilde a$ with Leibniz's rule?  
> **username_0**: Right, that would be a (quite computationally intensive) numerical approximation. I could do that, but if all else fails.  
> **username_2**: You can look at http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula  
> **username_2**: which is usually estimated using Monte-Carlo/Quasi-MC methods. To my knowledge, this is how we solve it  
> **username_0**: @VicAche I don't think my formula qualifies for Feynman-Kac: I have the final integral term $\int q \partial_a F da'$ that does not appear in your reference.  
> **username_2**: Will look into it :(  
> **username_2**: http://quant.stackexchange.com/questions/10359/is-there-an-intuitive-explanation-for-the-feynman-kac-theorem maybe you should ask your question there if this is not enough, actually. They seem far more confident that I am about the magic :)  
> **username_1**: Is there a difference between $m(a,t)$ and $f(a,t)$ used here?  

---

## 1 answers

### Answer by username_1 on 2015-05-18. Score: 4

Here's my best guess. I haven't checked to thoroughly if this is right, but maybe it will help.
Evolution of population density
I understand the model as follows. $f(a,t)$
is the density of people of age $a$ at time $t$. Suppose at time $t=0$, the density of the population is $f_0(a)$. To model the aging process as well as the mortality rate, the density $f$ must evolve over time so that it satisfies the condition that you derived. Thus, $m$ must satisfy the following partial differential equation
$$
\frac{\partial f}{\partial t} + \frac{\partial f}{\partial a} + d(a, t) f(a,t) = 0
$$
and the initial condition
$$
f(a,0) = f_0(a).
$$
I believe that PDEs of this form can have relatively straightforward solutions depending on the functional form chosen for $d$. Here are some cases.
Case 1: Constant mortality rate, $d(a,t) = d_0$.
Suppose a constant mortality rate, $d(a,t) = d_0$. Then (using Mathematica),
DSolve[
 {D[f[a, t], t] + D[f[a, t], a] + d0 f[a, t] == 0, 
  f[a, 0] ==  f0[a]},
 f[a, t], {a, t}]

results in 
{{f[a, t] -> E^(-a d0 + d0 (a - t)) f0[a - t]}}

So, as we can see, this gives a simple solution
$$
f(a,t) = \exp\{-a d_0 + d_0 (a-t)\} \cdot f_0(a-t)
$$
Case 2: Log Mortality, $d(a,t) = \log(a+1)$.
Here we have
DSolve[
 {D[f[a, t], t] + D[f[a, t], a] + Log[a+1] f[a, t] == 0, 
  f[a, 0] ==  f0[a]},
 f[a, t], {a, t}]

which results in 
{{f[a, t] -> (1 + a)^(-1 - a) E^t (1 + a - t)^(1 + a - t) f0[a - t]}}

This also gives a simple solution
$$
f(a,t) = (1+a)^{-a-1} e^t (1+a-t)^{1+a-t} \cdot f_0(a-t)
$$
Case 3: The general case
For the case where we do not yet specify $d(a,t)$,
DSolve[
 {D[f[a, t], t] + D[f[a, t], a] + d[a, t] f[a, t] == 0, 
  f[a, 0] ==  f0[a]},
 f[a, t], {a, t}]

gives us (in TeXForm for clarity)
$$
\begin{align*}
\left\{\left\{f(a,t)\to \\
   \text{f0}(a-t) \cdot \\
   \exp \left(\int_1^a
   -d(K[1],K[1]-a+t) \, dK[1]-\int_1^{a-t} -d(K[1],K[1]-a+t) \,
   dK[1]\right)\right\}\right\}
\end{align*}
$$

> **Comments:**
> **username_2**: As stated in the comments of the question above, they are (Monte-Carlo) approaches that are pretty efficient on the generic case. Why would you assume log or constant MR?  
> **username_1**: I used those two cases as simple examples. I've put the general case in now. I don't quite see how you would use MC methods. I'll have to look into it more. I understand that Feynman-Kac gives an equivalence between certain PDEs and stochastic processes and that this allows us to solve these PDEs by approximated a conditional expectation with MC methods. However, as @FooBar mentioned, I'm having trouble seeing how to fit this problem into that framework.  
> **username_1**: Also, as for efficiency, I believe that the strength of Feynman-Kac is that you can avoid Monte Carlo methods. As I understand it, solution by MC  methods is reliable but not particularly efficient. Shreve, in Stochastic Calculus for Finance II (p.268), says  "the Euler methods described (an MC method)... for determining this function converges slowly and gives the function value for only one pair (t,x). Numerical algorithms for solving [the PDE] converge quickly in the case of one dimensional $x$ being considered here... and given the function g(t,x) for all values simultaneously." (abridged)  
> **username_2**: Your answer is much more convincing in its current form. You get the upvote :)  
> **username_2**: tho you don't quite answer the question I believe. You're giving the integral we need to compute, not a proposed method (except "using mathematica") to compute it.  
> **username_1**: You're right. I'm not sure where to go from there, but I figured this is maybe a helpful start. All I can suggest is to look into various quadrature methods. It would be nice to know more about the particular functional forms in order to see how to proceed. Also, this gives us a solution for $f$ and not for $\partial_t F$ as the OP requested. But, I suppose giving $f$ is helpful anyway.  

---

