Average life expectancy..exponential function

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.

• 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/… – Ethan Bolker Mar 27 '18 at 23:43
• 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. – J.Moriarty Mar 27 '18 at 23:48
• Thank you Karn for formatting. – J.Moriarty Mar 27 '18 at 23:54

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 $n>1$,

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

• 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. – J.Moriarty Mar 28 '18 at 0:18
• @J.Moriarty You are welcome :) I just edited the answer to be more rigorous now please check the edited ver out. – Karn Watcharasupat Mar 28 '18 at 0:19
• Yep this is better. Thank you! – J.Moriarty Mar 28 '18 at 0:23