I've recently come across an equation for the expected mean number of breeding seasons after the first breeding season, as a function of the annual survival rate (S) and the probability of breeding,

$$ \mathbb{E}(\#\text{ of breeding seasons}) = \dfrac{1}{-\ln(S)} \times \text{breeding probability} $$

I'm having a hard time understanding what the term $1 / -\ln(S)$ represents. Any ideas?


It's just a continuous version of the discrete calculation. The discrete version is the (infinite series) sum

$$ \sum_{i = 0}^\infty S^i \cdot bp $$

adding up every (chance of survival to season $i$) x (breeding probability given that survival).

Making this continuous converts the equation to

$$ \begin{split} & \int_0^\infty bp \cdot S^i \, di \\ = \enspace & bp \int e^{i \ln S} \, di \end{split} $$

integrating gives

$$ \left.bp \cdot \frac{e^{i \, \ln(S)} }{\ln(S)} \right\vert_{0}^{\infty} $$

Evaluating gives $$ = \frac{bp \cdot e^{-\infty} }{\ln(S)} - \frac{bp \cdot e^{0} }{\ln(S)} $$ (remembering that $0 <= S < 1$, so $\ln(S) < 0$)

$$ = bp \cdot 0 - bp \cdot \frac{1}{\ln(S)} = bp \cdot \left(0 - \frac{1}{\ln(S)}\right) = bp \cdot \frac{-1}{\ln(S)} $$

  • $\begingroup$ This makes sense, thanks! $\endgroup$
    – user13317
    Aug 16 '21 at 19:02
  • 2
    $\begingroup$ @BenBolker Thank you so much for the example that LaTeX works here! I will work on completing the formatting. $\endgroup$
    – Armand
    Aug 16 '21 at 19:10

I think your explanation is correct. The expected value of the exponential distribution is:

$$t \sim \text{e}^{-\lambda t} \implies \langle t \rangle = \int_0^\infty t \ \text{e}^{-\lambda t} \; \text{d}t = 1/\lambda.$$

For the exponential survival function, we have to identify the parameter $S$. Since $S$ is the number of survived individuals after one year, we derive:

$$S = \text{e}^{-\lambda} \implies \lambda = - \ln S.$$

Therefore, the life expectancy is simply: $$\langle t \rangle = -\frac{1}{\ln S}.$$

Because the probability of a breeding to happen each year is independent, we can multiply the life expectancy by the probability that the breeding happens in a one year period. This gives us the final result:

$$\langle \# \text{ breedings} \rangle = \langle t \rangle \cdot P(\text{breeding}) = -\frac{1}{\ln S}\ P(\text{breeding}). $$

  • $\begingroup$ This is also a good explanation! Thanks for the help. $\endgroup$
    – user13317
    Aug 16 '21 at 19:09

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