Donald Hobson comments on An optimal stopping paradox

[Donald Hobson]

Differentiating the expected reward over time.

$$\frac{d}{dt}\beta t{e}^{-\alpha t}=\beta e^{-\alpha t}-\alpha \beta t{e}^{-\alpha t}$$

So the best time to sell is when $t=1/\alpha$.

if you have already waited time $c$ then the reward becomes $\beta (t+c)e^{-\alpha (t+c)}=e^{-\alpha c}\beta (t+c)e^{-\alpha t}$ The stopping time becomes

$$\frac{d}{dt}\beta (t+c)e^{-\alpha (t+c)}=\beta e^{-\alpha (t+c)}-\alpha \beta (t+c)e^{-\alpha (t+c)}$$

With a solution at $t=\frac{1}{\alpha }-c$. Nothing wierd is going on here, a plot of expected value vs sell time looks like this.

Suppose the exponential decay term was 1 day. After 1 second, waiting another second makes sense, it will double your value and the chance of a fail is tiny. After a week, you already have a large pot of value that you are risking. It is no longer worth waiting.

[Pattern]

looks like this.

That link doesn’t work.

[Donald Hobson]

Fixed.