Strategy in a waiting time game: fisherman's dilemma

Solution 1:

Let $f$ be the probability density function, $t$ be the recast time and $T$ be the resulting average catch time. We have: $$ \begin{align} T &= \int_0^t x f(x)\ \mathrm{d}x + \left(\int_t^\infty f(x)\ \mathrm{d}x \right)(t + T) \\ T &= \frac{\displaystyle \int_0^t x f(x)\ \mathrm{d}x + t \int_t^\infty f(x)\ \mathrm{d}x}{\displaystyle 1 - \int_t^\infty f(x)\ \mathrm{d}x} \label{1}\tag{1} \end{align} $$ The minimum is either at $t = 0$ (recast as soon as you can), $t = \infty$ (never recast), at $\frac{\mathrm{d}T}{\mathrm{d}t} = 0$ or at a discontinuity of $f$. The case $t = \infty$ is simply the expected value. For $t = 0$ we take the limit to 0 and need to use L'Hôpital's rule. $$ \lim_{t \downarrow 0}\ T = \lim_{t \downarrow 0}\ \frac{\displaystyle \int_t^\infty f(x)\ \mathrm{d}x}{f(t)} = \frac{1}{f(0)} $$ Now the case $\frac{\mathrm{d}T}{\mathrm{d}t} = 0$: $$ \begin{align} \frac{\mathrm{d}T}{\mathrm{d}t} &= \frac{ \left( \int_t^\infty f(x)\ \mathrm{d}x \right) \left( 1 - \int_t^\infty f(x)\ \mathrm{d}x \right) - \left( \int_0^t x f(x)\ \mathrm{d}x + t \int_t^\infty f(x)\ \mathrm{d}x \right) f(t) }{ \left( 1 - \int_t^\infty f(x)\ \mathrm{d}x \right)^2 } \\ 0 &= \left( \int_t^\infty f(x)\ \mathrm{d}x \right) \left( 1 - t f(t) - \int_t^\infty f(x)\ \mathrm{d}x \right) - f(t) \int_0^t x f(x)\ \mathrm{d}x \end{align} $$ Some examples:

• The exponential distribution $f(x) = \lambda e^{-\lambda x}$ has $T = \frac{1}{\lambda}$. It's the only distribution in which $T$ does not depend on $t$.
• In a delayed exponential distribution $t = \infty$ gives the minimum value for $T$. $$ f(x) = \begin{cases} e^{1-x} & x \ge 1 \\ 0 & x < 1 \end{cases} $$
• With the following distribution the minimum is at $t = 1$. $$ f(x) = \begin{cases} \frac{1}{2} & x \in [0,1] \vee x \in [3,4] \\ 0 & \text{otherwise} \end{cases} $$
• This distribution has minima at $t = 1$ and $t = \frac52$: $$ f(x) = \begin{cases} \frac{1}{3} & x \in [0,1] \vee x \in [\frac32, \frac52] \vee x \in [5, 6] \\ 0 & \text{otherwise} \end{cases} $$
• This gives a minimum at any $t \in [0, 1]$: $$ f(x) = \begin{cases} e^{-x} & x \in [0,1] \\ \frac{1}{e} & x \in [2,3] \\ 0 & \text{otherwise} \end{cases} $$
• This gives a minimum at $t = 0$: $$ f(x) = \begin{cases} e^{-x}-\frac{1}{e} & x \in [0,1] \\ \frac{2}{e} & x \in [2,3] \\ 0 & \text{otherwise} \end{cases} $$
• The following distribution has expected value $\infty$ and has a minimum at $t = 0$. $$ f(x) = \frac{1}{(x+1)^2} $$

To your question if the fisherman should ever recast if the expected value of the distribution is $\infty$: Yes, he should always do that. Recasting gives a finite value for $T$ while not recasting will give $T = \infty$.