1
$\begingroup$

Let $$N_0 = \text{initial number of AIDS patients}$$ $$N= \text{number of patients left}$$

The equation is given by: $$N=N_0\exp(-kt)$$ What is the average life expectancy of one person? (The answer is $t= \frac1k$)

How did we get to this answer without using expected value and probably/statistics analysis? (differential equations problem) Thanks in advance.

Edit: I know how to come up with the answer using expected value, but the problem is presented as a differential equations one.

$\endgroup$
  • $\begingroup$ Hint. Do you know the integral that computes the expected value of a continuous distribution? And please format with mathjax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Ethan Bolker Mar 27 '18 at 23:43
  • 1
    $\begingroup$ Hello.. i know how to come up with the answer using expected value, but the problem is presented as a differential equations one. I am asking for a friend currently taking a DE course.. sorry about mathjax but i cannot format using my phone. $\endgroup$ – J.Moriarty Mar 27 '18 at 23:48
  • 1
    $\begingroup$ Thank you Karn for formatting. $\endgroup$ – J.Moriarty Mar 27 '18 at 23:54
0
$\begingroup$

The answer is not very rigorous since, as you know, the DE itself is derived using expectation and average out everything but I think you will get the overall idea.


Let $N(\tau)=N_0-n$, $N(\tau+\delta)=N_0-(n+1)$. Hence, one life has lapsed in time $\delta$.

$$\tau=\frac{1}{k}\ln\frac{N_0}{N_0-n}$$ $$\tau+\delta=\frac{1}{k}\ln\frac{N_0}{N_0-(n+1)}$$

$$\delta=\frac{1}{k}\left(\ln\frac{N_0}{N_0-(n+1)}-\ln\frac{N_0}{N_0-n}\right)=\frac{1}{k}\left(\ln\frac{N_0-n}{N_0-(n+1)}\right)$$

Since this must hold true for all $n>1$,

$$\text{life expectancy}=\lim_{n\to \infty}\delta=\lim_{n\to \infty}\frac{1}{k}\left(\ln\frac{N_0-n}{N_0-(n+1)}\right)=\frac1k$$

$\endgroup$
  • 1
    $\begingroup$ Thank you very much Karn. I understand the error was in presenting the problem as a differential equations one, as it really is more than that. Thanks for the answer. $\endgroup$ – J.Moriarty Mar 28 '18 at 0:18
  • $\begingroup$ @J.Moriarty You are welcome :) I just edited the answer to be more rigorous now please check the edited ver out. $\endgroup$ – Karn Watcharasupat Mar 28 '18 at 0:19
  • $\begingroup$ Yep this is better. Thank you! $\endgroup$ – J.Moriarty Mar 28 '18 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.